Can every proof by contradiction also be shown without contradiction?












311












$begingroup$


Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantage/disadvantages of proving by contradiction?



As an aside, how is proving by contradiction viewed in general by 'advanced' mathematicians, is it a bit of an 'easy way out' when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn't fond of proof by contradiction.










share|cite|improve this question











$endgroup$








  • 263




    $begingroup$
    Let us assume every proof by contradiction can also be shown without contradiction...
    $endgroup$
    – Graphth
    Nov 24 '12 at 18:21






  • 11




    $begingroup$
    Since this topic came up in some of the answers and comments, Andrej Bauer's blog post Proof of negation and proof by contradiction seems relevant.
    $endgroup$
    – user50545
    Nov 24 '12 at 18:21








  • 4




    $begingroup$
    ... and Gowers's blog post: When is proof by contradiction necessary?.
    $endgroup$
    – user50546
    Nov 24 '12 at 18:29








  • 4




    $begingroup$
    I should know this, but I am not sure. You are basically asking whether intuitionistic logic is equivalent to classical logic? I am pretty sure it is not.
    $endgroup$
    – Panayiotis Karabassis
    Nov 25 '12 at 0:36






  • 82




    $begingroup$
    It would be mathematically ironic if a question about the weakness of proof by contradiction would be closed as not constructive. [Insert a rimshot sound here]
    $endgroup$
    – Asaf Karagila
    Nov 25 '12 at 16:01
















311












$begingroup$


Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantage/disadvantages of proving by contradiction?



As an aside, how is proving by contradiction viewed in general by 'advanced' mathematicians, is it a bit of an 'easy way out' when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn't fond of proof by contradiction.










share|cite|improve this question











$endgroup$








  • 263




    $begingroup$
    Let us assume every proof by contradiction can also be shown without contradiction...
    $endgroup$
    – Graphth
    Nov 24 '12 at 18:21






  • 11




    $begingroup$
    Since this topic came up in some of the answers and comments, Andrej Bauer's blog post Proof of negation and proof by contradiction seems relevant.
    $endgroup$
    – user50545
    Nov 24 '12 at 18:21








  • 4




    $begingroup$
    ... and Gowers's blog post: When is proof by contradiction necessary?.
    $endgroup$
    – user50546
    Nov 24 '12 at 18:29








  • 4




    $begingroup$
    I should know this, but I am not sure. You are basically asking whether intuitionistic logic is equivalent to classical logic? I am pretty sure it is not.
    $endgroup$
    – Panayiotis Karabassis
    Nov 25 '12 at 0:36






  • 82




    $begingroup$
    It would be mathematically ironic if a question about the weakness of proof by contradiction would be closed as not constructive. [Insert a rimshot sound here]
    $endgroup$
    – Asaf Karagila
    Nov 25 '12 at 16:01














311












311








311


165



$begingroup$


Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantage/disadvantages of proving by contradiction?



As an aside, how is proving by contradiction viewed in general by 'advanced' mathematicians, is it a bit of an 'easy way out' when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn't fond of proof by contradiction.










share|cite|improve this question











$endgroup$




Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantage/disadvantages of proving by contradiction?



As an aside, how is proving by contradiction viewed in general by 'advanced' mathematicians, is it a bit of an 'easy way out' when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn't fond of proof by contradiction.







logic proof-writing propositional-calculus proof-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 1 '18 at 10:35









Namaste

1




1










asked Nov 24 '12 at 15:31









sonicboomsonicboom

3,73082853




3,73082853








  • 263




    $begingroup$
    Let us assume every proof by contradiction can also be shown without contradiction...
    $endgroup$
    – Graphth
    Nov 24 '12 at 18:21






  • 11




    $begingroup$
    Since this topic came up in some of the answers and comments, Andrej Bauer's blog post Proof of negation and proof by contradiction seems relevant.
    $endgroup$
    – user50545
    Nov 24 '12 at 18:21








  • 4




    $begingroup$
    ... and Gowers's blog post: When is proof by contradiction necessary?.
    $endgroup$
    – user50546
    Nov 24 '12 at 18:29








  • 4




    $begingroup$
    I should know this, but I am not sure. You are basically asking whether intuitionistic logic is equivalent to classical logic? I am pretty sure it is not.
    $endgroup$
    – Panayiotis Karabassis
    Nov 25 '12 at 0:36






  • 82




    $begingroup$
    It would be mathematically ironic if a question about the weakness of proof by contradiction would be closed as not constructive. [Insert a rimshot sound here]
    $endgroup$
    – Asaf Karagila
    Nov 25 '12 at 16:01














  • 263




    $begingroup$
    Let us assume every proof by contradiction can also be shown without contradiction...
    $endgroup$
    – Graphth
    Nov 24 '12 at 18:21






  • 11




    $begingroup$
    Since this topic came up in some of the answers and comments, Andrej Bauer's blog post Proof of negation and proof by contradiction seems relevant.
    $endgroup$
    – user50545
    Nov 24 '12 at 18:21








  • 4




    $begingroup$
    ... and Gowers's blog post: When is proof by contradiction necessary?.
    $endgroup$
    – user50546
    Nov 24 '12 at 18:29








  • 4




    $begingroup$
    I should know this, but I am not sure. You are basically asking whether intuitionistic logic is equivalent to classical logic? I am pretty sure it is not.
    $endgroup$
    – Panayiotis Karabassis
    Nov 25 '12 at 0:36






  • 82




    $begingroup$
    It would be mathematically ironic if a question about the weakness of proof by contradiction would be closed as not constructive. [Insert a rimshot sound here]
    $endgroup$
    – Asaf Karagila
    Nov 25 '12 at 16:01








263




263




$begingroup$
Let us assume every proof by contradiction can also be shown without contradiction...
$endgroup$
– Graphth
Nov 24 '12 at 18:21




$begingroup$
Let us assume every proof by contradiction can also be shown without contradiction...
$endgroup$
– Graphth
Nov 24 '12 at 18:21




11




11




$begingroup$
Since this topic came up in some of the answers and comments, Andrej Bauer's blog post Proof of negation and proof by contradiction seems relevant.
$endgroup$
– user50545
Nov 24 '12 at 18:21






$begingroup$
Since this topic came up in some of the answers and comments, Andrej Bauer's blog post Proof of negation and proof by contradiction seems relevant.
$endgroup$
– user50545
Nov 24 '12 at 18:21






4




4




$begingroup$
... and Gowers's blog post: When is proof by contradiction necessary?.
$endgroup$
– user50546
Nov 24 '12 at 18:29






$begingroup$
... and Gowers's blog post: When is proof by contradiction necessary?.
$endgroup$
– user50546
Nov 24 '12 at 18:29






4




4




$begingroup$
I should know this, but I am not sure. You are basically asking whether intuitionistic logic is equivalent to classical logic? I am pretty sure it is not.
$endgroup$
– Panayiotis Karabassis
Nov 25 '12 at 0:36




$begingroup$
I should know this, but I am not sure. You are basically asking whether intuitionistic logic is equivalent to classical logic? I am pretty sure it is not.
$endgroup$
– Panayiotis Karabassis
Nov 25 '12 at 0:36




82




82




$begingroup$
It would be mathematically ironic if a question about the weakness of proof by contradiction would be closed as not constructive. [Insert a rimshot sound here]
$endgroup$
– Asaf Karagila
Nov 25 '12 at 16:01




$begingroup$
It would be mathematically ironic if a question about the weakness of proof by contradiction would be closed as not constructive. [Insert a rimshot sound here]
$endgroup$
– Asaf Karagila
Nov 25 '12 at 16:01










14 Answers
14






active

oldest

votes


















247












$begingroup$

To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $bot$ represent an identically false proposition. Then $lnot A$, the negation of $A$, is equivalent to $A to bot$, and we take the latter to be the definition of the former in terms of $bot$.



There are two key logical principles that express different parts of what we call "proof by contradiction":




  1. The principle of explosion: for any statement $A$, we can take "$bot$ implies $A$" as an axiom. This is also called ex falso quodlibet.


  2. The law of the excluded middle: for any statement $A$, we can take "$A$ or $lnot A$" as an axiom.



In proof theory, there are three well known systems:




  • Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose.


  • Intuitionistic logic includes minimal logic and the principle of explosion


  • Classical logic includes intuitionistic logic and the law of the excluded middle



It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.



The scheme "If $A$ implies a contradiction, then $lnot A$ must hold" is true even in intuitionistic logic, because $lnot A$ is just an abbreviation for $A to bot$, and so that scheme just says "if $A to bot$ then $A to bot$". But in intuitionistic logic, if we prove $lnot A to bot$, this only shows that $lnot lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $lnot lnot A$ implies $A$, which means that in classical logic if we can prove $lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.






share|cite|improve this answer











$endgroup$









  • 38




    $begingroup$
    +1 Great explanation. Can you emphasize the affirmative answer?: So there are statements that are provable by contradiction that are not provable directly.
    $endgroup$
    – ypercubeᵀᴹ
    Nov 29 '12 at 20:10








  • 4




    $begingroup$
    What is "identically false proposition"?
    $endgroup$
    – SasQ
    Jul 11 '13 at 17:08






  • 3




    $begingroup$
    @SasQ: saying that a statement is "true" or "false" is a claim about some specific model. Saying that a statement is identically false indicates that the statement is disprovable, so it is false in all models.
    $endgroup$
    – Carl Mummert
    Jul 12 '13 at 11:53






  • 6




    $begingroup$
    Note: "identically false/true" is often called "valid/invalid" instead.
    $endgroup$
    – Noldorin
    Apr 21 '14 at 14:50






  • 3




    $begingroup$
    An excellent. It's great to see a positive characterisation of intuitionistic and classical logic, rather than the usual meaningless 'intuitionistic logic is classical logic without the law of excluded middle'.
    $endgroup$
    – Miles Rout
    Sep 12 '16 at 22:43



















56












$begingroup$

If a statements says "not $X$" then it is perfectly fine to assume $X$, arrive at a contradiction and conclude "not $X$". However, in many occasions a proof by contradiction is presented while it is really not used (let alone necessary). The reasoning then goes as follows:




Proof of $X$: Suppose not $X$. Then ... complete proof of $X$ follows here... This is a contradiction and therefore $X$.




A famous example is Euclid's proof of the infinitude of primes. It is often stated as follows (not by Euclid by the way):




Suppose there is only a finite number of primes. Then ... construction of new prime follows ... This is a contradiction so there are infinitely many primes.




Without the contradiction part, you'd be left with a perfectly fine argument. Namely, given a finite set of primes, a new prime can be constructed.



This kind of presentation is really something that you should learn to avoid. Once you're aware of this pattern its amazing how often you'll encounter it, including here on math.se.






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    Another example of this is the common presentation of the proof that there is no surjection from $Bbb N$ to $(0,1)$. The usual presentation begins "suppose we have such a surjection…" and concludes "therefore it is not a surjection, and we have a contradiction". A simpler presentation simply lets $f$ be an arbitrary function $Bbb Nto (0,1)$, follows the same argument, and concludes "…therefore, $f$ is not surjective".
    $endgroup$
    – MJD
    Nov 24 '12 at 22:08








  • 10




    $begingroup$
    Could you detail the part "construction of new prime follows" in your Euclid example please? I'd bet your version of the proof is either flawed, or uses contradiction. The key point being that "construction of new prime follows" normally relies on the assumption that only a finite number of primes exist.
    $endgroup$
    – Axel
    Nov 24 '12 at 23:16






  • 6




    $begingroup$
    @Axel: The construction is: given a (possibly empty) finite set of primes, we can construct a number larger than one which is not divisible by any prime in the set. Therefore a prime exists which is not in the set. Therefore the set of all primes must be larger than any finite set, i.e., infinite.
    $endgroup$
    – Dietrich Epp
    Nov 25 '12 at 1:34








  • 4




    $begingroup$
    @Axel: Where is the contradiction? The point illustrated in this answer is that the contradiction comes from saying "suppose that that the set of all primes is finite", but this is unnecessary for the proof. What I said is "suppose a set of primes is finite". No contradiction there.
    $endgroup$
    – Dietrich Epp
    Nov 25 '12 at 7:54






  • 3




    $begingroup$
    @DietrichEpp: You're simply not there yet. You have shown that any finite set of primes does not contain all primes. Thus the set of all primes cannot be a finite set. Fine. Now show that the set of all primes is infinite. The key part is still missing.
    $endgroup$
    – Axel
    Nov 25 '12 at 8:19





















46












$begingroup$

It somewhat depends on whether you are intuitionist or not (or both? or neither? Who knows without the law of excluded middle). According to the Wikipedia article even intuitionists accept some versions of what one could call indirect proof, but reject most. In that sense, a direct proof would be preferable (and is often even a bit more elegant).





An example:



Theorem. There exist irrational numbers $a,b$ such that $a^b$ is rational.



Proof: Assume that $a,bnotin mathbb Q$ always implies $a^bnotin mathbb Q$. Then $u:=sqrt 2^{sqrt 2}notin mathbb Q$ and $u^sqrt 2=sqrt 2^{sqrt 2cdotsqrt 2}=sqrt 2^2=2notin mathbb Q$ - contradiction!



Indeed, an intuitionist would complain that we do not exhibit a pair $(a,b)$ with $a,bnotin mathbb Q$ and $a^bin mathbb Q$. Instead, we only show that either $(sqrt 2,sqrt 2)$ or $(u,sqrt 2)$ is such a pair.
Converting the proof given above to a direct and constructive proof would in fact require you to actually prove one of the two possible options $uin mathbb Q$ or $unotinmathbb Q$.






share|cite|improve this answer











$endgroup$









  • 14




    $begingroup$
    Of course you are correct the intuitionist would complain that the proof does not exhibit a specific pair. But an intuitionist would not agree that the proof produces two pairs such that either the first pair is an example or the second is an example. The intuitionistic reading of that claimed conclusion would say that, to prove the "or", we would have to prove which pair is the example, which is exactly what the proof does not do. The intuitionist would accept the proof shown here as showing "it is not the case that for all $a,b not in mathbb{Q}$, $a^b not in mathbb{Q}$".
    $endgroup$
    – Carl Mummert
    Nov 24 '12 at 21:20








  • 6




    $begingroup$
    By Gelfond–Schneider $sqrt2^{sqrt2}$ is transcendental. Voilà! No more proof by contradiction ;-)
    $endgroup$
    – kahen
    Nov 24 '12 at 23:11








  • 4




    $begingroup$
    @kahen Well, the proof of Gelfond Schneider referenced in the wikipedia article you referenced seems to contain a step which is proven by contradiction... (if Delta not 0 ... conclude that Delta is 0)
    $endgroup$
    – mkl
    Nov 25 '12 at 22:54



















27












$begingroup$

See this post: Are proofs by contradiction weaker than other proofs?.
There are some wonderful answers related to your question - and addresses, directly, your "aside": See, in particular, what JDH writes.





One of the advantageous to constructing direct proofs of propositions, when this is feasible, is that one can discover other useful propositions in the process. That is, direct proofs help clarify the necessary and sufficient conditions that make a theorem true, and provide a structure demonstrating how these conditions relate, and how the chain of implications imply the conclusion.



Indirect proofs, on the other hand (aka "proofs by contradiction") only tell us that supposing a proposition to be otherwise leads to a contradiction at some point. But such a proof doesn't really provide the sort of insight that can be gained from direct proofs.



That is not to say that indirect proofs don't have their place (e.g., they come in handy when asked to prove propositions during a time-limited exam!). They often help "rule out" certain propositions on the basis that they contradict well established axioms or theorems. Also, indirect proofs are sometimes more intuitive than direct proofs. For example, proving that $sqrt{2}$ is not rational using a proof by contradiction is clean, and intuitive.



Sometimes an indirect proof will emerge first, after which one can seek to proceed with trying to construct a direct proof to prove the same proposition. That is, providing an indirect proof of a proposition often motivates the construction of direct proofs.





Edit:



I found this blog entry (Gowers's Weblog) When is a proof by contradiction necessary.
from which I'll quote an introductory remark:




It seems to be possible to classify theorems into three types: ones where it would be ridiculous to use contradiction, ones where there are equally sensible proofs using contradiction or not using contradiction, and ones where contradiction seems forced. But what is it that puts a theorem into one of these three categories?




The post follows immediately with a nice reply from Terence Tao.






share|cite|improve this answer











$endgroup$









  • 7




    $begingroup$
    The definition of "irrational" is "not rational", so the statement "$sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational").
    $endgroup$
    – Zhen Lin
    Nov 24 '12 at 16:48








  • 2




    $begingroup$
    What if you prove that for $n,minmathbb Z$ and $mne0$, the distance between $n/m$ and $sqrt{2}$ is at least $1/(3m^2)$? Could that qualify as a "direct" proof of irrationality?
    $endgroup$
    – Michael Hardy
    Nov 24 '12 at 17:34






  • 1




    $begingroup$
    @Michael Hardy: in fact some constuctivists actually take that to be the definition of irrational, so that this is stronger to them than just "not rational". But the classical definition of "irrational" is just "not rational", and that is the perspective of my previous comment in this thread.
    $endgroup$
    – Carl Mummert
    Nov 24 '12 at 18:28






  • 2




    $begingroup$
    @sonicboom mathoverflow.net/questions/32011/direct-proof-of-irrationality math.stackexchange.com/questions/20567/…
    $endgroup$
    – MJD
    Nov 24 '12 at 22:10






  • 3




    $begingroup$
    @ZhenLin Don't you need something more specific than "not rational" to define "irrational"? Are imaginary numbers, matrixes, dogs, and mattresses rational? Are they irrational?
    $endgroup$
    – Peter Olson
    Nov 25 '12 at 4:15



















12












$begingroup$

A few points from my (limited) experience:




  • I love proof by contradiction and I have used it in graduate level classes and no one seemed to mind so long as the logic was infallible.

  • For me, it's much easier to think about a proposition in terms of "What if this wasn't true?". That is usually my first instinct, this makes proof by contradiction the natural first choice. For instance, if I were to be asked to prove something like "Prove that a non-singular matrix has a unique inverse". My first instinct would be "What if a non-singular matrix had 2 inverses?" and from then on, the proof follows cleanly.

  • Sometimes, however, contradictions don't come cleanly and proof by simple logical deductions would probably take 5 lines whereas contradiction will take millions. I could point you to specific proofs but I'll have to do some digging. Further, if you look at every proof and try using Proof by Contradiction, another problem you will face is that sometimes, you will state your intended contradiction but never use it. In other words, solve using direct proof.

  • Another aspect about Proof by Contradiction (IMHO) is that you really must know all definitions and their equivalent statements fairly well to come up with a nice contradiction. Else, you will end up proving several lemmas on the way which looks clean in a direct proof but not so much in a Proof by Contradiction, but again, this might be a personal choice.


In summary, if you find it easier to think in terms of "What if not" then go ahead, use it but make sure your proof skills using other strategies are as good because $exists$ nail that you cannot hit with the PbC hammer that you'll carry.






share|cite|improve this answer









$endgroup$





















    12












    $begingroup$

    What is a proof by contradiction? This is actually quite difficult to answer in a satisfactory way, but usually what people mean is something like this: given a statement $phi$, a proof of $phi$ by contradiction is a derivation of a contradiction from the assumption $lnot phi$. In order to analyse this, it is very important to distinguish between the statement $phi$ and the statement $lnot lnot phi$; the two statements are formally distinct (as obvious from the fact that their written forms are different!) even though they always have the same truth value in classical logic.



    Let $bot$ denote contradiction. When we show a contradiction assuming $lnot phi$, what we have is a conditional proof of $bot$ from $lnot phi$. This can then be transformed into a proof of the statement $lnot phi to bot$, which is the long form of $lnot lnot phi$ – in other words, we have a proof that "it is not the case that $lnot phi$". This, strictly speaking, is not a complete proof of $phi$: we must still write down the last step deducing $phi$ from $lnot lnot phi$. This is the point of contention between constructivists and non-constructivists: in the constructive interpretation of logic, $lnot lnot phi$ is not only formally distinct from $phi$ but also semantically distinct; in particular, constructivists reject the principle that $phi$ can be deduced from $lnot lnot phi$ (though they may accept some limited instances of this rule).



    There is one case where proof by contradiction is always acceptable to constructivists (or at least intuitionists): this is when the statement $phi$ to be proven is itself of the form $lnot psi$. This is because it is a theorem of intuitionistic logic that $lnot lnot lnot psi$ holds if and only if $lnot psi$. On the other hand, it is also in principle possible to give a "direct" proof of $lnot psi$ in the following sense: we simply have to derive a contradiction by assuming $psi$. Any proof of $lnot psi$ by contradiction can thus be transformed into a "direct" proof because one can always derive $lnot lnot psi$ from $psi$; so if we can obtain a contradiction by assuming $lnot lnot psi$, we can certainly derive a contradiction by assuming $psi$.



    Ultimately, both of the above methods involve making a counterfactual assumption and deriving a contradiction. However, it is sometimes possible to "push" the negation inward and even eliminate it. For example, if $phi$ is the statement "there exists an $x$ such that $theta (x)$ holds", then $lnot phi$ can be deduced from the statement "$theta (x)$ does not hold for any $x$". In particular, if $theta (x)$ is itself a negative statement, say $lnot sigma (x)$, then $lnot phi$ can be deduced from the statement "$sigma (x)$ holds for all $x$". Thus, proving "there does not exist an $x$ such that $sigma (x)$ does not hold" by showing "$sigma (x)$ holds for all $x$" might be considered a more "direct" proof than either of the two previously-mentioned approaches.



    Can all proofs by contradiction be transformed into direct proofs? In some sense the answer has to be no: intuitionistic logic is known to be weaker than classical logic, i.e. there are statements have proofs in classical logic but not intuitionistic logic. The only difference between classical logic and intuitionistic logic is the principle that $phi$ is deducible from $lnot lnot phi$, so this (in some sense) implies that there are theorems that can only be proven by contradiction.



    So what are the advantages of proof by contradiction? Well, it makes proofs easier. So much so that one algorithm for automatically proving theorems in propositional logic is based on it. But it also has its disadvantages: a proof by contradiction can be more confusing (because it has counterfactual assumptions floating around!), and in a precise technical sense it is less satisfactory because it generally cannot be (re)used in constructive contexts. But most mathematicians don't worry about the latter problem.






    share|cite|improve this answer









    $endgroup$





















      8












      $begingroup$

      As "Inquest"'s answer mentions, it's often easier to find a proof by contradiction than a direct proof. But after you do that, you can often make the proof simpler by rearranging it into a direct proof. It is not good to make a proof appear more complicated than it really is.



      To see another disadvantage of some proofs by contradiction, consider this:



      Proof: To prove $A$, assume not $A$. [insert 50 pages of argument here] We have reached a contradiction. Therefore $A$. End of proof



      Now ask yourself: Which of the propositions proved in those 50 pages are erroneous and could be proved only because one relied on the false assumption that not $A$, and which are validly proved, and which are true but not validly proved because the assumption that not $A$ was relied on? It's not so easy to tell without a lot more work. And if you remember a proof of one of those propositions, you might just mistakenly think that it's been proved and is therefore known to be true. So it might be far better to limit the use of proof by contradiction to some portions of those 50 pages where no other method works.



      Perhaps proofs of non-existence can be done only by contradiction. Here I might offer as an example the various proofs of the irrationality of $sqrt{2}$, but for the fact that I've seen it asserted that if $m$, $n$ are integers, than $m/n$ differs from $sqrt{2}$ by at least an amount that depends on $n$ --- I think it might have been $1/(3n^2)$. Here's another example: How would one prove the non-existence of a non-trivial (i.e. $>1$) common divisor of $n$ and $n+1$?



      I've seen a book on logic asserting that a proof by contradiction of a non-existence assertion does not constitute an "indirect proof", since the assertion is inherently negative. I don't know how conventional that is.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Formally, a negative statement such as $lnot phi$ is an abbreviation for $phi to bot$, i.e. it is the statement that $phi$ leads to an absurdity; accordingly, to prove $lnot phi$, one must show that assuming $phi$ leads to contradiction. This is completely orthodox logic.
        $endgroup$
        – Zhen Lin
        Nov 24 '12 at 16:46










      • $begingroup$
        But when must a statement be written in the form $lnotvarphi$ ?
        $endgroup$
        – Michael Hardy
        Nov 24 '12 at 16:50












      • $begingroup$
        There isn't any "must" about it. If $phi$ is a compound formula such as $psi lor theta$ you can just push the $lnot$ inward and get $(lnot psi) land (lnot theta)$. But that just causes the number of negative statements to multiply...
        $endgroup$
        – Zhen Lin
        Nov 24 '12 at 17:01










      • $begingroup$
        So when must it be written in a form that has negative statements? Does the answer depend on which formal language you write it in?
        $endgroup$
        – Michael Hardy
        Nov 24 '12 at 17:31






      • 2




        $begingroup$
        Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $ne$? Sometimes constructivists take an "apartness" relation $mathrel{#}$ to be primitive, in which case $=$ becomes the negation of $mathrel{#}$!)
        $endgroup$
        – Zhen Lin
        Nov 24 '12 at 18:05





















      7












      $begingroup$

      Another example of a contradiction proof that provides no idea on a constructive proof is the strategy-stealing argument. For certain symmetric games, the second player cannot have a winning strategy. If he did, the first player could "pretend" to be the second player and steal his winning strategy, stealing it from him, a contradiction.



      An interesting example is the game Hex. It is easy to show that Hex cannot end in a tie, and the strategy-stealing argument does apply to it. Therefore, it is a first player win. But for symmetric $n$ x $n$, the actual winning strategy is still not known. Thus, this is an example of something that has been proven using contradiction and not constructively (yet).






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        I would formulate it as: Let $S$ be any strategy for the second player. Then by symmetry the first player can apply $S$ too. Since not both players can win and both are applying $S$, $S$ is not a winning strategy.
        $endgroup$
        – WimC
        Nov 27 '12 at 21:31



















      5












      $begingroup$

      There is nothing wrong with proof by contradiction. You can show that they work using a truth table. In the end, that's all that really matters, right?



      As far as I know, you can't know for certain that something is not provable by a direct proof. However, a proof by contradiction might be an easier way to prove some things, like the irrationality of certain numbers. For example, I have never seen a direct proof of the irrationality of $sqrt{2}$.



      EDIT: As Carl Mummert said in his answer, the above part in italics is not true. There are propositions which are only provable by contradiction.



      A proof by contradiction can be also be formulated as a proof by contrapositive. If we know $Q$ is false, if we can show $PRightarrow Q$ then we have proved that $P$ is false. Whether you view this as "proof without contradiction" or not is up to you. In any case, they are logically equivalent.






      share|cite|improve this answer











      $endgroup$









      • 4




        $begingroup$
        As I explained in my answer, we do know that there are things that are not provable directly modulo the usual formalization of proofs. The usual proof that $sqrt{2}$ is not rational is a direct proof when it is formalized in the usual way, although it is a direct proof of a negative statement, which can be deceptive at first.
        $endgroup$
        – Carl Mummert
        Nov 24 '12 at 19:10








      • 1




        $begingroup$
        I found the link I had lost, which explains some of this: math.andrej.com/2010/03/29/…
        $endgroup$
        – Carl Mummert
        Nov 24 '12 at 22:43






      • 1




        $begingroup$
        Thank you for pointing this out, and for the reference. Both your comment and the article were very informative and enlightening!
        $endgroup$
        – Espen Nielsen
        Nov 24 '12 at 23:12



















      3












      $begingroup$

      I do believe there are some proofs that are only demonstrable through contradiction, and I'm going to attempt to describe them logically:



      Let X be a logical statement such that: X $rightarrow$ y, where y is a known contradiction (such as 2+2=5 in the normal arithmetic structure). Without knowing anything else of X, $neg$X implies nothing and nothing implies $neg$X (and hence is not provable). But, of course, assuming X implies a contradiction, and thus, $neg$X.



      This form of statement X is isolated, in that it only relates to itself and the contradiction. I do believe they can be constructed though, for it seems it they can be described.



      With that said, in real math and logic, or in general real world scenarios, I don't believe any statements of this form exist, except possibly ones that are constructed to meet this criteria and otherwise meaningless. The proof of the primes was eventually proved without contradiction, to my understanding; until math had been more developed, I think that the statement "the number of primes is $infty$" was basically an isolated logical statement at Euclid's time and for many years after probably, in that there were no other things known to imply it and it didn't imply anything else useful towards its proof.






      share|cite|improve this answer









      $endgroup$





















        3












        $begingroup$

        First of all, this is not an answer to the title but to the aside question and is just an example of why you would prefer a constructive proof to a proof by contradiction. Consider the example below,



        Prove that $x^2 = 1$ has a root.



        Proof by contradiction: Assume that $x^2 = 1$ has no root. Let $f(x) = x^2 - 1 $ then $x^2 = 1$ has a root if and only if $f(x_0) = 0$ for some $x_0$. By assumption $x^2 =1$ has no root and thus, $f(x) neq 0$ for every $x$. Note that $f$ is continuous and $f(0) = -1$ and $f(2) = 3$. Hence, by the intermediate value theorem, $exists x_0$ such that $f(x_0) = 0$ which is a contradiction. Therefore, $x^2=1$ has a root.



        Constructive proof: For $x^2=1$ if and only if $x^2-1=0$ iff $(x-1)(x+1)=0$. Hence, for $x=pm 1$ the equation is satisfied, namely the roots are $-1$ and $1$.



        The difference is not about the length of the proofs but the information you have. In constructive proof, you know what the roots are but not in the proof by contradiction. Of course, in proof by contradiction, you could have said "let $x_0 = 1$, then $x_0^2=1$ which is a contradiction since $1$ is a root." but then, it is not a clear distinction between the two types of proofs.






        share|cite|improve this answer









        $endgroup$





















          3












          $begingroup$

          My non-mathematical response.



          A == B equals !(A != B)



          You always end up with a binary decision, is or is not. And in any language is = !(is not).



          But I guess it is too simple to be ok.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            is = !(is not) does not hold in minimal and intuitionistic logic. See Carl's answer.
            $endgroup$
            – user26486
            Feb 24 '15 at 18:03





















          2












          $begingroup$

          Whether a proof is "by contradiction" really just depends on the statement you started with. If your inital statement is $P rightarrow Q$, then showing the equivalent $neg Q rightarrow neg P$ is "proof by contradiction". But in reality, the "direct" proof for $ P rightarrow Q$ is just a proof "by contradiction" for $neg Q rightarrow neg P$. The only reason why we started with $P rightarrow Q$ instead of $neg Q rightarrow neg P$ is our intuition.



          This is just my opinion, but also remember that sometimes, it is also very valuable to know what holds if $Q$ does not hold.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It really depends by what you accept as proof. Every logical system has certain logical axioms, nobody is forcing you to accept them. $neg neg A leftrightarrow A$ is in fact not provable in intuitionistic logic.
            $endgroup$
            – Panayiotis Karabassis
            Nov 25 '12 at 0:41










          • $begingroup$
            ok fine. while i don't know what intuitionistic logic is, i was assuming that the law of excluded middle would hold.
            $endgroup$
            – kutschkem
            Nov 27 '12 at 11:21



















          2












          $begingroup$

          An interesting example of this is the entire study of Smooth Infinitesimal Analysis. It relies on not having the law of the excluded middle (i.e. no proof by contradictions are accepted) in order to be valid. Thus if everything provable by contradiction was also able to be proven directly, then there could not be smooth infinitesimal analysis! Look at Bell's book for more details, though the wiki gives a good example.






          share|cite|improve this answer









          $endgroup$












            protected by user99914 Feb 7 '18 at 2:40



            Thank you for your interest in this question.
            Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



            Would you like to answer one of these unanswered questions instead?














            14 Answers
            14






            active

            oldest

            votes








            14 Answers
            14






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            247












            $begingroup$

            To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $bot$ represent an identically false proposition. Then $lnot A$, the negation of $A$, is equivalent to $A to bot$, and we take the latter to be the definition of the former in terms of $bot$.



            There are two key logical principles that express different parts of what we call "proof by contradiction":




            1. The principle of explosion: for any statement $A$, we can take "$bot$ implies $A$" as an axiom. This is also called ex falso quodlibet.


            2. The law of the excluded middle: for any statement $A$, we can take "$A$ or $lnot A$" as an axiom.



            In proof theory, there are three well known systems:




            • Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose.


            • Intuitionistic logic includes minimal logic and the principle of explosion


            • Classical logic includes intuitionistic logic and the law of the excluded middle



            It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.



            The scheme "If $A$ implies a contradiction, then $lnot A$ must hold" is true even in intuitionistic logic, because $lnot A$ is just an abbreviation for $A to bot$, and so that scheme just says "if $A to bot$ then $A to bot$". But in intuitionistic logic, if we prove $lnot A to bot$, this only shows that $lnot lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $lnot lnot A$ implies $A$, which means that in classical logic if we can prove $lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.






            share|cite|improve this answer











            $endgroup$









            • 38




              $begingroup$
              +1 Great explanation. Can you emphasize the affirmative answer?: So there are statements that are provable by contradiction that are not provable directly.
              $endgroup$
              – ypercubeᵀᴹ
              Nov 29 '12 at 20:10








            • 4




              $begingroup$
              What is "identically false proposition"?
              $endgroup$
              – SasQ
              Jul 11 '13 at 17:08






            • 3




              $begingroup$
              @SasQ: saying that a statement is "true" or "false" is a claim about some specific model. Saying that a statement is identically false indicates that the statement is disprovable, so it is false in all models.
              $endgroup$
              – Carl Mummert
              Jul 12 '13 at 11:53






            • 6




              $begingroup$
              Note: "identically false/true" is often called "valid/invalid" instead.
              $endgroup$
              – Noldorin
              Apr 21 '14 at 14:50






            • 3




              $begingroup$
              An excellent. It's great to see a positive characterisation of intuitionistic and classical logic, rather than the usual meaningless 'intuitionistic logic is classical logic without the law of excluded middle'.
              $endgroup$
              – Miles Rout
              Sep 12 '16 at 22:43
















            247












            $begingroup$

            To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $bot$ represent an identically false proposition. Then $lnot A$, the negation of $A$, is equivalent to $A to bot$, and we take the latter to be the definition of the former in terms of $bot$.



            There are two key logical principles that express different parts of what we call "proof by contradiction":




            1. The principle of explosion: for any statement $A$, we can take "$bot$ implies $A$" as an axiom. This is also called ex falso quodlibet.


            2. The law of the excluded middle: for any statement $A$, we can take "$A$ or $lnot A$" as an axiom.



            In proof theory, there are three well known systems:




            • Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose.


            • Intuitionistic logic includes minimal logic and the principle of explosion


            • Classical logic includes intuitionistic logic and the law of the excluded middle



            It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.



            The scheme "If $A$ implies a contradiction, then $lnot A$ must hold" is true even in intuitionistic logic, because $lnot A$ is just an abbreviation for $A to bot$, and so that scheme just says "if $A to bot$ then $A to bot$". But in intuitionistic logic, if we prove $lnot A to bot$, this only shows that $lnot lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $lnot lnot A$ implies $A$, which means that in classical logic if we can prove $lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.






            share|cite|improve this answer











            $endgroup$









            • 38




              $begingroup$
              +1 Great explanation. Can you emphasize the affirmative answer?: So there are statements that are provable by contradiction that are not provable directly.
              $endgroup$
              – ypercubeᵀᴹ
              Nov 29 '12 at 20:10








            • 4




              $begingroup$
              What is "identically false proposition"?
              $endgroup$
              – SasQ
              Jul 11 '13 at 17:08






            • 3




              $begingroup$
              @SasQ: saying that a statement is "true" or "false" is a claim about some specific model. Saying that a statement is identically false indicates that the statement is disprovable, so it is false in all models.
              $endgroup$
              – Carl Mummert
              Jul 12 '13 at 11:53






            • 6




              $begingroup$
              Note: "identically false/true" is often called "valid/invalid" instead.
              $endgroup$
              – Noldorin
              Apr 21 '14 at 14:50






            • 3




              $begingroup$
              An excellent. It's great to see a positive characterisation of intuitionistic and classical logic, rather than the usual meaningless 'intuitionistic logic is classical logic without the law of excluded middle'.
              $endgroup$
              – Miles Rout
              Sep 12 '16 at 22:43














            247












            247








            247





            $begingroup$

            To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $bot$ represent an identically false proposition. Then $lnot A$, the negation of $A$, is equivalent to $A to bot$, and we take the latter to be the definition of the former in terms of $bot$.



            There are two key logical principles that express different parts of what we call "proof by contradiction":




            1. The principle of explosion: for any statement $A$, we can take "$bot$ implies $A$" as an axiom. This is also called ex falso quodlibet.


            2. The law of the excluded middle: for any statement $A$, we can take "$A$ or $lnot A$" as an axiom.



            In proof theory, there are three well known systems:




            • Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose.


            • Intuitionistic logic includes minimal logic and the principle of explosion


            • Classical logic includes intuitionistic logic and the law of the excluded middle



            It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.



            The scheme "If $A$ implies a contradiction, then $lnot A$ must hold" is true even in intuitionistic logic, because $lnot A$ is just an abbreviation for $A to bot$, and so that scheme just says "if $A to bot$ then $A to bot$". But in intuitionistic logic, if we prove $lnot A to bot$, this only shows that $lnot lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $lnot lnot A$ implies $A$, which means that in classical logic if we can prove $lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.






            share|cite|improve this answer











            $endgroup$



            To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $bot$ represent an identically false proposition. Then $lnot A$, the negation of $A$, is equivalent to $A to bot$, and we take the latter to be the definition of the former in terms of $bot$.



            There are two key logical principles that express different parts of what we call "proof by contradiction":




            1. The principle of explosion: for any statement $A$, we can take "$bot$ implies $A$" as an axiom. This is also called ex falso quodlibet.


            2. The law of the excluded middle: for any statement $A$, we can take "$A$ or $lnot A$" as an axiom.



            In proof theory, there are three well known systems:




            • Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose.


            • Intuitionistic logic includes minimal logic and the principle of explosion


            • Classical logic includes intuitionistic logic and the law of the excluded middle



            It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.



            The scheme "If $A$ implies a contradiction, then $lnot A$ must hold" is true even in intuitionistic logic, because $lnot A$ is just an abbreviation for $A to bot$, and so that scheme just says "if $A to bot$ then $A to bot$". But in intuitionistic logic, if we prove $lnot A to bot$, this only shows that $lnot lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $lnot lnot A$ implies $A$, which means that in classical logic if we can prove $lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 24 '12 at 18:29

























            answered Nov 24 '12 at 18:08









            Carl MummertCarl Mummert

            67.5k7133251




            67.5k7133251








            • 38




              $begingroup$
              +1 Great explanation. Can you emphasize the affirmative answer?: So there are statements that are provable by contradiction that are not provable directly.
              $endgroup$
              – ypercubeᵀᴹ
              Nov 29 '12 at 20:10








            • 4




              $begingroup$
              What is "identically false proposition"?
              $endgroup$
              – SasQ
              Jul 11 '13 at 17:08






            • 3




              $begingroup$
              @SasQ: saying that a statement is "true" or "false" is a claim about some specific model. Saying that a statement is identically false indicates that the statement is disprovable, so it is false in all models.
              $endgroup$
              – Carl Mummert
              Jul 12 '13 at 11:53






            • 6




              $begingroup$
              Note: "identically false/true" is often called "valid/invalid" instead.
              $endgroup$
              – Noldorin
              Apr 21 '14 at 14:50






            • 3




              $begingroup$
              An excellent. It's great to see a positive characterisation of intuitionistic and classical logic, rather than the usual meaningless 'intuitionistic logic is classical logic without the law of excluded middle'.
              $endgroup$
              – Miles Rout
              Sep 12 '16 at 22:43














            • 38




              $begingroup$
              +1 Great explanation. Can you emphasize the affirmative answer?: So there are statements that are provable by contradiction that are not provable directly.
              $endgroup$
              – ypercubeᵀᴹ
              Nov 29 '12 at 20:10








            • 4




              $begingroup$
              What is "identically false proposition"?
              $endgroup$
              – SasQ
              Jul 11 '13 at 17:08






            • 3




              $begingroup$
              @SasQ: saying that a statement is "true" or "false" is a claim about some specific model. Saying that a statement is identically false indicates that the statement is disprovable, so it is false in all models.
              $endgroup$
              – Carl Mummert
              Jul 12 '13 at 11:53






            • 6




              $begingroup$
              Note: "identically false/true" is often called "valid/invalid" instead.
              $endgroup$
              – Noldorin
              Apr 21 '14 at 14:50






            • 3




              $begingroup$
              An excellent. It's great to see a positive characterisation of intuitionistic and classical logic, rather than the usual meaningless 'intuitionistic logic is classical logic without the law of excluded middle'.
              $endgroup$
              – Miles Rout
              Sep 12 '16 at 22:43








            38




            38




            $begingroup$
            +1 Great explanation. Can you emphasize the affirmative answer?: So there are statements that are provable by contradiction that are not provable directly.
            $endgroup$
            – ypercubeᵀᴹ
            Nov 29 '12 at 20:10






            $begingroup$
            +1 Great explanation. Can you emphasize the affirmative answer?: So there are statements that are provable by contradiction that are not provable directly.
            $endgroup$
            – ypercubeᵀᴹ
            Nov 29 '12 at 20:10






            4




            4




            $begingroup$
            What is "identically false proposition"?
            $endgroup$
            – SasQ
            Jul 11 '13 at 17:08




            $begingroup$
            What is "identically false proposition"?
            $endgroup$
            – SasQ
            Jul 11 '13 at 17:08




            3




            3




            $begingroup$
            @SasQ: saying that a statement is "true" or "false" is a claim about some specific model. Saying that a statement is identically false indicates that the statement is disprovable, so it is false in all models.
            $endgroup$
            – Carl Mummert
            Jul 12 '13 at 11:53




            $begingroup$
            @SasQ: saying that a statement is "true" or "false" is a claim about some specific model. Saying that a statement is identically false indicates that the statement is disprovable, so it is false in all models.
            $endgroup$
            – Carl Mummert
            Jul 12 '13 at 11:53




            6




            6




            $begingroup$
            Note: "identically false/true" is often called "valid/invalid" instead.
            $endgroup$
            – Noldorin
            Apr 21 '14 at 14:50




            $begingroup$
            Note: "identically false/true" is often called "valid/invalid" instead.
            $endgroup$
            – Noldorin
            Apr 21 '14 at 14:50




            3




            3




            $begingroup$
            An excellent. It's great to see a positive characterisation of intuitionistic and classical logic, rather than the usual meaningless 'intuitionistic logic is classical logic without the law of excluded middle'.
            $endgroup$
            – Miles Rout
            Sep 12 '16 at 22:43




            $begingroup$
            An excellent. It's great to see a positive characterisation of intuitionistic and classical logic, rather than the usual meaningless 'intuitionistic logic is classical logic without the law of excluded middle'.
            $endgroup$
            – Miles Rout
            Sep 12 '16 at 22:43











            56












            $begingroup$

            If a statements says "not $X$" then it is perfectly fine to assume $X$, arrive at a contradiction and conclude "not $X$". However, in many occasions a proof by contradiction is presented while it is really not used (let alone necessary). The reasoning then goes as follows:




            Proof of $X$: Suppose not $X$. Then ... complete proof of $X$ follows here... This is a contradiction and therefore $X$.




            A famous example is Euclid's proof of the infinitude of primes. It is often stated as follows (not by Euclid by the way):




            Suppose there is only a finite number of primes. Then ... construction of new prime follows ... This is a contradiction so there are infinitely many primes.




            Without the contradiction part, you'd be left with a perfectly fine argument. Namely, given a finite set of primes, a new prime can be constructed.



            This kind of presentation is really something that you should learn to avoid. Once you're aware of this pattern its amazing how often you'll encounter it, including here on math.se.






            share|cite|improve this answer









            $endgroup$









            • 4




              $begingroup$
              Another example of this is the common presentation of the proof that there is no surjection from $Bbb N$ to $(0,1)$. The usual presentation begins "suppose we have such a surjection…" and concludes "therefore it is not a surjection, and we have a contradiction". A simpler presentation simply lets $f$ be an arbitrary function $Bbb Nto (0,1)$, follows the same argument, and concludes "…therefore, $f$ is not surjective".
              $endgroup$
              – MJD
              Nov 24 '12 at 22:08








            • 10




              $begingroup$
              Could you detail the part "construction of new prime follows" in your Euclid example please? I'd bet your version of the proof is either flawed, or uses contradiction. The key point being that "construction of new prime follows" normally relies on the assumption that only a finite number of primes exist.
              $endgroup$
              – Axel
              Nov 24 '12 at 23:16






            • 6




              $begingroup$
              @Axel: The construction is: given a (possibly empty) finite set of primes, we can construct a number larger than one which is not divisible by any prime in the set. Therefore a prime exists which is not in the set. Therefore the set of all primes must be larger than any finite set, i.e., infinite.
              $endgroup$
              – Dietrich Epp
              Nov 25 '12 at 1:34








            • 4




              $begingroup$
              @Axel: Where is the contradiction? The point illustrated in this answer is that the contradiction comes from saying "suppose that that the set of all primes is finite", but this is unnecessary for the proof. What I said is "suppose a set of primes is finite". No contradiction there.
              $endgroup$
              – Dietrich Epp
              Nov 25 '12 at 7:54






            • 3




              $begingroup$
              @DietrichEpp: You're simply not there yet. You have shown that any finite set of primes does not contain all primes. Thus the set of all primes cannot be a finite set. Fine. Now show that the set of all primes is infinite. The key part is still missing.
              $endgroup$
              – Axel
              Nov 25 '12 at 8:19


















            56












            $begingroup$

            If a statements says "not $X$" then it is perfectly fine to assume $X$, arrive at a contradiction and conclude "not $X$". However, in many occasions a proof by contradiction is presented while it is really not used (let alone necessary). The reasoning then goes as follows:




            Proof of $X$: Suppose not $X$. Then ... complete proof of $X$ follows here... This is a contradiction and therefore $X$.




            A famous example is Euclid's proof of the infinitude of primes. It is often stated as follows (not by Euclid by the way):




            Suppose there is only a finite number of primes. Then ... construction of new prime follows ... This is a contradiction so there are infinitely many primes.




            Without the contradiction part, you'd be left with a perfectly fine argument. Namely, given a finite set of primes, a new prime can be constructed.



            This kind of presentation is really something that you should learn to avoid. Once you're aware of this pattern its amazing how often you'll encounter it, including here on math.se.






            share|cite|improve this answer









            $endgroup$









            • 4




              $begingroup$
              Another example of this is the common presentation of the proof that there is no surjection from $Bbb N$ to $(0,1)$. The usual presentation begins "suppose we have such a surjection…" and concludes "therefore it is not a surjection, and we have a contradiction". A simpler presentation simply lets $f$ be an arbitrary function $Bbb Nto (0,1)$, follows the same argument, and concludes "…therefore, $f$ is not surjective".
              $endgroup$
              – MJD
              Nov 24 '12 at 22:08








            • 10




              $begingroup$
              Could you detail the part "construction of new prime follows" in your Euclid example please? I'd bet your version of the proof is either flawed, or uses contradiction. The key point being that "construction of new prime follows" normally relies on the assumption that only a finite number of primes exist.
              $endgroup$
              – Axel
              Nov 24 '12 at 23:16






            • 6




              $begingroup$
              @Axel: The construction is: given a (possibly empty) finite set of primes, we can construct a number larger than one which is not divisible by any prime in the set. Therefore a prime exists which is not in the set. Therefore the set of all primes must be larger than any finite set, i.e., infinite.
              $endgroup$
              – Dietrich Epp
              Nov 25 '12 at 1:34








            • 4




              $begingroup$
              @Axel: Where is the contradiction? The point illustrated in this answer is that the contradiction comes from saying "suppose that that the set of all primes is finite", but this is unnecessary for the proof. What I said is "suppose a set of primes is finite". No contradiction there.
              $endgroup$
              – Dietrich Epp
              Nov 25 '12 at 7:54






            • 3




              $begingroup$
              @DietrichEpp: You're simply not there yet. You have shown that any finite set of primes does not contain all primes. Thus the set of all primes cannot be a finite set. Fine. Now show that the set of all primes is infinite. The key part is still missing.
              $endgroup$
              – Axel
              Nov 25 '12 at 8:19
















            56












            56








            56





            $begingroup$

            If a statements says "not $X$" then it is perfectly fine to assume $X$, arrive at a contradiction and conclude "not $X$". However, in many occasions a proof by contradiction is presented while it is really not used (let alone necessary). The reasoning then goes as follows:




            Proof of $X$: Suppose not $X$. Then ... complete proof of $X$ follows here... This is a contradiction and therefore $X$.




            A famous example is Euclid's proof of the infinitude of primes. It is often stated as follows (not by Euclid by the way):




            Suppose there is only a finite number of primes. Then ... construction of new prime follows ... This is a contradiction so there are infinitely many primes.




            Without the contradiction part, you'd be left with a perfectly fine argument. Namely, given a finite set of primes, a new prime can be constructed.



            This kind of presentation is really something that you should learn to avoid. Once you're aware of this pattern its amazing how often you'll encounter it, including here on math.se.






            share|cite|improve this answer









            $endgroup$



            If a statements says "not $X$" then it is perfectly fine to assume $X$, arrive at a contradiction and conclude "not $X$". However, in many occasions a proof by contradiction is presented while it is really not used (let alone necessary). The reasoning then goes as follows:




            Proof of $X$: Suppose not $X$. Then ... complete proof of $X$ follows here... This is a contradiction and therefore $X$.




            A famous example is Euclid's proof of the infinitude of primes. It is often stated as follows (not by Euclid by the way):




            Suppose there is only a finite number of primes. Then ... construction of new prime follows ... This is a contradiction so there are infinitely many primes.




            Without the contradiction part, you'd be left with a perfectly fine argument. Namely, given a finite set of primes, a new prime can be constructed.



            This kind of presentation is really something that you should learn to avoid. Once you're aware of this pattern its amazing how often you'll encounter it, including here on math.se.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 '12 at 16:06









            WimCWimC

            24.3k23063




            24.3k23063








            • 4




              $begingroup$
              Another example of this is the common presentation of the proof that there is no surjection from $Bbb N$ to $(0,1)$. The usual presentation begins "suppose we have such a surjection…" and concludes "therefore it is not a surjection, and we have a contradiction". A simpler presentation simply lets $f$ be an arbitrary function $Bbb Nto (0,1)$, follows the same argument, and concludes "…therefore, $f$ is not surjective".
              $endgroup$
              – MJD
              Nov 24 '12 at 22:08








            • 10




              $begingroup$
              Could you detail the part "construction of new prime follows" in your Euclid example please? I'd bet your version of the proof is either flawed, or uses contradiction. The key point being that "construction of new prime follows" normally relies on the assumption that only a finite number of primes exist.
              $endgroup$
              – Axel
              Nov 24 '12 at 23:16






            • 6




              $begingroup$
              @Axel: The construction is: given a (possibly empty) finite set of primes, we can construct a number larger than one which is not divisible by any prime in the set. Therefore a prime exists which is not in the set. Therefore the set of all primes must be larger than any finite set, i.e., infinite.
              $endgroup$
              – Dietrich Epp
              Nov 25 '12 at 1:34








            • 4




              $begingroup$
              @Axel: Where is the contradiction? The point illustrated in this answer is that the contradiction comes from saying "suppose that that the set of all primes is finite", but this is unnecessary for the proof. What I said is "suppose a set of primes is finite". No contradiction there.
              $endgroup$
              – Dietrich Epp
              Nov 25 '12 at 7:54






            • 3




              $begingroup$
              @DietrichEpp: You're simply not there yet. You have shown that any finite set of primes does not contain all primes. Thus the set of all primes cannot be a finite set. Fine. Now show that the set of all primes is infinite. The key part is still missing.
              $endgroup$
              – Axel
              Nov 25 '12 at 8:19
















            • 4




              $begingroup$
              Another example of this is the common presentation of the proof that there is no surjection from $Bbb N$ to $(0,1)$. The usual presentation begins "suppose we have such a surjection…" and concludes "therefore it is not a surjection, and we have a contradiction". A simpler presentation simply lets $f$ be an arbitrary function $Bbb Nto (0,1)$, follows the same argument, and concludes "…therefore, $f$ is not surjective".
              $endgroup$
              – MJD
              Nov 24 '12 at 22:08








            • 10




              $begingroup$
              Could you detail the part "construction of new prime follows" in your Euclid example please? I'd bet your version of the proof is either flawed, or uses contradiction. The key point being that "construction of new prime follows" normally relies on the assumption that only a finite number of primes exist.
              $endgroup$
              – Axel
              Nov 24 '12 at 23:16






            • 6




              $begingroup$
              @Axel: The construction is: given a (possibly empty) finite set of primes, we can construct a number larger than one which is not divisible by any prime in the set. Therefore a prime exists which is not in the set. Therefore the set of all primes must be larger than any finite set, i.e., infinite.
              $endgroup$
              – Dietrich Epp
              Nov 25 '12 at 1:34








            • 4




              $begingroup$
              @Axel: Where is the contradiction? The point illustrated in this answer is that the contradiction comes from saying "suppose that that the set of all primes is finite", but this is unnecessary for the proof. What I said is "suppose a set of primes is finite". No contradiction there.
              $endgroup$
              – Dietrich Epp
              Nov 25 '12 at 7:54






            • 3




              $begingroup$
              @DietrichEpp: You're simply not there yet. You have shown that any finite set of primes does not contain all primes. Thus the set of all primes cannot be a finite set. Fine. Now show that the set of all primes is infinite. The key part is still missing.
              $endgroup$
              – Axel
              Nov 25 '12 at 8:19










            4




            4




            $begingroup$
            Another example of this is the common presentation of the proof that there is no surjection from $Bbb N$ to $(0,1)$. The usual presentation begins "suppose we have such a surjection…" and concludes "therefore it is not a surjection, and we have a contradiction". A simpler presentation simply lets $f$ be an arbitrary function $Bbb Nto (0,1)$, follows the same argument, and concludes "…therefore, $f$ is not surjective".
            $endgroup$
            – MJD
            Nov 24 '12 at 22:08






            $begingroup$
            Another example of this is the common presentation of the proof that there is no surjection from $Bbb N$ to $(0,1)$. The usual presentation begins "suppose we have such a surjection…" and concludes "therefore it is not a surjection, and we have a contradiction". A simpler presentation simply lets $f$ be an arbitrary function $Bbb Nto (0,1)$, follows the same argument, and concludes "…therefore, $f$ is not surjective".
            $endgroup$
            – MJD
            Nov 24 '12 at 22:08






            10




            10




            $begingroup$
            Could you detail the part "construction of new prime follows" in your Euclid example please? I'd bet your version of the proof is either flawed, or uses contradiction. The key point being that "construction of new prime follows" normally relies on the assumption that only a finite number of primes exist.
            $endgroup$
            – Axel
            Nov 24 '12 at 23:16




            $begingroup$
            Could you detail the part "construction of new prime follows" in your Euclid example please? I'd bet your version of the proof is either flawed, or uses contradiction. The key point being that "construction of new prime follows" normally relies on the assumption that only a finite number of primes exist.
            $endgroup$
            – Axel
            Nov 24 '12 at 23:16




            6




            6




            $begingroup$
            @Axel: The construction is: given a (possibly empty) finite set of primes, we can construct a number larger than one which is not divisible by any prime in the set. Therefore a prime exists which is not in the set. Therefore the set of all primes must be larger than any finite set, i.e., infinite.
            $endgroup$
            – Dietrich Epp
            Nov 25 '12 at 1:34






            $begingroup$
            @Axel: The construction is: given a (possibly empty) finite set of primes, we can construct a number larger than one which is not divisible by any prime in the set. Therefore a prime exists which is not in the set. Therefore the set of all primes must be larger than any finite set, i.e., infinite.
            $endgroup$
            – Dietrich Epp
            Nov 25 '12 at 1:34






            4




            4




            $begingroup$
            @Axel: Where is the contradiction? The point illustrated in this answer is that the contradiction comes from saying "suppose that that the set of all primes is finite", but this is unnecessary for the proof. What I said is "suppose a set of primes is finite". No contradiction there.
            $endgroup$
            – Dietrich Epp
            Nov 25 '12 at 7:54




            $begingroup$
            @Axel: Where is the contradiction? The point illustrated in this answer is that the contradiction comes from saying "suppose that that the set of all primes is finite", but this is unnecessary for the proof. What I said is "suppose a set of primes is finite". No contradiction there.
            $endgroup$
            – Dietrich Epp
            Nov 25 '12 at 7:54




            3




            3




            $begingroup$
            @DietrichEpp: You're simply not there yet. You have shown that any finite set of primes does not contain all primes. Thus the set of all primes cannot be a finite set. Fine. Now show that the set of all primes is infinite. The key part is still missing.
            $endgroup$
            – Axel
            Nov 25 '12 at 8:19






            $begingroup$
            @DietrichEpp: You're simply not there yet. You have shown that any finite set of primes does not contain all primes. Thus the set of all primes cannot be a finite set. Fine. Now show that the set of all primes is infinite. The key part is still missing.
            $endgroup$
            – Axel
            Nov 25 '12 at 8:19













            46












            $begingroup$

            It somewhat depends on whether you are intuitionist or not (or both? or neither? Who knows without the law of excluded middle). According to the Wikipedia article even intuitionists accept some versions of what one could call indirect proof, but reject most. In that sense, a direct proof would be preferable (and is often even a bit more elegant).





            An example:



            Theorem. There exist irrational numbers $a,b$ such that $a^b$ is rational.



            Proof: Assume that $a,bnotin mathbb Q$ always implies $a^bnotin mathbb Q$. Then $u:=sqrt 2^{sqrt 2}notin mathbb Q$ and $u^sqrt 2=sqrt 2^{sqrt 2cdotsqrt 2}=sqrt 2^2=2notin mathbb Q$ - contradiction!



            Indeed, an intuitionist would complain that we do not exhibit a pair $(a,b)$ with $a,bnotin mathbb Q$ and $a^bin mathbb Q$. Instead, we only show that either $(sqrt 2,sqrt 2)$ or $(u,sqrt 2)$ is such a pair.
            Converting the proof given above to a direct and constructive proof would in fact require you to actually prove one of the two possible options $uin mathbb Q$ or $unotinmathbb Q$.






            share|cite|improve this answer











            $endgroup$









            • 14




              $begingroup$
              Of course you are correct the intuitionist would complain that the proof does not exhibit a specific pair. But an intuitionist would not agree that the proof produces two pairs such that either the first pair is an example or the second is an example. The intuitionistic reading of that claimed conclusion would say that, to prove the "or", we would have to prove which pair is the example, which is exactly what the proof does not do. The intuitionist would accept the proof shown here as showing "it is not the case that for all $a,b not in mathbb{Q}$, $a^b not in mathbb{Q}$".
              $endgroup$
              – Carl Mummert
              Nov 24 '12 at 21:20








            • 6




              $begingroup$
              By Gelfond–Schneider $sqrt2^{sqrt2}$ is transcendental. Voilà! No more proof by contradiction ;-)
              $endgroup$
              – kahen
              Nov 24 '12 at 23:11








            • 4




              $begingroup$
              @kahen Well, the proof of Gelfond Schneider referenced in the wikipedia article you referenced seems to contain a step which is proven by contradiction... (if Delta not 0 ... conclude that Delta is 0)
              $endgroup$
              – mkl
              Nov 25 '12 at 22:54
















            46












            $begingroup$

            It somewhat depends on whether you are intuitionist or not (or both? or neither? Who knows without the law of excluded middle). According to the Wikipedia article even intuitionists accept some versions of what one could call indirect proof, but reject most. In that sense, a direct proof would be preferable (and is often even a bit more elegant).





            An example:



            Theorem. There exist irrational numbers $a,b$ such that $a^b$ is rational.



            Proof: Assume that $a,bnotin mathbb Q$ always implies $a^bnotin mathbb Q$. Then $u:=sqrt 2^{sqrt 2}notin mathbb Q$ and $u^sqrt 2=sqrt 2^{sqrt 2cdotsqrt 2}=sqrt 2^2=2notin mathbb Q$ - contradiction!



            Indeed, an intuitionist would complain that we do not exhibit a pair $(a,b)$ with $a,bnotin mathbb Q$ and $a^bin mathbb Q$. Instead, we only show that either $(sqrt 2,sqrt 2)$ or $(u,sqrt 2)$ is such a pair.
            Converting the proof given above to a direct and constructive proof would in fact require you to actually prove one of the two possible options $uin mathbb Q$ or $unotinmathbb Q$.






            share|cite|improve this answer











            $endgroup$









            • 14




              $begingroup$
              Of course you are correct the intuitionist would complain that the proof does not exhibit a specific pair. But an intuitionist would not agree that the proof produces two pairs such that either the first pair is an example or the second is an example. The intuitionistic reading of that claimed conclusion would say that, to prove the "or", we would have to prove which pair is the example, which is exactly what the proof does not do. The intuitionist would accept the proof shown here as showing "it is not the case that for all $a,b not in mathbb{Q}$, $a^b not in mathbb{Q}$".
              $endgroup$
              – Carl Mummert
              Nov 24 '12 at 21:20








            • 6




              $begingroup$
              By Gelfond–Schneider $sqrt2^{sqrt2}$ is transcendental. Voilà! No more proof by contradiction ;-)
              $endgroup$
              – kahen
              Nov 24 '12 at 23:11








            • 4




              $begingroup$
              @kahen Well, the proof of Gelfond Schneider referenced in the wikipedia article you referenced seems to contain a step which is proven by contradiction... (if Delta not 0 ... conclude that Delta is 0)
              $endgroup$
              – mkl
              Nov 25 '12 at 22:54














            46












            46








            46





            $begingroup$

            It somewhat depends on whether you are intuitionist or not (or both? or neither? Who knows without the law of excluded middle). According to the Wikipedia article even intuitionists accept some versions of what one could call indirect proof, but reject most. In that sense, a direct proof would be preferable (and is often even a bit more elegant).





            An example:



            Theorem. There exist irrational numbers $a,b$ such that $a^b$ is rational.



            Proof: Assume that $a,bnotin mathbb Q$ always implies $a^bnotin mathbb Q$. Then $u:=sqrt 2^{sqrt 2}notin mathbb Q$ and $u^sqrt 2=sqrt 2^{sqrt 2cdotsqrt 2}=sqrt 2^2=2notin mathbb Q$ - contradiction!



            Indeed, an intuitionist would complain that we do not exhibit a pair $(a,b)$ with $a,bnotin mathbb Q$ and $a^bin mathbb Q$. Instead, we only show that either $(sqrt 2,sqrt 2)$ or $(u,sqrt 2)$ is such a pair.
            Converting the proof given above to a direct and constructive proof would in fact require you to actually prove one of the two possible options $uin mathbb Q$ or $unotinmathbb Q$.






            share|cite|improve this answer











            $endgroup$



            It somewhat depends on whether you are intuitionist or not (or both? or neither? Who knows without the law of excluded middle). According to the Wikipedia article even intuitionists accept some versions of what one could call indirect proof, but reject most. In that sense, a direct proof would be preferable (and is often even a bit more elegant).





            An example:



            Theorem. There exist irrational numbers $a,b$ such that $a^b$ is rational.



            Proof: Assume that $a,bnotin mathbb Q$ always implies $a^bnotin mathbb Q$. Then $u:=sqrt 2^{sqrt 2}notin mathbb Q$ and $u^sqrt 2=sqrt 2^{sqrt 2cdotsqrt 2}=sqrt 2^2=2notin mathbb Q$ - contradiction!



            Indeed, an intuitionist would complain that we do not exhibit a pair $(a,b)$ with $a,bnotin mathbb Q$ and $a^bin mathbb Q$. Instead, we only show that either $(sqrt 2,sqrt 2)$ or $(u,sqrt 2)$ is such a pair.
            Converting the proof given above to a direct and constructive proof would in fact require you to actually prove one of the two possible options $uin mathbb Q$ or $unotinmathbb Q$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 24 '12 at 15:59

























            answered Nov 24 '12 at 15:49









            Hagen von EitzenHagen von Eitzen

            282k23272507




            282k23272507








            • 14




              $begingroup$
              Of course you are correct the intuitionist would complain that the proof does not exhibit a specific pair. But an intuitionist would not agree that the proof produces two pairs such that either the first pair is an example or the second is an example. The intuitionistic reading of that claimed conclusion would say that, to prove the "or", we would have to prove which pair is the example, which is exactly what the proof does not do. The intuitionist would accept the proof shown here as showing "it is not the case that for all $a,b not in mathbb{Q}$, $a^b not in mathbb{Q}$".
              $endgroup$
              – Carl Mummert
              Nov 24 '12 at 21:20








            • 6




              $begingroup$
              By Gelfond–Schneider $sqrt2^{sqrt2}$ is transcendental. Voilà! No more proof by contradiction ;-)
              $endgroup$
              – kahen
              Nov 24 '12 at 23:11








            • 4




              $begingroup$
              @kahen Well, the proof of Gelfond Schneider referenced in the wikipedia article you referenced seems to contain a step which is proven by contradiction... (if Delta not 0 ... conclude that Delta is 0)
              $endgroup$
              – mkl
              Nov 25 '12 at 22:54














            • 14




              $begingroup$
              Of course you are correct the intuitionist would complain that the proof does not exhibit a specific pair. But an intuitionist would not agree that the proof produces two pairs such that either the first pair is an example or the second is an example. The intuitionistic reading of that claimed conclusion would say that, to prove the "or", we would have to prove which pair is the example, which is exactly what the proof does not do. The intuitionist would accept the proof shown here as showing "it is not the case that for all $a,b not in mathbb{Q}$, $a^b not in mathbb{Q}$".
              $endgroup$
              – Carl Mummert
              Nov 24 '12 at 21:20








            • 6




              $begingroup$
              By Gelfond–Schneider $sqrt2^{sqrt2}$ is transcendental. Voilà! No more proof by contradiction ;-)
              $endgroup$
              – kahen
              Nov 24 '12 at 23:11








            • 4




              $begingroup$
              @kahen Well, the proof of Gelfond Schneider referenced in the wikipedia article you referenced seems to contain a step which is proven by contradiction... (if Delta not 0 ... conclude that Delta is 0)
              $endgroup$
              – mkl
              Nov 25 '12 at 22:54








            14




            14




            $begingroup$
            Of course you are correct the intuitionist would complain that the proof does not exhibit a specific pair. But an intuitionist would not agree that the proof produces two pairs such that either the first pair is an example or the second is an example. The intuitionistic reading of that claimed conclusion would say that, to prove the "or", we would have to prove which pair is the example, which is exactly what the proof does not do. The intuitionist would accept the proof shown here as showing "it is not the case that for all $a,b not in mathbb{Q}$, $a^b not in mathbb{Q}$".
            $endgroup$
            – Carl Mummert
            Nov 24 '12 at 21:20






            $begingroup$
            Of course you are correct the intuitionist would complain that the proof does not exhibit a specific pair. But an intuitionist would not agree that the proof produces two pairs such that either the first pair is an example or the second is an example. The intuitionistic reading of that claimed conclusion would say that, to prove the "or", we would have to prove which pair is the example, which is exactly what the proof does not do. The intuitionist would accept the proof shown here as showing "it is not the case that for all $a,b not in mathbb{Q}$, $a^b not in mathbb{Q}$".
            $endgroup$
            – Carl Mummert
            Nov 24 '12 at 21:20






            6




            6




            $begingroup$
            By Gelfond–Schneider $sqrt2^{sqrt2}$ is transcendental. Voilà! No more proof by contradiction ;-)
            $endgroup$
            – kahen
            Nov 24 '12 at 23:11






            $begingroup$
            By Gelfond–Schneider $sqrt2^{sqrt2}$ is transcendental. Voilà! No more proof by contradiction ;-)
            $endgroup$
            – kahen
            Nov 24 '12 at 23:11






            4




            4




            $begingroup$
            @kahen Well, the proof of Gelfond Schneider referenced in the wikipedia article you referenced seems to contain a step which is proven by contradiction... (if Delta not 0 ... conclude that Delta is 0)
            $endgroup$
            – mkl
            Nov 25 '12 at 22:54




            $begingroup$
            @kahen Well, the proof of Gelfond Schneider referenced in the wikipedia article you referenced seems to contain a step which is proven by contradiction... (if Delta not 0 ... conclude that Delta is 0)
            $endgroup$
            – mkl
            Nov 25 '12 at 22:54











            27












            $begingroup$

            See this post: Are proofs by contradiction weaker than other proofs?.
            There are some wonderful answers related to your question - and addresses, directly, your "aside": See, in particular, what JDH writes.





            One of the advantageous to constructing direct proofs of propositions, when this is feasible, is that one can discover other useful propositions in the process. That is, direct proofs help clarify the necessary and sufficient conditions that make a theorem true, and provide a structure demonstrating how these conditions relate, and how the chain of implications imply the conclusion.



            Indirect proofs, on the other hand (aka "proofs by contradiction") only tell us that supposing a proposition to be otherwise leads to a contradiction at some point. But such a proof doesn't really provide the sort of insight that can be gained from direct proofs.



            That is not to say that indirect proofs don't have their place (e.g., they come in handy when asked to prove propositions during a time-limited exam!). They often help "rule out" certain propositions on the basis that they contradict well established axioms or theorems. Also, indirect proofs are sometimes more intuitive than direct proofs. For example, proving that $sqrt{2}$ is not rational using a proof by contradiction is clean, and intuitive.



            Sometimes an indirect proof will emerge first, after which one can seek to proceed with trying to construct a direct proof to prove the same proposition. That is, providing an indirect proof of a proposition often motivates the construction of direct proofs.





            Edit:



            I found this blog entry (Gowers's Weblog) When is a proof by contradiction necessary.
            from which I'll quote an introductory remark:




            It seems to be possible to classify theorems into three types: ones where it would be ridiculous to use contradiction, ones where there are equally sensible proofs using contradiction or not using contradiction, and ones where contradiction seems forced. But what is it that puts a theorem into one of these three categories?




            The post follows immediately with a nice reply from Terence Tao.






            share|cite|improve this answer











            $endgroup$









            • 7




              $begingroup$
              The definition of "irrational" is "not rational", so the statement "$sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational").
              $endgroup$
              – Zhen Lin
              Nov 24 '12 at 16:48








            • 2




              $begingroup$
              What if you prove that for $n,minmathbb Z$ and $mne0$, the distance between $n/m$ and $sqrt{2}$ is at least $1/(3m^2)$? Could that qualify as a "direct" proof of irrationality?
              $endgroup$
              – Michael Hardy
              Nov 24 '12 at 17:34






            • 1




              $begingroup$
              @Michael Hardy: in fact some constuctivists actually take that to be the definition of irrational, so that this is stronger to them than just "not rational". But the classical definition of "irrational" is just "not rational", and that is the perspective of my previous comment in this thread.
              $endgroup$
              – Carl Mummert
              Nov 24 '12 at 18:28






            • 2




              $begingroup$
              @sonicboom mathoverflow.net/questions/32011/direct-proof-of-irrationality math.stackexchange.com/questions/20567/…
              $endgroup$
              – MJD
              Nov 24 '12 at 22:10






            • 3




              $begingroup$
              @ZhenLin Don't you need something more specific than "not rational" to define "irrational"? Are imaginary numbers, matrixes, dogs, and mattresses rational? Are they irrational?
              $endgroup$
              – Peter Olson
              Nov 25 '12 at 4:15
















            27












            $begingroup$

            See this post: Are proofs by contradiction weaker than other proofs?.
            There are some wonderful answers related to your question - and addresses, directly, your "aside": See, in particular, what JDH writes.





            One of the advantageous to constructing direct proofs of propositions, when this is feasible, is that one can discover other useful propositions in the process. That is, direct proofs help clarify the necessary and sufficient conditions that make a theorem true, and provide a structure demonstrating how these conditions relate, and how the chain of implications imply the conclusion.



            Indirect proofs, on the other hand (aka "proofs by contradiction") only tell us that supposing a proposition to be otherwise leads to a contradiction at some point. But such a proof doesn't really provide the sort of insight that can be gained from direct proofs.



            That is not to say that indirect proofs don't have their place (e.g., they come in handy when asked to prove propositions during a time-limited exam!). They often help "rule out" certain propositions on the basis that they contradict well established axioms or theorems. Also, indirect proofs are sometimes more intuitive than direct proofs. For example, proving that $sqrt{2}$ is not rational using a proof by contradiction is clean, and intuitive.



            Sometimes an indirect proof will emerge first, after which one can seek to proceed with trying to construct a direct proof to prove the same proposition. That is, providing an indirect proof of a proposition often motivates the construction of direct proofs.





            Edit:



            I found this blog entry (Gowers's Weblog) When is a proof by contradiction necessary.
            from which I'll quote an introductory remark:




            It seems to be possible to classify theorems into three types: ones where it would be ridiculous to use contradiction, ones where there are equally sensible proofs using contradiction or not using contradiction, and ones where contradiction seems forced. But what is it that puts a theorem into one of these three categories?




            The post follows immediately with a nice reply from Terence Tao.






            share|cite|improve this answer











            $endgroup$









            • 7




              $begingroup$
              The definition of "irrational" is "not rational", so the statement "$sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational").
              $endgroup$
              – Zhen Lin
              Nov 24 '12 at 16:48








            • 2




              $begingroup$
              What if you prove that for $n,minmathbb Z$ and $mne0$, the distance between $n/m$ and $sqrt{2}$ is at least $1/(3m^2)$? Could that qualify as a "direct" proof of irrationality?
              $endgroup$
              – Michael Hardy
              Nov 24 '12 at 17:34






            • 1




              $begingroup$
              @Michael Hardy: in fact some constuctivists actually take that to be the definition of irrational, so that this is stronger to them than just "not rational". But the classical definition of "irrational" is just "not rational", and that is the perspective of my previous comment in this thread.
              $endgroup$
              – Carl Mummert
              Nov 24 '12 at 18:28






            • 2




              $begingroup$
              @sonicboom mathoverflow.net/questions/32011/direct-proof-of-irrationality math.stackexchange.com/questions/20567/…
              $endgroup$
              – MJD
              Nov 24 '12 at 22:10






            • 3




              $begingroup$
              @ZhenLin Don't you need something more specific than "not rational" to define "irrational"? Are imaginary numbers, matrixes, dogs, and mattresses rational? Are they irrational?
              $endgroup$
              – Peter Olson
              Nov 25 '12 at 4:15














            27












            27








            27





            $begingroup$

            See this post: Are proofs by contradiction weaker than other proofs?.
            There are some wonderful answers related to your question - and addresses, directly, your "aside": See, in particular, what JDH writes.





            One of the advantageous to constructing direct proofs of propositions, when this is feasible, is that one can discover other useful propositions in the process. That is, direct proofs help clarify the necessary and sufficient conditions that make a theorem true, and provide a structure demonstrating how these conditions relate, and how the chain of implications imply the conclusion.



            Indirect proofs, on the other hand (aka "proofs by contradiction") only tell us that supposing a proposition to be otherwise leads to a contradiction at some point. But such a proof doesn't really provide the sort of insight that can be gained from direct proofs.



            That is not to say that indirect proofs don't have their place (e.g., they come in handy when asked to prove propositions during a time-limited exam!). They often help "rule out" certain propositions on the basis that they contradict well established axioms or theorems. Also, indirect proofs are sometimes more intuitive than direct proofs. For example, proving that $sqrt{2}$ is not rational using a proof by contradiction is clean, and intuitive.



            Sometimes an indirect proof will emerge first, after which one can seek to proceed with trying to construct a direct proof to prove the same proposition. That is, providing an indirect proof of a proposition often motivates the construction of direct proofs.





            Edit:



            I found this blog entry (Gowers's Weblog) When is a proof by contradiction necessary.
            from which I'll quote an introductory remark:




            It seems to be possible to classify theorems into three types: ones where it would be ridiculous to use contradiction, ones where there are equally sensible proofs using contradiction or not using contradiction, and ones where contradiction seems forced. But what is it that puts a theorem into one of these three categories?




            The post follows immediately with a nice reply from Terence Tao.






            share|cite|improve this answer











            $endgroup$



            See this post: Are proofs by contradiction weaker than other proofs?.
            There are some wonderful answers related to your question - and addresses, directly, your "aside": See, in particular, what JDH writes.





            One of the advantageous to constructing direct proofs of propositions, when this is feasible, is that one can discover other useful propositions in the process. That is, direct proofs help clarify the necessary and sufficient conditions that make a theorem true, and provide a structure demonstrating how these conditions relate, and how the chain of implications imply the conclusion.



            Indirect proofs, on the other hand (aka "proofs by contradiction") only tell us that supposing a proposition to be otherwise leads to a contradiction at some point. But such a proof doesn't really provide the sort of insight that can be gained from direct proofs.



            That is not to say that indirect proofs don't have their place (e.g., they come in handy when asked to prove propositions during a time-limited exam!). They often help "rule out" certain propositions on the basis that they contradict well established axioms or theorems. Also, indirect proofs are sometimes more intuitive than direct proofs. For example, proving that $sqrt{2}$ is not rational using a proof by contradiction is clean, and intuitive.



            Sometimes an indirect proof will emerge first, after which one can seek to proceed with trying to construct a direct proof to prove the same proposition. That is, providing an indirect proof of a proposition often motivates the construction of direct proofs.





            Edit:



            I found this blog entry (Gowers's Weblog) When is a proof by contradiction necessary.
            from which I'll quote an introductory remark:




            It seems to be possible to classify theorems into three types: ones where it would be ridiculous to use contradiction, ones where there are equally sensible proofs using contradiction or not using contradiction, and ones where contradiction seems forced. But what is it that puts a theorem into one of these three categories?




            The post follows immediately with a nice reply from Terence Tao.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 13 '17 at 12:21









            Community

            1




            1










            answered Nov 24 '12 at 15:49









            NamasteNamaste

            1




            1








            • 7




              $begingroup$
              The definition of "irrational" is "not rational", so the statement "$sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational").
              $endgroup$
              – Zhen Lin
              Nov 24 '12 at 16:48








            • 2




              $begingroup$
              What if you prove that for $n,minmathbb Z$ and $mne0$, the distance between $n/m$ and $sqrt{2}$ is at least $1/(3m^2)$? Could that qualify as a "direct" proof of irrationality?
              $endgroup$
              – Michael Hardy
              Nov 24 '12 at 17:34






            • 1




              $begingroup$
              @Michael Hardy: in fact some constuctivists actually take that to be the definition of irrational, so that this is stronger to them than just "not rational". But the classical definition of "irrational" is just "not rational", and that is the perspective of my previous comment in this thread.
              $endgroup$
              – Carl Mummert
              Nov 24 '12 at 18:28






            • 2




              $begingroup$
              @sonicboom mathoverflow.net/questions/32011/direct-proof-of-irrationality math.stackexchange.com/questions/20567/…
              $endgroup$
              – MJD
              Nov 24 '12 at 22:10






            • 3




              $begingroup$
              @ZhenLin Don't you need something more specific than "not rational" to define "irrational"? Are imaginary numbers, matrixes, dogs, and mattresses rational? Are they irrational?
              $endgroup$
              – Peter Olson
              Nov 25 '12 at 4:15














            • 7




              $begingroup$
              The definition of "irrational" is "not rational", so the statement "$sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational").
              $endgroup$
              – Zhen Lin
              Nov 24 '12 at 16:48








            • 2




              $begingroup$
              What if you prove that for $n,minmathbb Z$ and $mne0$, the distance between $n/m$ and $sqrt{2}$ is at least $1/(3m^2)$? Could that qualify as a "direct" proof of irrationality?
              $endgroup$
              – Michael Hardy
              Nov 24 '12 at 17:34






            • 1




              $begingroup$
              @Michael Hardy: in fact some constuctivists actually take that to be the definition of irrational, so that this is stronger to them than just "not rational". But the classical definition of "irrational" is just "not rational", and that is the perspective of my previous comment in this thread.
              $endgroup$
              – Carl Mummert
              Nov 24 '12 at 18:28






            • 2




              $begingroup$
              @sonicboom mathoverflow.net/questions/32011/direct-proof-of-irrationality math.stackexchange.com/questions/20567/…
              $endgroup$
              – MJD
              Nov 24 '12 at 22:10






            • 3




              $begingroup$
              @ZhenLin Don't you need something more specific than "not rational" to define "irrational"? Are imaginary numbers, matrixes, dogs, and mattresses rational? Are they irrational?
              $endgroup$
              – Peter Olson
              Nov 25 '12 at 4:15








            7




            7




            $begingroup$
            The definition of "irrational" is "not rational", so the statement "$sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational").
            $endgroup$
            – Zhen Lin
            Nov 24 '12 at 16:48






            $begingroup$
            The definition of "irrational" is "not rational", so the statement "$sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational").
            $endgroup$
            – Zhen Lin
            Nov 24 '12 at 16:48






            2




            2




            $begingroup$
            What if you prove that for $n,minmathbb Z$ and $mne0$, the distance between $n/m$ and $sqrt{2}$ is at least $1/(3m^2)$? Could that qualify as a "direct" proof of irrationality?
            $endgroup$
            – Michael Hardy
            Nov 24 '12 at 17:34




            $begingroup$
            What if you prove that for $n,minmathbb Z$ and $mne0$, the distance between $n/m$ and $sqrt{2}$ is at least $1/(3m^2)$? Could that qualify as a "direct" proof of irrationality?
            $endgroup$
            – Michael Hardy
            Nov 24 '12 at 17:34




            1




            1




            $begingroup$
            @Michael Hardy: in fact some constuctivists actually take that to be the definition of irrational, so that this is stronger to them than just "not rational". But the classical definition of "irrational" is just "not rational", and that is the perspective of my previous comment in this thread.
            $endgroup$
            – Carl Mummert
            Nov 24 '12 at 18:28




            $begingroup$
            @Michael Hardy: in fact some constuctivists actually take that to be the definition of irrational, so that this is stronger to them than just "not rational". But the classical definition of "irrational" is just "not rational", and that is the perspective of my previous comment in this thread.
            $endgroup$
            – Carl Mummert
            Nov 24 '12 at 18:28




            2




            2




            $begingroup$
            @sonicboom mathoverflow.net/questions/32011/direct-proof-of-irrationality math.stackexchange.com/questions/20567/…
            $endgroup$
            – MJD
            Nov 24 '12 at 22:10




            $begingroup$
            @sonicboom mathoverflow.net/questions/32011/direct-proof-of-irrationality math.stackexchange.com/questions/20567/…
            $endgroup$
            – MJD
            Nov 24 '12 at 22:10




            3




            3




            $begingroup$
            @ZhenLin Don't you need something more specific than "not rational" to define "irrational"? Are imaginary numbers, matrixes, dogs, and mattresses rational? Are they irrational?
            $endgroup$
            – Peter Olson
            Nov 25 '12 at 4:15




            $begingroup$
            @ZhenLin Don't you need something more specific than "not rational" to define "irrational"? Are imaginary numbers, matrixes, dogs, and mattresses rational? Are they irrational?
            $endgroup$
            – Peter Olson
            Nov 25 '12 at 4:15











            12












            $begingroup$

            A few points from my (limited) experience:




            • I love proof by contradiction and I have used it in graduate level classes and no one seemed to mind so long as the logic was infallible.

            • For me, it's much easier to think about a proposition in terms of "What if this wasn't true?". That is usually my first instinct, this makes proof by contradiction the natural first choice. For instance, if I were to be asked to prove something like "Prove that a non-singular matrix has a unique inverse". My first instinct would be "What if a non-singular matrix had 2 inverses?" and from then on, the proof follows cleanly.

            • Sometimes, however, contradictions don't come cleanly and proof by simple logical deductions would probably take 5 lines whereas contradiction will take millions. I could point you to specific proofs but I'll have to do some digging. Further, if you look at every proof and try using Proof by Contradiction, another problem you will face is that sometimes, you will state your intended contradiction but never use it. In other words, solve using direct proof.

            • Another aspect about Proof by Contradiction (IMHO) is that you really must know all definitions and their equivalent statements fairly well to come up with a nice contradiction. Else, you will end up proving several lemmas on the way which looks clean in a direct proof but not so much in a Proof by Contradiction, but again, this might be a personal choice.


            In summary, if you find it easier to think in terms of "What if not" then go ahead, use it but make sure your proof skills using other strategies are as good because $exists$ nail that you cannot hit with the PbC hammer that you'll carry.






            share|cite|improve this answer









            $endgroup$


















              12












              $begingroup$

              A few points from my (limited) experience:




              • I love proof by contradiction and I have used it in graduate level classes and no one seemed to mind so long as the logic was infallible.

              • For me, it's much easier to think about a proposition in terms of "What if this wasn't true?". That is usually my first instinct, this makes proof by contradiction the natural first choice. For instance, if I were to be asked to prove something like "Prove that a non-singular matrix has a unique inverse". My first instinct would be "What if a non-singular matrix had 2 inverses?" and from then on, the proof follows cleanly.

              • Sometimes, however, contradictions don't come cleanly and proof by simple logical deductions would probably take 5 lines whereas contradiction will take millions. I could point you to specific proofs but I'll have to do some digging. Further, if you look at every proof and try using Proof by Contradiction, another problem you will face is that sometimes, you will state your intended contradiction but never use it. In other words, solve using direct proof.

              • Another aspect about Proof by Contradiction (IMHO) is that you really must know all definitions and their equivalent statements fairly well to come up with a nice contradiction. Else, you will end up proving several lemmas on the way which looks clean in a direct proof but not so much in a Proof by Contradiction, but again, this might be a personal choice.


              In summary, if you find it easier to think in terms of "What if not" then go ahead, use it but make sure your proof skills using other strategies are as good because $exists$ nail that you cannot hit with the PbC hammer that you'll carry.






              share|cite|improve this answer









              $endgroup$
















                12












                12








                12





                $begingroup$

                A few points from my (limited) experience:




                • I love proof by contradiction and I have used it in graduate level classes and no one seemed to mind so long as the logic was infallible.

                • For me, it's much easier to think about a proposition in terms of "What if this wasn't true?". That is usually my first instinct, this makes proof by contradiction the natural first choice. For instance, if I were to be asked to prove something like "Prove that a non-singular matrix has a unique inverse". My first instinct would be "What if a non-singular matrix had 2 inverses?" and from then on, the proof follows cleanly.

                • Sometimes, however, contradictions don't come cleanly and proof by simple logical deductions would probably take 5 lines whereas contradiction will take millions. I could point you to specific proofs but I'll have to do some digging. Further, if you look at every proof and try using Proof by Contradiction, another problem you will face is that sometimes, you will state your intended contradiction but never use it. In other words, solve using direct proof.

                • Another aspect about Proof by Contradiction (IMHO) is that you really must know all definitions and their equivalent statements fairly well to come up with a nice contradiction. Else, you will end up proving several lemmas on the way which looks clean in a direct proof but not so much in a Proof by Contradiction, but again, this might be a personal choice.


                In summary, if you find it easier to think in terms of "What if not" then go ahead, use it but make sure your proof skills using other strategies are as good because $exists$ nail that you cannot hit with the PbC hammer that you'll carry.






                share|cite|improve this answer









                $endgroup$



                A few points from my (limited) experience:




                • I love proof by contradiction and I have used it in graduate level classes and no one seemed to mind so long as the logic was infallible.

                • For me, it's much easier to think about a proposition in terms of "What if this wasn't true?". That is usually my first instinct, this makes proof by contradiction the natural first choice. For instance, if I were to be asked to prove something like "Prove that a non-singular matrix has a unique inverse". My first instinct would be "What if a non-singular matrix had 2 inverses?" and from then on, the proof follows cleanly.

                • Sometimes, however, contradictions don't come cleanly and proof by simple logical deductions would probably take 5 lines whereas contradiction will take millions. I could point you to specific proofs but I'll have to do some digging. Further, if you look at every proof and try using Proof by Contradiction, another problem you will face is that sometimes, you will state your intended contradiction but never use it. In other words, solve using direct proof.

                • Another aspect about Proof by Contradiction (IMHO) is that you really must know all definitions and their equivalent statements fairly well to come up with a nice contradiction. Else, you will end up proving several lemmas on the way which looks clean in a direct proof but not so much in a Proof by Contradiction, but again, this might be a personal choice.


                In summary, if you find it easier to think in terms of "What if not" then go ahead, use it but make sure your proof skills using other strategies are as good because $exists$ nail that you cannot hit with the PbC hammer that you'll carry.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 24 '12 at 15:50









                InquestInquest

                5,01722448




                5,01722448























                    12












                    $begingroup$

                    What is a proof by contradiction? This is actually quite difficult to answer in a satisfactory way, but usually what people mean is something like this: given a statement $phi$, a proof of $phi$ by contradiction is a derivation of a contradiction from the assumption $lnot phi$. In order to analyse this, it is very important to distinguish between the statement $phi$ and the statement $lnot lnot phi$; the two statements are formally distinct (as obvious from the fact that their written forms are different!) even though they always have the same truth value in classical logic.



                    Let $bot$ denote contradiction. When we show a contradiction assuming $lnot phi$, what we have is a conditional proof of $bot$ from $lnot phi$. This can then be transformed into a proof of the statement $lnot phi to bot$, which is the long form of $lnot lnot phi$ – in other words, we have a proof that "it is not the case that $lnot phi$". This, strictly speaking, is not a complete proof of $phi$: we must still write down the last step deducing $phi$ from $lnot lnot phi$. This is the point of contention between constructivists and non-constructivists: in the constructive interpretation of logic, $lnot lnot phi$ is not only formally distinct from $phi$ but also semantically distinct; in particular, constructivists reject the principle that $phi$ can be deduced from $lnot lnot phi$ (though they may accept some limited instances of this rule).



                    There is one case where proof by contradiction is always acceptable to constructivists (or at least intuitionists): this is when the statement $phi$ to be proven is itself of the form $lnot psi$. This is because it is a theorem of intuitionistic logic that $lnot lnot lnot psi$ holds if and only if $lnot psi$. On the other hand, it is also in principle possible to give a "direct" proof of $lnot psi$ in the following sense: we simply have to derive a contradiction by assuming $psi$. Any proof of $lnot psi$ by contradiction can thus be transformed into a "direct" proof because one can always derive $lnot lnot psi$ from $psi$; so if we can obtain a contradiction by assuming $lnot lnot psi$, we can certainly derive a contradiction by assuming $psi$.



                    Ultimately, both of the above methods involve making a counterfactual assumption and deriving a contradiction. However, it is sometimes possible to "push" the negation inward and even eliminate it. For example, if $phi$ is the statement "there exists an $x$ such that $theta (x)$ holds", then $lnot phi$ can be deduced from the statement "$theta (x)$ does not hold for any $x$". In particular, if $theta (x)$ is itself a negative statement, say $lnot sigma (x)$, then $lnot phi$ can be deduced from the statement "$sigma (x)$ holds for all $x$". Thus, proving "there does not exist an $x$ such that $sigma (x)$ does not hold" by showing "$sigma (x)$ holds for all $x$" might be considered a more "direct" proof than either of the two previously-mentioned approaches.



                    Can all proofs by contradiction be transformed into direct proofs? In some sense the answer has to be no: intuitionistic logic is known to be weaker than classical logic, i.e. there are statements have proofs in classical logic but not intuitionistic logic. The only difference between classical logic and intuitionistic logic is the principle that $phi$ is deducible from $lnot lnot phi$, so this (in some sense) implies that there are theorems that can only be proven by contradiction.



                    So what are the advantages of proof by contradiction? Well, it makes proofs easier. So much so that one algorithm for automatically proving theorems in propositional logic is based on it. But it also has its disadvantages: a proof by contradiction can be more confusing (because it has counterfactual assumptions floating around!), and in a precise technical sense it is less satisfactory because it generally cannot be (re)used in constructive contexts. But most mathematicians don't worry about the latter problem.






                    share|cite|improve this answer









                    $endgroup$


















                      12












                      $begingroup$

                      What is a proof by contradiction? This is actually quite difficult to answer in a satisfactory way, but usually what people mean is something like this: given a statement $phi$, a proof of $phi$ by contradiction is a derivation of a contradiction from the assumption $lnot phi$. In order to analyse this, it is very important to distinguish between the statement $phi$ and the statement $lnot lnot phi$; the two statements are formally distinct (as obvious from the fact that their written forms are different!) even though they always have the same truth value in classical logic.



                      Let $bot$ denote contradiction. When we show a contradiction assuming $lnot phi$, what we have is a conditional proof of $bot$ from $lnot phi$. This can then be transformed into a proof of the statement $lnot phi to bot$, which is the long form of $lnot lnot phi$ – in other words, we have a proof that "it is not the case that $lnot phi$". This, strictly speaking, is not a complete proof of $phi$: we must still write down the last step deducing $phi$ from $lnot lnot phi$. This is the point of contention between constructivists and non-constructivists: in the constructive interpretation of logic, $lnot lnot phi$ is not only formally distinct from $phi$ but also semantically distinct; in particular, constructivists reject the principle that $phi$ can be deduced from $lnot lnot phi$ (though they may accept some limited instances of this rule).



                      There is one case where proof by contradiction is always acceptable to constructivists (or at least intuitionists): this is when the statement $phi$ to be proven is itself of the form $lnot psi$. This is because it is a theorem of intuitionistic logic that $lnot lnot lnot psi$ holds if and only if $lnot psi$. On the other hand, it is also in principle possible to give a "direct" proof of $lnot psi$ in the following sense: we simply have to derive a contradiction by assuming $psi$. Any proof of $lnot psi$ by contradiction can thus be transformed into a "direct" proof because one can always derive $lnot lnot psi$ from $psi$; so if we can obtain a contradiction by assuming $lnot lnot psi$, we can certainly derive a contradiction by assuming $psi$.



                      Ultimately, both of the above methods involve making a counterfactual assumption and deriving a contradiction. However, it is sometimes possible to "push" the negation inward and even eliminate it. For example, if $phi$ is the statement "there exists an $x$ such that $theta (x)$ holds", then $lnot phi$ can be deduced from the statement "$theta (x)$ does not hold for any $x$". In particular, if $theta (x)$ is itself a negative statement, say $lnot sigma (x)$, then $lnot phi$ can be deduced from the statement "$sigma (x)$ holds for all $x$". Thus, proving "there does not exist an $x$ such that $sigma (x)$ does not hold" by showing "$sigma (x)$ holds for all $x$" might be considered a more "direct" proof than either of the two previously-mentioned approaches.



                      Can all proofs by contradiction be transformed into direct proofs? In some sense the answer has to be no: intuitionistic logic is known to be weaker than classical logic, i.e. there are statements have proofs in classical logic but not intuitionistic logic. The only difference between classical logic and intuitionistic logic is the principle that $phi$ is deducible from $lnot lnot phi$, so this (in some sense) implies that there are theorems that can only be proven by contradiction.



                      So what are the advantages of proof by contradiction? Well, it makes proofs easier. So much so that one algorithm for automatically proving theorems in propositional logic is based on it. But it also has its disadvantages: a proof by contradiction can be more confusing (because it has counterfactual assumptions floating around!), and in a precise technical sense it is less satisfactory because it generally cannot be (re)used in constructive contexts. But most mathematicians don't worry about the latter problem.






                      share|cite|improve this answer









                      $endgroup$
















                        12












                        12








                        12





                        $begingroup$

                        What is a proof by contradiction? This is actually quite difficult to answer in a satisfactory way, but usually what people mean is something like this: given a statement $phi$, a proof of $phi$ by contradiction is a derivation of a contradiction from the assumption $lnot phi$. In order to analyse this, it is very important to distinguish between the statement $phi$ and the statement $lnot lnot phi$; the two statements are formally distinct (as obvious from the fact that their written forms are different!) even though they always have the same truth value in classical logic.



                        Let $bot$ denote contradiction. When we show a contradiction assuming $lnot phi$, what we have is a conditional proof of $bot$ from $lnot phi$. This can then be transformed into a proof of the statement $lnot phi to bot$, which is the long form of $lnot lnot phi$ – in other words, we have a proof that "it is not the case that $lnot phi$". This, strictly speaking, is not a complete proof of $phi$: we must still write down the last step deducing $phi$ from $lnot lnot phi$. This is the point of contention between constructivists and non-constructivists: in the constructive interpretation of logic, $lnot lnot phi$ is not only formally distinct from $phi$ but also semantically distinct; in particular, constructivists reject the principle that $phi$ can be deduced from $lnot lnot phi$ (though they may accept some limited instances of this rule).



                        There is one case where proof by contradiction is always acceptable to constructivists (or at least intuitionists): this is when the statement $phi$ to be proven is itself of the form $lnot psi$. This is because it is a theorem of intuitionistic logic that $lnot lnot lnot psi$ holds if and only if $lnot psi$. On the other hand, it is also in principle possible to give a "direct" proof of $lnot psi$ in the following sense: we simply have to derive a contradiction by assuming $psi$. Any proof of $lnot psi$ by contradiction can thus be transformed into a "direct" proof because one can always derive $lnot lnot psi$ from $psi$; so if we can obtain a contradiction by assuming $lnot lnot psi$, we can certainly derive a contradiction by assuming $psi$.



                        Ultimately, both of the above methods involve making a counterfactual assumption and deriving a contradiction. However, it is sometimes possible to "push" the negation inward and even eliminate it. For example, if $phi$ is the statement "there exists an $x$ such that $theta (x)$ holds", then $lnot phi$ can be deduced from the statement "$theta (x)$ does not hold for any $x$". In particular, if $theta (x)$ is itself a negative statement, say $lnot sigma (x)$, then $lnot phi$ can be deduced from the statement "$sigma (x)$ holds for all $x$". Thus, proving "there does not exist an $x$ such that $sigma (x)$ does not hold" by showing "$sigma (x)$ holds for all $x$" might be considered a more "direct" proof than either of the two previously-mentioned approaches.



                        Can all proofs by contradiction be transformed into direct proofs? In some sense the answer has to be no: intuitionistic logic is known to be weaker than classical logic, i.e. there are statements have proofs in classical logic but not intuitionistic logic. The only difference between classical logic and intuitionistic logic is the principle that $phi$ is deducible from $lnot lnot phi$, so this (in some sense) implies that there are theorems that can only be proven by contradiction.



                        So what are the advantages of proof by contradiction? Well, it makes proofs easier. So much so that one algorithm for automatically proving theorems in propositional logic is based on it. But it also has its disadvantages: a proof by contradiction can be more confusing (because it has counterfactual assumptions floating around!), and in a precise technical sense it is less satisfactory because it generally cannot be (re)used in constructive contexts. But most mathematicians don't worry about the latter problem.






                        share|cite|improve this answer









                        $endgroup$



                        What is a proof by contradiction? This is actually quite difficult to answer in a satisfactory way, but usually what people mean is something like this: given a statement $phi$, a proof of $phi$ by contradiction is a derivation of a contradiction from the assumption $lnot phi$. In order to analyse this, it is very important to distinguish between the statement $phi$ and the statement $lnot lnot phi$; the two statements are formally distinct (as obvious from the fact that their written forms are different!) even though they always have the same truth value in classical logic.



                        Let $bot$ denote contradiction. When we show a contradiction assuming $lnot phi$, what we have is a conditional proof of $bot$ from $lnot phi$. This can then be transformed into a proof of the statement $lnot phi to bot$, which is the long form of $lnot lnot phi$ – in other words, we have a proof that "it is not the case that $lnot phi$". This, strictly speaking, is not a complete proof of $phi$: we must still write down the last step deducing $phi$ from $lnot lnot phi$. This is the point of contention between constructivists and non-constructivists: in the constructive interpretation of logic, $lnot lnot phi$ is not only formally distinct from $phi$ but also semantically distinct; in particular, constructivists reject the principle that $phi$ can be deduced from $lnot lnot phi$ (though they may accept some limited instances of this rule).



                        There is one case where proof by contradiction is always acceptable to constructivists (or at least intuitionists): this is when the statement $phi$ to be proven is itself of the form $lnot psi$. This is because it is a theorem of intuitionistic logic that $lnot lnot lnot psi$ holds if and only if $lnot psi$. On the other hand, it is also in principle possible to give a "direct" proof of $lnot psi$ in the following sense: we simply have to derive a contradiction by assuming $psi$. Any proof of $lnot psi$ by contradiction can thus be transformed into a "direct" proof because one can always derive $lnot lnot psi$ from $psi$; so if we can obtain a contradiction by assuming $lnot lnot psi$, we can certainly derive a contradiction by assuming $psi$.



                        Ultimately, both of the above methods involve making a counterfactual assumption and deriving a contradiction. However, it is sometimes possible to "push" the negation inward and even eliminate it. For example, if $phi$ is the statement "there exists an $x$ such that $theta (x)$ holds", then $lnot phi$ can be deduced from the statement "$theta (x)$ does not hold for any $x$". In particular, if $theta (x)$ is itself a negative statement, say $lnot sigma (x)$, then $lnot phi$ can be deduced from the statement "$sigma (x)$ holds for all $x$". Thus, proving "there does not exist an $x$ such that $sigma (x)$ does not hold" by showing "$sigma (x)$ holds for all $x$" might be considered a more "direct" proof than either of the two previously-mentioned approaches.



                        Can all proofs by contradiction be transformed into direct proofs? In some sense the answer has to be no: intuitionistic logic is known to be weaker than classical logic, i.e. there are statements have proofs in classical logic but not intuitionistic logic. The only difference between classical logic and intuitionistic logic is the principle that $phi$ is deducible from $lnot lnot phi$, so this (in some sense) implies that there are theorems that can only be proven by contradiction.



                        So what are the advantages of proof by contradiction? Well, it makes proofs easier. So much so that one algorithm for automatically proving theorems in propositional logic is based on it. But it also has its disadvantages: a proof by contradiction can be more confusing (because it has counterfactual assumptions floating around!), and in a precise technical sense it is less satisfactory because it generally cannot be (re)used in constructive contexts. But most mathematicians don't worry about the latter problem.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 24 '12 at 18:00









                        Zhen LinZhen Lin

                        60.7k4110227




                        60.7k4110227























                            8












                            $begingroup$

                            As "Inquest"'s answer mentions, it's often easier to find a proof by contradiction than a direct proof. But after you do that, you can often make the proof simpler by rearranging it into a direct proof. It is not good to make a proof appear more complicated than it really is.



                            To see another disadvantage of some proofs by contradiction, consider this:



                            Proof: To prove $A$, assume not $A$. [insert 50 pages of argument here] We have reached a contradiction. Therefore $A$. End of proof



                            Now ask yourself: Which of the propositions proved in those 50 pages are erroneous and could be proved only because one relied on the false assumption that not $A$, and which are validly proved, and which are true but not validly proved because the assumption that not $A$ was relied on? It's not so easy to tell without a lot more work. And if you remember a proof of one of those propositions, you might just mistakenly think that it's been proved and is therefore known to be true. So it might be far better to limit the use of proof by contradiction to some portions of those 50 pages where no other method works.



                            Perhaps proofs of non-existence can be done only by contradiction. Here I might offer as an example the various proofs of the irrationality of $sqrt{2}$, but for the fact that I've seen it asserted that if $m$, $n$ are integers, than $m/n$ differs from $sqrt{2}$ by at least an amount that depends on $n$ --- I think it might have been $1/(3n^2)$. Here's another example: How would one prove the non-existence of a non-trivial (i.e. $>1$) common divisor of $n$ and $n+1$?



                            I've seen a book on logic asserting that a proof by contradiction of a non-existence assertion does not constitute an "indirect proof", since the assertion is inherently negative. I don't know how conventional that is.






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              Formally, a negative statement such as $lnot phi$ is an abbreviation for $phi to bot$, i.e. it is the statement that $phi$ leads to an absurdity; accordingly, to prove $lnot phi$, one must show that assuming $phi$ leads to contradiction. This is completely orthodox logic.
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 16:46










                            • $begingroup$
                              But when must a statement be written in the form $lnotvarphi$ ?
                              $endgroup$
                              – Michael Hardy
                              Nov 24 '12 at 16:50












                            • $begingroup$
                              There isn't any "must" about it. If $phi$ is a compound formula such as $psi lor theta$ you can just push the $lnot$ inward and get $(lnot psi) land (lnot theta)$. But that just causes the number of negative statements to multiply...
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 17:01










                            • $begingroup$
                              So when must it be written in a form that has negative statements? Does the answer depend on which formal language you write it in?
                              $endgroup$
                              – Michael Hardy
                              Nov 24 '12 at 17:31






                            • 2




                              $begingroup$
                              Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $ne$? Sometimes constructivists take an "apartness" relation $mathrel{#}$ to be primitive, in which case $=$ becomes the negation of $mathrel{#}$!)
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 18:05


















                            8












                            $begingroup$

                            As "Inquest"'s answer mentions, it's often easier to find a proof by contradiction than a direct proof. But after you do that, you can often make the proof simpler by rearranging it into a direct proof. It is not good to make a proof appear more complicated than it really is.



                            To see another disadvantage of some proofs by contradiction, consider this:



                            Proof: To prove $A$, assume not $A$. [insert 50 pages of argument here] We have reached a contradiction. Therefore $A$. End of proof



                            Now ask yourself: Which of the propositions proved in those 50 pages are erroneous and could be proved only because one relied on the false assumption that not $A$, and which are validly proved, and which are true but not validly proved because the assumption that not $A$ was relied on? It's not so easy to tell without a lot more work. And if you remember a proof of one of those propositions, you might just mistakenly think that it's been proved and is therefore known to be true. So it might be far better to limit the use of proof by contradiction to some portions of those 50 pages where no other method works.



                            Perhaps proofs of non-existence can be done only by contradiction. Here I might offer as an example the various proofs of the irrationality of $sqrt{2}$, but for the fact that I've seen it asserted that if $m$, $n$ are integers, than $m/n$ differs from $sqrt{2}$ by at least an amount that depends on $n$ --- I think it might have been $1/(3n^2)$. Here's another example: How would one prove the non-existence of a non-trivial (i.e. $>1$) common divisor of $n$ and $n+1$?



                            I've seen a book on logic asserting that a proof by contradiction of a non-existence assertion does not constitute an "indirect proof", since the assertion is inherently negative. I don't know how conventional that is.






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              Formally, a negative statement such as $lnot phi$ is an abbreviation for $phi to bot$, i.e. it is the statement that $phi$ leads to an absurdity; accordingly, to prove $lnot phi$, one must show that assuming $phi$ leads to contradiction. This is completely orthodox logic.
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 16:46










                            • $begingroup$
                              But when must a statement be written in the form $lnotvarphi$ ?
                              $endgroup$
                              – Michael Hardy
                              Nov 24 '12 at 16:50












                            • $begingroup$
                              There isn't any "must" about it. If $phi$ is a compound formula such as $psi lor theta$ you can just push the $lnot$ inward and get $(lnot psi) land (lnot theta)$. But that just causes the number of negative statements to multiply...
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 17:01










                            • $begingroup$
                              So when must it be written in a form that has negative statements? Does the answer depend on which formal language you write it in?
                              $endgroup$
                              – Michael Hardy
                              Nov 24 '12 at 17:31






                            • 2




                              $begingroup$
                              Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $ne$? Sometimes constructivists take an "apartness" relation $mathrel{#}$ to be primitive, in which case $=$ becomes the negation of $mathrel{#}$!)
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 18:05
















                            8












                            8








                            8





                            $begingroup$

                            As "Inquest"'s answer mentions, it's often easier to find a proof by contradiction than a direct proof. But after you do that, you can often make the proof simpler by rearranging it into a direct proof. It is not good to make a proof appear more complicated than it really is.



                            To see another disadvantage of some proofs by contradiction, consider this:



                            Proof: To prove $A$, assume not $A$. [insert 50 pages of argument here] We have reached a contradiction. Therefore $A$. End of proof



                            Now ask yourself: Which of the propositions proved in those 50 pages are erroneous and could be proved only because one relied on the false assumption that not $A$, and which are validly proved, and which are true but not validly proved because the assumption that not $A$ was relied on? It's not so easy to tell without a lot more work. And if you remember a proof of one of those propositions, you might just mistakenly think that it's been proved and is therefore known to be true. So it might be far better to limit the use of proof by contradiction to some portions of those 50 pages where no other method works.



                            Perhaps proofs of non-existence can be done only by contradiction. Here I might offer as an example the various proofs of the irrationality of $sqrt{2}$, but for the fact that I've seen it asserted that if $m$, $n$ are integers, than $m/n$ differs from $sqrt{2}$ by at least an amount that depends on $n$ --- I think it might have been $1/(3n^2)$. Here's another example: How would one prove the non-existence of a non-trivial (i.e. $>1$) common divisor of $n$ and $n+1$?



                            I've seen a book on logic asserting that a proof by contradiction of a non-existence assertion does not constitute an "indirect proof", since the assertion is inherently negative. I don't know how conventional that is.






                            share|cite|improve this answer









                            $endgroup$



                            As "Inquest"'s answer mentions, it's often easier to find a proof by contradiction than a direct proof. But after you do that, you can often make the proof simpler by rearranging it into a direct proof. It is not good to make a proof appear more complicated than it really is.



                            To see another disadvantage of some proofs by contradiction, consider this:



                            Proof: To prove $A$, assume not $A$. [insert 50 pages of argument here] We have reached a contradiction. Therefore $A$. End of proof



                            Now ask yourself: Which of the propositions proved in those 50 pages are erroneous and could be proved only because one relied on the false assumption that not $A$, and which are validly proved, and which are true but not validly proved because the assumption that not $A$ was relied on? It's not so easy to tell without a lot more work. And if you remember a proof of one of those propositions, you might just mistakenly think that it's been proved and is therefore known to be true. So it might be far better to limit the use of proof by contradiction to some portions of those 50 pages where no other method works.



                            Perhaps proofs of non-existence can be done only by contradiction. Here I might offer as an example the various proofs of the irrationality of $sqrt{2}$, but for the fact that I've seen it asserted that if $m$, $n$ are integers, than $m/n$ differs from $sqrt{2}$ by at least an amount that depends on $n$ --- I think it might have been $1/(3n^2)$. Here's another example: How would one prove the non-existence of a non-trivial (i.e. $>1$) common divisor of $n$ and $n+1$?



                            I've seen a book on logic asserting that a proof by contradiction of a non-existence assertion does not constitute an "indirect proof", since the assertion is inherently negative. I don't know how conventional that is.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 '12 at 16:08









                            Michael HardyMichael Hardy

                            1




                            1








                            • 1




                              $begingroup$
                              Formally, a negative statement such as $lnot phi$ is an abbreviation for $phi to bot$, i.e. it is the statement that $phi$ leads to an absurdity; accordingly, to prove $lnot phi$, one must show that assuming $phi$ leads to contradiction. This is completely orthodox logic.
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 16:46










                            • $begingroup$
                              But when must a statement be written in the form $lnotvarphi$ ?
                              $endgroup$
                              – Michael Hardy
                              Nov 24 '12 at 16:50












                            • $begingroup$
                              There isn't any "must" about it. If $phi$ is a compound formula such as $psi lor theta$ you can just push the $lnot$ inward and get $(lnot psi) land (lnot theta)$. But that just causes the number of negative statements to multiply...
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 17:01










                            • $begingroup$
                              So when must it be written in a form that has negative statements? Does the answer depend on which formal language you write it in?
                              $endgroup$
                              – Michael Hardy
                              Nov 24 '12 at 17:31






                            • 2




                              $begingroup$
                              Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $ne$? Sometimes constructivists take an "apartness" relation $mathrel{#}$ to be primitive, in which case $=$ becomes the negation of $mathrel{#}$!)
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 18:05
















                            • 1




                              $begingroup$
                              Formally, a negative statement such as $lnot phi$ is an abbreviation for $phi to bot$, i.e. it is the statement that $phi$ leads to an absurdity; accordingly, to prove $lnot phi$, one must show that assuming $phi$ leads to contradiction. This is completely orthodox logic.
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 16:46










                            • $begingroup$
                              But when must a statement be written in the form $lnotvarphi$ ?
                              $endgroup$
                              – Michael Hardy
                              Nov 24 '12 at 16:50












                            • $begingroup$
                              There isn't any "must" about it. If $phi$ is a compound formula such as $psi lor theta$ you can just push the $lnot$ inward and get $(lnot psi) land (lnot theta)$. But that just causes the number of negative statements to multiply...
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 17:01










                            • $begingroup$
                              So when must it be written in a form that has negative statements? Does the answer depend on which formal language you write it in?
                              $endgroup$
                              – Michael Hardy
                              Nov 24 '12 at 17:31






                            • 2




                              $begingroup$
                              Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $ne$? Sometimes constructivists take an "apartness" relation $mathrel{#}$ to be primitive, in which case $=$ becomes the negation of $mathrel{#}$!)
                              $endgroup$
                              – Zhen Lin
                              Nov 24 '12 at 18:05










                            1




                            1




                            $begingroup$
                            Formally, a negative statement such as $lnot phi$ is an abbreviation for $phi to bot$, i.e. it is the statement that $phi$ leads to an absurdity; accordingly, to prove $lnot phi$, one must show that assuming $phi$ leads to contradiction. This is completely orthodox logic.
                            $endgroup$
                            – Zhen Lin
                            Nov 24 '12 at 16:46




                            $begingroup$
                            Formally, a negative statement such as $lnot phi$ is an abbreviation for $phi to bot$, i.e. it is the statement that $phi$ leads to an absurdity; accordingly, to prove $lnot phi$, one must show that assuming $phi$ leads to contradiction. This is completely orthodox logic.
                            $endgroup$
                            – Zhen Lin
                            Nov 24 '12 at 16:46












                            $begingroup$
                            But when must a statement be written in the form $lnotvarphi$ ?
                            $endgroup$
                            – Michael Hardy
                            Nov 24 '12 at 16:50






                            $begingroup$
                            But when must a statement be written in the form $lnotvarphi$ ?
                            $endgroup$
                            – Michael Hardy
                            Nov 24 '12 at 16:50














                            $begingroup$
                            There isn't any "must" about it. If $phi$ is a compound formula such as $psi lor theta$ you can just push the $lnot$ inward and get $(lnot psi) land (lnot theta)$. But that just causes the number of negative statements to multiply...
                            $endgroup$
                            – Zhen Lin
                            Nov 24 '12 at 17:01




                            $begingroup$
                            There isn't any "must" about it. If $phi$ is a compound formula such as $psi lor theta$ you can just push the $lnot$ inward and get $(lnot psi) land (lnot theta)$. But that just causes the number of negative statements to multiply...
                            $endgroup$
                            – Zhen Lin
                            Nov 24 '12 at 17:01












                            $begingroup$
                            So when must it be written in a form that has negative statements? Does the answer depend on which formal language you write it in?
                            $endgroup$
                            – Michael Hardy
                            Nov 24 '12 at 17:31




                            $begingroup$
                            So when must it be written in a form that has negative statements? Does the answer depend on which formal language you write it in?
                            $endgroup$
                            – Michael Hardy
                            Nov 24 '12 at 17:31




                            2




                            2




                            $begingroup$
                            Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $ne$? Sometimes constructivists take an "apartness" relation $mathrel{#}$ to be primitive, in which case $=$ becomes the negation of $mathrel{#}$!)
                            $endgroup$
                            – Zhen Lin
                            Nov 24 '12 at 18:05






                            $begingroup$
                            Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $ne$? Sometimes constructivists take an "apartness" relation $mathrel{#}$ to be primitive, in which case $=$ becomes the negation of $mathrel{#}$!)
                            $endgroup$
                            – Zhen Lin
                            Nov 24 '12 at 18:05













                            7












                            $begingroup$

                            Another example of a contradiction proof that provides no idea on a constructive proof is the strategy-stealing argument. For certain symmetric games, the second player cannot have a winning strategy. If he did, the first player could "pretend" to be the second player and steal his winning strategy, stealing it from him, a contradiction.



                            An interesting example is the game Hex. It is easy to show that Hex cannot end in a tie, and the strategy-stealing argument does apply to it. Therefore, it is a first player win. But for symmetric $n$ x $n$, the actual winning strategy is still not known. Thus, this is an example of something that has been proven using contradiction and not constructively (yet).






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              I would formulate it as: Let $S$ be any strategy for the second player. Then by symmetry the first player can apply $S$ too. Since not both players can win and both are applying $S$, $S$ is not a winning strategy.
                              $endgroup$
                              – WimC
                              Nov 27 '12 at 21:31
















                            7












                            $begingroup$

                            Another example of a contradiction proof that provides no idea on a constructive proof is the strategy-stealing argument. For certain symmetric games, the second player cannot have a winning strategy. If he did, the first player could "pretend" to be the second player and steal his winning strategy, stealing it from him, a contradiction.



                            An interesting example is the game Hex. It is easy to show that Hex cannot end in a tie, and the strategy-stealing argument does apply to it. Therefore, it is a first player win. But for symmetric $n$ x $n$, the actual winning strategy is still not known. Thus, this is an example of something that has been proven using contradiction and not constructively (yet).






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              I would formulate it as: Let $S$ be any strategy for the second player. Then by symmetry the first player can apply $S$ too. Since not both players can win and both are applying $S$, $S$ is not a winning strategy.
                              $endgroup$
                              – WimC
                              Nov 27 '12 at 21:31














                            7












                            7








                            7





                            $begingroup$

                            Another example of a contradiction proof that provides no idea on a constructive proof is the strategy-stealing argument. For certain symmetric games, the second player cannot have a winning strategy. If he did, the first player could "pretend" to be the second player and steal his winning strategy, stealing it from him, a contradiction.



                            An interesting example is the game Hex. It is easy to show that Hex cannot end in a tie, and the strategy-stealing argument does apply to it. Therefore, it is a first player win. But for symmetric $n$ x $n$, the actual winning strategy is still not known. Thus, this is an example of something that has been proven using contradiction and not constructively (yet).






                            share|cite|improve this answer









                            $endgroup$



                            Another example of a contradiction proof that provides no idea on a constructive proof is the strategy-stealing argument. For certain symmetric games, the second player cannot have a winning strategy. If he did, the first player could "pretend" to be the second player and steal his winning strategy, stealing it from him, a contradiction.



                            An interesting example is the game Hex. It is easy to show that Hex cannot end in a tie, and the strategy-stealing argument does apply to it. Therefore, it is a first player win. But for symmetric $n$ x $n$, the actual winning strategy is still not known. Thus, this is an example of something that has been proven using contradiction and not constructively (yet).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 25 '12 at 2:10









                            asmeurerasmeurer

                            5,88042443




                            5,88042443








                            • 1




                              $begingroup$
                              I would formulate it as: Let $S$ be any strategy for the second player. Then by symmetry the first player can apply $S$ too. Since not both players can win and both are applying $S$, $S$ is not a winning strategy.
                              $endgroup$
                              – WimC
                              Nov 27 '12 at 21:31














                            • 1




                              $begingroup$
                              I would formulate it as: Let $S$ be any strategy for the second player. Then by symmetry the first player can apply $S$ too. Since not both players can win and both are applying $S$, $S$ is not a winning strategy.
                              $endgroup$
                              – WimC
                              Nov 27 '12 at 21:31








                            1




                            1




                            $begingroup$
                            I would formulate it as: Let $S$ be any strategy for the second player. Then by symmetry the first player can apply $S$ too. Since not both players can win and both are applying $S$, $S$ is not a winning strategy.
                            $endgroup$
                            – WimC
                            Nov 27 '12 at 21:31




                            $begingroup$
                            I would formulate it as: Let $S$ be any strategy for the second player. Then by symmetry the first player can apply $S$ too. Since not both players can win and both are applying $S$, $S$ is not a winning strategy.
                            $endgroup$
                            – WimC
                            Nov 27 '12 at 21:31











                            5












                            $begingroup$

                            There is nothing wrong with proof by contradiction. You can show that they work using a truth table. In the end, that's all that really matters, right?



                            As far as I know, you can't know for certain that something is not provable by a direct proof. However, a proof by contradiction might be an easier way to prove some things, like the irrationality of certain numbers. For example, I have never seen a direct proof of the irrationality of $sqrt{2}$.



                            EDIT: As Carl Mummert said in his answer, the above part in italics is not true. There are propositions which are only provable by contradiction.



                            A proof by contradiction can be also be formulated as a proof by contrapositive. If we know $Q$ is false, if we can show $PRightarrow Q$ then we have proved that $P$ is false. Whether you view this as "proof without contradiction" or not is up to you. In any case, they are logically equivalent.






                            share|cite|improve this answer











                            $endgroup$









                            • 4




                              $begingroup$
                              As I explained in my answer, we do know that there are things that are not provable directly modulo the usual formalization of proofs. The usual proof that $sqrt{2}$ is not rational is a direct proof when it is formalized in the usual way, although it is a direct proof of a negative statement, which can be deceptive at first.
                              $endgroup$
                              – Carl Mummert
                              Nov 24 '12 at 19:10








                            • 1




                              $begingroup$
                              I found the link I had lost, which explains some of this: math.andrej.com/2010/03/29/…
                              $endgroup$
                              – Carl Mummert
                              Nov 24 '12 at 22:43






                            • 1




                              $begingroup$
                              Thank you for pointing this out, and for the reference. Both your comment and the article were very informative and enlightening!
                              $endgroup$
                              – Espen Nielsen
                              Nov 24 '12 at 23:12
















                            5












                            $begingroup$

                            There is nothing wrong with proof by contradiction. You can show that they work using a truth table. In the end, that's all that really matters, right?



                            As far as I know, you can't know for certain that something is not provable by a direct proof. However, a proof by contradiction might be an easier way to prove some things, like the irrationality of certain numbers. For example, I have never seen a direct proof of the irrationality of $sqrt{2}$.



                            EDIT: As Carl Mummert said in his answer, the above part in italics is not true. There are propositions which are only provable by contradiction.



                            A proof by contradiction can be also be formulated as a proof by contrapositive. If we know $Q$ is false, if we can show $PRightarrow Q$ then we have proved that $P$ is false. Whether you view this as "proof without contradiction" or not is up to you. In any case, they are logically equivalent.






                            share|cite|improve this answer











                            $endgroup$









                            • 4




                              $begingroup$
                              As I explained in my answer, we do know that there are things that are not provable directly modulo the usual formalization of proofs. The usual proof that $sqrt{2}$ is not rational is a direct proof when it is formalized in the usual way, although it is a direct proof of a negative statement, which can be deceptive at first.
                              $endgroup$
                              – Carl Mummert
                              Nov 24 '12 at 19:10








                            • 1




                              $begingroup$
                              I found the link I had lost, which explains some of this: math.andrej.com/2010/03/29/…
                              $endgroup$
                              – Carl Mummert
                              Nov 24 '12 at 22:43






                            • 1




                              $begingroup$
                              Thank you for pointing this out, and for the reference. Both your comment and the article were very informative and enlightening!
                              $endgroup$
                              – Espen Nielsen
                              Nov 24 '12 at 23:12














                            5












                            5








                            5





                            $begingroup$

                            There is nothing wrong with proof by contradiction. You can show that they work using a truth table. In the end, that's all that really matters, right?



                            As far as I know, you can't know for certain that something is not provable by a direct proof. However, a proof by contradiction might be an easier way to prove some things, like the irrationality of certain numbers. For example, I have never seen a direct proof of the irrationality of $sqrt{2}$.



                            EDIT: As Carl Mummert said in his answer, the above part in italics is not true. There are propositions which are only provable by contradiction.



                            A proof by contradiction can be also be formulated as a proof by contrapositive. If we know $Q$ is false, if we can show $PRightarrow Q$ then we have proved that $P$ is false. Whether you view this as "proof without contradiction" or not is up to you. In any case, they are logically equivalent.






                            share|cite|improve this answer











                            $endgroup$



                            There is nothing wrong with proof by contradiction. You can show that they work using a truth table. In the end, that's all that really matters, right?



                            As far as I know, you can't know for certain that something is not provable by a direct proof. However, a proof by contradiction might be an easier way to prove some things, like the irrationality of certain numbers. For example, I have never seen a direct proof of the irrationality of $sqrt{2}$.



                            EDIT: As Carl Mummert said in his answer, the above part in italics is not true. There are propositions which are only provable by contradiction.



                            A proof by contradiction can be also be formulated as a proof by contrapositive. If we know $Q$ is false, if we can show $PRightarrow Q$ then we have proved that $P$ is false. Whether you view this as "proof without contradiction" or not is up to you. In any case, they are logically equivalent.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 24 '12 at 22:55

























                            answered Nov 24 '12 at 15:49









                            Espen NielsenEspen Nielsen

                            3,050926




                            3,050926








                            • 4




                              $begingroup$
                              As I explained in my answer, we do know that there are things that are not provable directly modulo the usual formalization of proofs. The usual proof that $sqrt{2}$ is not rational is a direct proof when it is formalized in the usual way, although it is a direct proof of a negative statement, which can be deceptive at first.
                              $endgroup$
                              – Carl Mummert
                              Nov 24 '12 at 19:10








                            • 1




                              $begingroup$
                              I found the link I had lost, which explains some of this: math.andrej.com/2010/03/29/…
                              $endgroup$
                              – Carl Mummert
                              Nov 24 '12 at 22:43






                            • 1




                              $begingroup$
                              Thank you for pointing this out, and for the reference. Both your comment and the article were very informative and enlightening!
                              $endgroup$
                              – Espen Nielsen
                              Nov 24 '12 at 23:12














                            • 4




                              $begingroup$
                              As I explained in my answer, we do know that there are things that are not provable directly modulo the usual formalization of proofs. The usual proof that $sqrt{2}$ is not rational is a direct proof when it is formalized in the usual way, although it is a direct proof of a negative statement, which can be deceptive at first.
                              $endgroup$
                              – Carl Mummert
                              Nov 24 '12 at 19:10








                            • 1




                              $begingroup$
                              I found the link I had lost, which explains some of this: math.andrej.com/2010/03/29/…
                              $endgroup$
                              – Carl Mummert
                              Nov 24 '12 at 22:43






                            • 1




                              $begingroup$
                              Thank you for pointing this out, and for the reference. Both your comment and the article were very informative and enlightening!
                              $endgroup$
                              – Espen Nielsen
                              Nov 24 '12 at 23:12








                            4




                            4




                            $begingroup$
                            As I explained in my answer, we do know that there are things that are not provable directly modulo the usual formalization of proofs. The usual proof that $sqrt{2}$ is not rational is a direct proof when it is formalized in the usual way, although it is a direct proof of a negative statement, which can be deceptive at first.
                            $endgroup$
                            – Carl Mummert
                            Nov 24 '12 at 19:10






                            $begingroup$
                            As I explained in my answer, we do know that there are things that are not provable directly modulo the usual formalization of proofs. The usual proof that $sqrt{2}$ is not rational is a direct proof when it is formalized in the usual way, although it is a direct proof of a negative statement, which can be deceptive at first.
                            $endgroup$
                            – Carl Mummert
                            Nov 24 '12 at 19:10






                            1




                            1




                            $begingroup$
                            I found the link I had lost, which explains some of this: math.andrej.com/2010/03/29/…
                            $endgroup$
                            – Carl Mummert
                            Nov 24 '12 at 22:43




                            $begingroup$
                            I found the link I had lost, which explains some of this: math.andrej.com/2010/03/29/…
                            $endgroup$
                            – Carl Mummert
                            Nov 24 '12 at 22:43




                            1




                            1




                            $begingroup$
                            Thank you for pointing this out, and for the reference. Both your comment and the article were very informative and enlightening!
                            $endgroup$
                            – Espen Nielsen
                            Nov 24 '12 at 23:12




                            $begingroup$
                            Thank you for pointing this out, and for the reference. Both your comment and the article were very informative and enlightening!
                            $endgroup$
                            – Espen Nielsen
                            Nov 24 '12 at 23:12











                            3












                            $begingroup$

                            I do believe there are some proofs that are only demonstrable through contradiction, and I'm going to attempt to describe them logically:



                            Let X be a logical statement such that: X $rightarrow$ y, where y is a known contradiction (such as 2+2=5 in the normal arithmetic structure). Without knowing anything else of X, $neg$X implies nothing and nothing implies $neg$X (and hence is not provable). But, of course, assuming X implies a contradiction, and thus, $neg$X.



                            This form of statement X is isolated, in that it only relates to itself and the contradiction. I do believe they can be constructed though, for it seems it they can be described.



                            With that said, in real math and logic, or in general real world scenarios, I don't believe any statements of this form exist, except possibly ones that are constructed to meet this criteria and otherwise meaningless. The proof of the primes was eventually proved without contradiction, to my understanding; until math had been more developed, I think that the statement "the number of primes is $infty$" was basically an isolated logical statement at Euclid's time and for many years after probably, in that there were no other things known to imply it and it didn't imply anything else useful towards its proof.






                            share|cite|improve this answer









                            $endgroup$


















                              3












                              $begingroup$

                              I do believe there are some proofs that are only demonstrable through contradiction, and I'm going to attempt to describe them logically:



                              Let X be a logical statement such that: X $rightarrow$ y, where y is a known contradiction (such as 2+2=5 in the normal arithmetic structure). Without knowing anything else of X, $neg$X implies nothing and nothing implies $neg$X (and hence is not provable). But, of course, assuming X implies a contradiction, and thus, $neg$X.



                              This form of statement X is isolated, in that it only relates to itself and the contradiction. I do believe they can be constructed though, for it seems it they can be described.



                              With that said, in real math and logic, or in general real world scenarios, I don't believe any statements of this form exist, except possibly ones that are constructed to meet this criteria and otherwise meaningless. The proof of the primes was eventually proved without contradiction, to my understanding; until math had been more developed, I think that the statement "the number of primes is $infty$" was basically an isolated logical statement at Euclid's time and for many years after probably, in that there were no other things known to imply it and it didn't imply anything else useful towards its proof.






                              share|cite|improve this answer









                              $endgroup$
















                                3












                                3








                                3





                                $begingroup$

                                I do believe there are some proofs that are only demonstrable through contradiction, and I'm going to attempt to describe them logically:



                                Let X be a logical statement such that: X $rightarrow$ y, where y is a known contradiction (such as 2+2=5 in the normal arithmetic structure). Without knowing anything else of X, $neg$X implies nothing and nothing implies $neg$X (and hence is not provable). But, of course, assuming X implies a contradiction, and thus, $neg$X.



                                This form of statement X is isolated, in that it only relates to itself and the contradiction. I do believe they can be constructed though, for it seems it they can be described.



                                With that said, in real math and logic, or in general real world scenarios, I don't believe any statements of this form exist, except possibly ones that are constructed to meet this criteria and otherwise meaningless. The proof of the primes was eventually proved without contradiction, to my understanding; until math had been more developed, I think that the statement "the number of primes is $infty$" was basically an isolated logical statement at Euclid's time and for many years after probably, in that there were no other things known to imply it and it didn't imply anything else useful towards its proof.






                                share|cite|improve this answer









                                $endgroup$



                                I do believe there are some proofs that are only demonstrable through contradiction, and I'm going to attempt to describe them logically:



                                Let X be a logical statement such that: X $rightarrow$ y, where y is a known contradiction (such as 2+2=5 in the normal arithmetic structure). Without knowing anything else of X, $neg$X implies nothing and nothing implies $neg$X (and hence is not provable). But, of course, assuming X implies a contradiction, and thus, $neg$X.



                                This form of statement X is isolated, in that it only relates to itself and the contradiction. I do believe they can be constructed though, for it seems it they can be described.



                                With that said, in real math and logic, or in general real world scenarios, I don't believe any statements of this form exist, except possibly ones that are constructed to meet this criteria and otherwise meaningless. The proof of the primes was eventually proved without contradiction, to my understanding; until math had been more developed, I think that the statement "the number of primes is $infty$" was basically an isolated logical statement at Euclid's time and for many years after probably, in that there were no other things known to imply it and it didn't imply anything else useful towards its proof.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 25 '12 at 8:51









                                roflsrofls

                                1392




                                1392























                                    3












                                    $begingroup$

                                    First of all, this is not an answer to the title but to the aside question and is just an example of why you would prefer a constructive proof to a proof by contradiction. Consider the example below,



                                    Prove that $x^2 = 1$ has a root.



                                    Proof by contradiction: Assume that $x^2 = 1$ has no root. Let $f(x) = x^2 - 1 $ then $x^2 = 1$ has a root if and only if $f(x_0) = 0$ for some $x_0$. By assumption $x^2 =1$ has no root and thus, $f(x) neq 0$ for every $x$. Note that $f$ is continuous and $f(0) = -1$ and $f(2) = 3$. Hence, by the intermediate value theorem, $exists x_0$ such that $f(x_0) = 0$ which is a contradiction. Therefore, $x^2=1$ has a root.



                                    Constructive proof: For $x^2=1$ if and only if $x^2-1=0$ iff $(x-1)(x+1)=0$. Hence, for $x=pm 1$ the equation is satisfied, namely the roots are $-1$ and $1$.



                                    The difference is not about the length of the proofs but the information you have. In constructive proof, you know what the roots are but not in the proof by contradiction. Of course, in proof by contradiction, you could have said "let $x_0 = 1$, then $x_0^2=1$ which is a contradiction since $1$ is a root." but then, it is not a clear distinction between the two types of proofs.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      3












                                      $begingroup$

                                      First of all, this is not an answer to the title but to the aside question and is just an example of why you would prefer a constructive proof to a proof by contradiction. Consider the example below,



                                      Prove that $x^2 = 1$ has a root.



                                      Proof by contradiction: Assume that $x^2 = 1$ has no root. Let $f(x) = x^2 - 1 $ then $x^2 = 1$ has a root if and only if $f(x_0) = 0$ for some $x_0$. By assumption $x^2 =1$ has no root and thus, $f(x) neq 0$ for every $x$. Note that $f$ is continuous and $f(0) = -1$ and $f(2) = 3$. Hence, by the intermediate value theorem, $exists x_0$ such that $f(x_0) = 0$ which is a contradiction. Therefore, $x^2=1$ has a root.



                                      Constructive proof: For $x^2=1$ if and only if $x^2-1=0$ iff $(x-1)(x+1)=0$. Hence, for $x=pm 1$ the equation is satisfied, namely the roots are $-1$ and $1$.



                                      The difference is not about the length of the proofs but the information you have. In constructive proof, you know what the roots are but not in the proof by contradiction. Of course, in proof by contradiction, you could have said "let $x_0 = 1$, then $x_0^2=1$ which is a contradiction since $1$ is a root." but then, it is not a clear distinction between the two types of proofs.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        3












                                        3








                                        3





                                        $begingroup$

                                        First of all, this is not an answer to the title but to the aside question and is just an example of why you would prefer a constructive proof to a proof by contradiction. Consider the example below,



                                        Prove that $x^2 = 1$ has a root.



                                        Proof by contradiction: Assume that $x^2 = 1$ has no root. Let $f(x) = x^2 - 1 $ then $x^2 = 1$ has a root if and only if $f(x_0) = 0$ for some $x_0$. By assumption $x^2 =1$ has no root and thus, $f(x) neq 0$ for every $x$. Note that $f$ is continuous and $f(0) = -1$ and $f(2) = 3$. Hence, by the intermediate value theorem, $exists x_0$ such that $f(x_0) = 0$ which is a contradiction. Therefore, $x^2=1$ has a root.



                                        Constructive proof: For $x^2=1$ if and only if $x^2-1=0$ iff $(x-1)(x+1)=0$. Hence, for $x=pm 1$ the equation is satisfied, namely the roots are $-1$ and $1$.



                                        The difference is not about the length of the proofs but the information you have. In constructive proof, you know what the roots are but not in the proof by contradiction. Of course, in proof by contradiction, you could have said "let $x_0 = 1$, then $x_0^2=1$ which is a contradiction since $1$ is a root." but then, it is not a clear distinction between the two types of proofs.






                                        share|cite|improve this answer









                                        $endgroup$



                                        First of all, this is not an answer to the title but to the aside question and is just an example of why you would prefer a constructive proof to a proof by contradiction. Consider the example below,



                                        Prove that $x^2 = 1$ has a root.



                                        Proof by contradiction: Assume that $x^2 = 1$ has no root. Let $f(x) = x^2 - 1 $ then $x^2 = 1$ has a root if and only if $f(x_0) = 0$ for some $x_0$. By assumption $x^2 =1$ has no root and thus, $f(x) neq 0$ for every $x$. Note that $f$ is continuous and $f(0) = -1$ and $f(2) = 3$. Hence, by the intermediate value theorem, $exists x_0$ such that $f(x_0) = 0$ which is a contradiction. Therefore, $x^2=1$ has a root.



                                        Constructive proof: For $x^2=1$ if and only if $x^2-1=0$ iff $(x-1)(x+1)=0$. Hence, for $x=pm 1$ the equation is satisfied, namely the roots are $-1$ and $1$.



                                        The difference is not about the length of the proofs but the information you have. In constructive proof, you know what the roots are but not in the proof by contradiction. Of course, in proof by contradiction, you could have said "let $x_0 = 1$, then $x_0^2=1$ which is a contradiction since $1$ is a root." but then, it is not a clear distinction between the two types of proofs.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Nov 27 '12 at 20:38







                                        user21965






























                                            3












                                            $begingroup$

                                            My non-mathematical response.



                                            A == B equals !(A != B)



                                            You always end up with a binary decision, is or is not. And in any language is = !(is not).



                                            But I guess it is too simple to be ok.






                                            share|cite|improve this answer











                                            $endgroup$









                                            • 3




                                              $begingroup$
                                              is = !(is not) does not hold in minimal and intuitionistic logic. See Carl's answer.
                                              $endgroup$
                                              – user26486
                                              Feb 24 '15 at 18:03


















                                            3












                                            $begingroup$

                                            My non-mathematical response.



                                            A == B equals !(A != B)



                                            You always end up with a binary decision, is or is not. And in any language is = !(is not).



                                            But I guess it is too simple to be ok.






                                            share|cite|improve this answer











                                            $endgroup$









                                            • 3




                                              $begingroup$
                                              is = !(is not) does not hold in minimal and intuitionistic logic. See Carl's answer.
                                              $endgroup$
                                              – user26486
                                              Feb 24 '15 at 18:03
















                                            3












                                            3








                                            3





                                            $begingroup$

                                            My non-mathematical response.



                                            A == B equals !(A != B)



                                            You always end up with a binary decision, is or is not. And in any language is = !(is not).



                                            But I guess it is too simple to be ok.






                                            share|cite|improve this answer











                                            $endgroup$



                                            My non-mathematical response.



                                            A == B equals !(A != B)



                                            You always end up with a binary decision, is or is not. And in any language is = !(is not).



                                            But I guess it is too simple to be ok.







                                            share|cite|improve this answer














                                            share|cite|improve this answer



                                            share|cite|improve this answer








                                            edited Nov 28 '12 at 15:29









                                            ctype.h

                                            6111618




                                            6111618










                                            answered Nov 26 '12 at 4:02









                                            Bart CalixtoBart Calixto

                                            1574




                                            1574








                                            • 3




                                              $begingroup$
                                              is = !(is not) does not hold in minimal and intuitionistic logic. See Carl's answer.
                                              $endgroup$
                                              – user26486
                                              Feb 24 '15 at 18:03
















                                            • 3




                                              $begingroup$
                                              is = !(is not) does not hold in minimal and intuitionistic logic. See Carl's answer.
                                              $endgroup$
                                              – user26486
                                              Feb 24 '15 at 18:03










                                            3




                                            3




                                            $begingroup$
                                            is = !(is not) does not hold in minimal and intuitionistic logic. See Carl's answer.
                                            $endgroup$
                                            – user26486
                                            Feb 24 '15 at 18:03






                                            $begingroup$
                                            is = !(is not) does not hold in minimal and intuitionistic logic. See Carl's answer.
                                            $endgroup$
                                            – user26486
                                            Feb 24 '15 at 18:03













                                            2












                                            $begingroup$

                                            Whether a proof is "by contradiction" really just depends on the statement you started with. If your inital statement is $P rightarrow Q$, then showing the equivalent $neg Q rightarrow neg P$ is "proof by contradiction". But in reality, the "direct" proof for $ P rightarrow Q$ is just a proof "by contradiction" for $neg Q rightarrow neg P$. The only reason why we started with $P rightarrow Q$ instead of $neg Q rightarrow neg P$ is our intuition.



                                            This is just my opinion, but also remember that sometimes, it is also very valuable to know what holds if $Q$ does not hold.






                                            share|cite|improve this answer









                                            $endgroup$













                                            • $begingroup$
                                              It really depends by what you accept as proof. Every logical system has certain logical axioms, nobody is forcing you to accept them. $neg neg A leftrightarrow A$ is in fact not provable in intuitionistic logic.
                                              $endgroup$
                                              – Panayiotis Karabassis
                                              Nov 25 '12 at 0:41










                                            • $begingroup$
                                              ok fine. while i don't know what intuitionistic logic is, i was assuming that the law of excluded middle would hold.
                                              $endgroup$
                                              – kutschkem
                                              Nov 27 '12 at 11:21
















                                            2












                                            $begingroup$

                                            Whether a proof is "by contradiction" really just depends on the statement you started with. If your inital statement is $P rightarrow Q$, then showing the equivalent $neg Q rightarrow neg P$ is "proof by contradiction". But in reality, the "direct" proof for $ P rightarrow Q$ is just a proof "by contradiction" for $neg Q rightarrow neg P$. The only reason why we started with $P rightarrow Q$ instead of $neg Q rightarrow neg P$ is our intuition.



                                            This is just my opinion, but also remember that sometimes, it is also very valuable to know what holds if $Q$ does not hold.






                                            share|cite|improve this answer









                                            $endgroup$













                                            • $begingroup$
                                              It really depends by what you accept as proof. Every logical system has certain logical axioms, nobody is forcing you to accept them. $neg neg A leftrightarrow A$ is in fact not provable in intuitionistic logic.
                                              $endgroup$
                                              – Panayiotis Karabassis
                                              Nov 25 '12 at 0:41










                                            • $begingroup$
                                              ok fine. while i don't know what intuitionistic logic is, i was assuming that the law of excluded middle would hold.
                                              $endgroup$
                                              – kutschkem
                                              Nov 27 '12 at 11:21














                                            2












                                            2








                                            2





                                            $begingroup$

                                            Whether a proof is "by contradiction" really just depends on the statement you started with. If your inital statement is $P rightarrow Q$, then showing the equivalent $neg Q rightarrow neg P$ is "proof by contradiction". But in reality, the "direct" proof for $ P rightarrow Q$ is just a proof "by contradiction" for $neg Q rightarrow neg P$. The only reason why we started with $P rightarrow Q$ instead of $neg Q rightarrow neg P$ is our intuition.



                                            This is just my opinion, but also remember that sometimes, it is also very valuable to know what holds if $Q$ does not hold.






                                            share|cite|improve this answer









                                            $endgroup$



                                            Whether a proof is "by contradiction" really just depends on the statement you started with. If your inital statement is $P rightarrow Q$, then showing the equivalent $neg Q rightarrow neg P$ is "proof by contradiction". But in reality, the "direct" proof for $ P rightarrow Q$ is just a proof "by contradiction" for $neg Q rightarrow neg P$. The only reason why we started with $P rightarrow Q$ instead of $neg Q rightarrow neg P$ is our intuition.



                                            This is just my opinion, but also remember that sometimes, it is also very valuable to know what holds if $Q$ does not hold.







                                            share|cite|improve this answer












                                            share|cite|improve this answer



                                            share|cite|improve this answer










                                            answered Nov 24 '12 at 17:47









                                            kutschkemkutschkem

                                            3161312




                                            3161312












                                            • $begingroup$
                                              It really depends by what you accept as proof. Every logical system has certain logical axioms, nobody is forcing you to accept them. $neg neg A leftrightarrow A$ is in fact not provable in intuitionistic logic.
                                              $endgroup$
                                              – Panayiotis Karabassis
                                              Nov 25 '12 at 0:41










                                            • $begingroup$
                                              ok fine. while i don't know what intuitionistic logic is, i was assuming that the law of excluded middle would hold.
                                              $endgroup$
                                              – kutschkem
                                              Nov 27 '12 at 11:21


















                                            • $begingroup$
                                              It really depends by what you accept as proof. Every logical system has certain logical axioms, nobody is forcing you to accept them. $neg neg A leftrightarrow A$ is in fact not provable in intuitionistic logic.
                                              $endgroup$
                                              – Panayiotis Karabassis
                                              Nov 25 '12 at 0:41










                                            • $begingroup$
                                              ok fine. while i don't know what intuitionistic logic is, i was assuming that the law of excluded middle would hold.
                                              $endgroup$
                                              – kutschkem
                                              Nov 27 '12 at 11:21
















                                            $begingroup$
                                            It really depends by what you accept as proof. Every logical system has certain logical axioms, nobody is forcing you to accept them. $neg neg A leftrightarrow A$ is in fact not provable in intuitionistic logic.
                                            $endgroup$
                                            – Panayiotis Karabassis
                                            Nov 25 '12 at 0:41




                                            $begingroup$
                                            It really depends by what you accept as proof. Every logical system has certain logical axioms, nobody is forcing you to accept them. $neg neg A leftrightarrow A$ is in fact not provable in intuitionistic logic.
                                            $endgroup$
                                            – Panayiotis Karabassis
                                            Nov 25 '12 at 0:41












                                            $begingroup$
                                            ok fine. while i don't know what intuitionistic logic is, i was assuming that the law of excluded middle would hold.
                                            $endgroup$
                                            – kutschkem
                                            Nov 27 '12 at 11:21




                                            $begingroup$
                                            ok fine. while i don't know what intuitionistic logic is, i was assuming that the law of excluded middle would hold.
                                            $endgroup$
                                            – kutschkem
                                            Nov 27 '12 at 11:21











                                            2












                                            $begingroup$

                                            An interesting example of this is the entire study of Smooth Infinitesimal Analysis. It relies on not having the law of the excluded middle (i.e. no proof by contradictions are accepted) in order to be valid. Thus if everything provable by contradiction was also able to be proven directly, then there could not be smooth infinitesimal analysis! Look at Bell's book for more details, though the wiki gives a good example.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              2












                                              $begingroup$

                                              An interesting example of this is the entire study of Smooth Infinitesimal Analysis. It relies on not having the law of the excluded middle (i.e. no proof by contradictions are accepted) in order to be valid. Thus if everything provable by contradiction was also able to be proven directly, then there could not be smooth infinitesimal analysis! Look at Bell's book for more details, though the wiki gives a good example.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                2












                                                2








                                                2





                                                $begingroup$

                                                An interesting example of this is the entire study of Smooth Infinitesimal Analysis. It relies on not having the law of the excluded middle (i.e. no proof by contradictions are accepted) in order to be valid. Thus if everything provable by contradiction was also able to be proven directly, then there could not be smooth infinitesimal analysis! Look at Bell's book for more details, though the wiki gives a good example.






                                                share|cite|improve this answer









                                                $endgroup$



                                                An interesting example of this is the entire study of Smooth Infinitesimal Analysis. It relies on not having the law of the excluded middle (i.e. no proof by contradictions are accepted) in order to be valid. Thus if everything provable by contradiction was also able to be proven directly, then there could not be smooth infinitesimal analysis! Look at Bell's book for more details, though the wiki gives a good example.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Mar 31 '16 at 6:01









                                                Chris RackauckasChris Rackauckas

                                                536312




                                                536312

















                                                    protected by user99914 Feb 7 '18 at 2:40



                                                    Thank you for your interest in this question.
                                                    Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



                                                    Would you like to answer one of these unanswered questions instead?



                                                    Popular posts from this blog

                                                    How to change which sound is reproduced for terminal bell?

                                                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                                                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents