What is the difference between “for some $k inmathbb{R}$” and “$forall: k inmathbb{R}$”?
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When proving by induction, in the IH you're supposed to write something like "suppose $a_k < a_{k+1} < 15$ for some $k in Bbb R$. If I instead write for all $k in Bbb R$ is that considered wrong?
abstract-algebra
edited Nov 19 at 8:11
Hirak
2,56411336
asked Nov 19 at 7:57
ming
1725
add a comment |
up vote
-1
down vote
favorite
When proving by induction, in the IH you're supposed to write something like "suppose $a_k < a_{k+1} < 15$ for some $k in Bbb R$. If I instead write for all $k in Bbb R$ is that considered wrong?
abstract-algebra
edited Nov 19 at 8:11
Hirak
2,56411336
asked Nov 19 at 7:57
ming
1725
3
Are you sure you don't mean $kinBbb N$?
– J.G.
Nov 19 at 7:58
In general, in a proof by induction we have to prove that $P(k)$ holds for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:02
The proof runs as follows : (i) Prove that $P(0)$ holds. (ii) prove that, provided that $P(k)$ holds, also $P(k+1)$ holds, i.e. that $P(k) to P(k+1)$ holds, for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:03
In general, when you do a proof by induction, you work in a countable set, most often $mathbb{N}$, and not $mathbb{R}$. So can you provide more details? Why do you want to do induction in $mathbb{R}$?
– Todor Markov
Nov 19 at 8:17
Oh I just had to prove that a recursive sequence is non-decreasing so I just said suppose $a_k < a_{k+1} < 15$ for all $k epsilon Bbb R$ but I got marks taken off for saying "all" instead of "some"
– ming
Nov 19 at 8:32
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
When proving by induction, in the IH you're supposed to write something like "suppose $a_k < a_{k+1} < 15$ for some $k in Bbb R$. If I instead write for all $k in Bbb R$ is that considered wrong?
abstract-algebra
edited Nov 19 at 8:11
Hirak
2,56411336
asked Nov 19 at 7:57
ming
1725
When proving by induction, in the IH you're supposed to write something like "suppose $a_k < a_{k+1} < 15$ for some $k in Bbb R$. If I instead write for all $k in Bbb R$ is that considered wrong?
abstract-algebra
abstract-algebra
edited Nov 19 at 8:11
Hirak
2,56411336
asked Nov 19 at 7:57
ming
1725
edited Nov 19 at 8:11
Hirak
2,56411336
asked Nov 19 at 7:57
ming
1725
edited Nov 19 at 8:11
Hirak
2,56411336
edited Nov 19 at 8:11
Hirak
2,56411336
edited Nov 19 at 8:11
Hirak
2,56411336
2,56411336
asked Nov 19 at 7:57
ming
1725
asked Nov 19 at 7:57
ming
1725
asked Nov 19 at 7:57
ming
1725
1725
3
Are you sure you don't mean $kinBbb N$?
– J.G.
Nov 19 at 7:58
In general, in a proof by induction we have to prove that $P(k)$ holds for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:02
The proof runs as follows : (i) Prove that $P(0)$ holds. (ii) prove that, provided that $P(k)$ holds, also $P(k+1)$ holds, i.e. that $P(k) to P(k+1)$ holds, for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:03
In general, when you do a proof by induction, you work in a countable set, most often $mathbb{N}$, and not $mathbb{R}$. So can you provide more details? Why do you want to do induction in $mathbb{R}$?
– Todor Markov
Nov 19 at 8:17
Oh I just had to prove that a recursive sequence is non-decreasing so I just said suppose $a_k < a_{k+1} < 15$ for all $k epsilon Bbb R$ but I got marks taken off for saying "all" instead of "some"
– ming
Nov 19 at 8:32
add a comment |
3
Are you sure you don't mean $kinBbb N$?
– J.G.
Nov 19 at 7:58
In general, in a proof by induction we have to prove that $P(k)$ holds for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:02
The proof runs as follows : (i) Prove that $P(0)$ holds. (ii) prove that, provided that $P(k)$ holds, also $P(k+1)$ holds, i.e. that $P(k) to P(k+1)$ holds, for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:03
In general, when you do a proof by induction, you work in a countable set, most often $mathbb{N}$, and not $mathbb{R}$. So can you provide more details? Why do you want to do induction in $mathbb{R}$?
– Todor Markov
Nov 19 at 8:17
Oh I just had to prove that a recursive sequence is non-decreasing so I just said suppose $a_k < a_{k+1} < 15$ for all $k epsilon Bbb R$ but I got marks taken off for saying "all" instead of "some"
– ming
Nov 19 at 8:32
3
3
Are you sure you don't mean $kinBbb N$?
– J.G.
Nov 19 at 7:58
Are you sure you don't mean $kinBbb N$?
– J.G.
Nov 19 at 7:58
In general, in a proof by induction we have to prove that $P(k)$ holds for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:02
In general, in a proof by induction we have to prove that $P(k)$ holds for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:02
The proof runs as follows : (i) Prove that $P(0)$ holds. (ii) prove that, provided that $P(k)$ holds, also $P(k+1)$ holds, i.e. that $P(k) to P(k+1)$ holds, for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:03
The proof runs as follows : (i) Prove that $P(0)$ holds. (ii) prove that, provided that $P(k)$ holds, also $P(k+1)$ holds, i.e. that $P(k) to P(k+1)$ holds, for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:03
In general, when you do a proof by induction, you work in a countable set, most often $mathbb{N}$, and not $mathbb{R}$. So can you provide more details? Why do you want to do induction in $mathbb{R}$?
– Todor Markov
Nov 19 at 8:17
In general, when you do a proof by induction, you work in a countable set, most often $mathbb{N}$, and not $mathbb{R}$. So can you provide more details? Why do you want to do induction in $mathbb{R}$?
– Todor Markov
Nov 19 at 8:17
Oh I just had to prove that a recursive sequence is non-decreasing so I just said suppose $a_k < a_{k+1} < 15$ for all $k epsilon Bbb R$ but I got marks taken off for saying "all" instead of "some"
– ming
Nov 19 at 8:32
Oh I just had to prove that a recursive sequence is non-decreasing so I just said suppose $a_k < a_{k+1} < 15$ for all $k epsilon Bbb R$ but I got marks taken off for saying "all" instead of "some"
– ming
Nov 19 at 8:32
add a comment |
1 Answer
1
active
oldest
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up vote
0
down vote
Consider the following situation: For every $ninmathbb{N}$ you have a statement $P(n)$, which is either true or false.
The induction principle says the following. You first explicitly show that $P(1)$ is true. Then you show that $P(n)implies P(n+1)$ for all $ngeq 1$. Now you get this nice domino effect: Since $P(1)$ is true and $P(1)implies P(2)$, $P(2)$ is true as well. Since $P(2)$ is true and $P(2)implies P(3)$, $P(3)$ is true. You can easily see that we now showed that $P(n)$ is true for all $n$.
Sometimes it is more convenient to show that $(P(1) wedge dots wedge P(n))implies P(n+1)$. In that case the left part is sometimes written as, $P(k)$ for all $kleq n$. You can show that a base step together with the above rule will also show that all $P(n)$ are true.
What you never should write is the following: $(forall nin mathbb{N}:P(n))implies P(n+1)$. This doesn't make sense for two reasons:
- You want to show $forall nin mathbb{N}:P(n)$, so why would it be your induction hypothesis? If you assume this, then obviously what you want to prove is correct, but what if it is false?
- the variable $n+1$ is not defined on the right side of the implication $(forall nin mathbb{N}:P(n))implies P(n+1)$. This is just bad mathematics.
answered Nov 19 at 8:23
Mathematician 42
8,50111438
Ohh so the reason I can't write for all k is we can't actually assume that it works for every k, we're just saying it works for "some" k, which is true because we proved the base case, and obviously if it works for one case, we know it actually works for AT LEAST 1 k, hence we can write "some"?
– ming
Nov 19 at 8:34
You definitely can't assume that it works for all $k$ as that is what you want to show. Usually when working with induction, you want to show that if $P(n)$ holds, so does $P(n+1)$.
– Mathematician 42
Nov 19 at 9:05
@ming, That comment is true, but there's a bit more to it. "Some" in your case refers to one number $k$, not a set of them. This is important, because only if you're talking about a specific $k$ you have a "next one" ($k + 1$). However, while it denotes a single number, it is not specified which. That is, you want to be able to substitute $k$ for any number, and get a true statement.
– Todor Markov
Nov 19 at 9:07
So.. then is it saying basically "we proved that ONE k works so far, the base case. Now let's prove that it works for k+1, k+2, k+3, etc.." And that one case is just denoted by "some"??
– ming
Nov 19 at 21:13
@ming No, you refer to one general (unknown) case. Let's rephrase the whole statement. What you want to prove is that "If it works for $k$, then it works for $k + 1$." This is not much different than the statement "If $n$ is even, then $n+2$ is even." This statement refers to one specific number $n$, but it can be any number. If you replace $n$ with a number, you can get "If 4 is even, then 6 is even", or "If 5 is even, then 7 is even" (both true). But for that statement to be useful, you need to know at least one even, say 4. Then it immediately gives you 6, 8, 10, etc. are also even.
– Todor Markov
Nov 20 at 13:40
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Consider the following situation: For every $ninmathbb{N}$ you have a statement $P(n)$, which is either true or false.
The induction principle says the following. You first explicitly show that $P(1)$ is true. Then you show that $P(n)implies P(n+1)$ for all $ngeq 1$. Now you get this nice domino effect: Since $P(1)$ is true and $P(1)implies P(2)$, $P(2)$ is true as well. Since $P(2)$ is true and $P(2)implies P(3)$, $P(3)$ is true. You can easily see that we now showed that $P(n)$ is true for all $n$.
Sometimes it is more convenient to show that $(P(1) wedge dots wedge P(n))implies P(n+1)$. In that case the left part is sometimes written as, $P(k)$ for all $kleq n$. You can show that a base step together with the above rule will also show that all $P(n)$ are true.
What you never should write is the following: $(forall nin mathbb{N}:P(n))implies P(n+1)$. This doesn't make sense for two reasons:
- You want to show $forall nin mathbb{N}:P(n)$, so why would it be your induction hypothesis? If you assume this, then obviously what you want to prove is correct, but what if it is false?
- the variable $n+1$ is not defined on the right side of the implication $(forall nin mathbb{N}:P(n))implies P(n+1)$. This is just bad mathematics.
answered Nov 19 at 8:23
Mathematician 42
8,50111438
Ohh so the reason I can't write for all k is we can't actually assume that it works for every k, we're just saying it works for "some" k, which is true because we proved the base case, and obviously if it works for one case, we know it actually works for AT LEAST 1 k, hence we can write "some"?
– ming
Nov 19 at 8:34
You definitely can't assume that it works for all $k$ as that is what you want to show. Usually when working with induction, you want to show that if $P(n)$ holds, so does $P(n+1)$.
– Mathematician 42
Nov 19 at 9:05
@ming, That comment is true, but there's a bit more to it. "Some" in your case refers to one number $k$, not a set of them. This is important, because only if you're talking about a specific $k$ you have a "next one" ($k + 1$). However, while it denotes a single number, it is not specified which. That is, you want to be able to substitute $k$ for any number, and get a true statement.
– Todor Markov
Nov 19 at 9:07
So.. then is it saying basically "we proved that ONE k works so far, the base case. Now let's prove that it works for k+1, k+2, k+3, etc.." And that one case is just denoted by "some"??
– ming
Nov 19 at 21:13
@ming No, you refer to one general (unknown) case. Let's rephrase the whole statement. What you want to prove is that "If it works for $k$, then it works for $k + 1$." This is not much different than the statement "If $n$ is even, then $n+2$ is even." This statement refers to one specific number $n$, but it can be any number. If you replace $n$ with a number, you can get "If 4 is even, then 6 is even", or "If 5 is even, then 7 is even" (both true). But for that statement to be useful, you need to know at least one even, say 4. Then it immediately gives you 6, 8, 10, etc. are also even.
– Todor Markov
Nov 20 at 13:40
add a comment |
up vote
0
down vote
Consider the following situation: For every $ninmathbb{N}$ you have a statement $P(n)$, which is either true or false.
The induction principle says the following. You first explicitly show that $P(1)$ is true. Then you show that $P(n)implies P(n+1)$ for all $ngeq 1$. Now you get this nice domino effect: Since $P(1)$ is true and $P(1)implies P(2)$, $P(2)$ is true as well. Since $P(2)$ is true and $P(2)implies P(3)$, $P(3)$ is true. You can easily see that we now showed that $P(n)$ is true for all $n$.
Sometimes it is more convenient to show that $(P(1) wedge dots wedge P(n))implies P(n+1)$. In that case the left part is sometimes written as, $P(k)$ for all $kleq n$. You can show that a base step together with the above rule will also show that all $P(n)$ are true.
What you never should write is the following: $(forall nin mathbb{N}:P(n))implies P(n+1)$. This doesn't make sense for two reasons:
- You want to show $forall nin mathbb{N}:P(n)$, so why would it be your induction hypothesis? If you assume this, then obviously what you want to prove is correct, but what if it is false?
- the variable $n+1$ is not defined on the right side of the implication $(forall nin mathbb{N}:P(n))implies P(n+1)$. This is just bad mathematics.
answered Nov 19 at 8:23
Mathematician 42
8,50111438
Ohh so the reason I can't write for all k is we can't actually assume that it works for every k, we're just saying it works for "some" k, which is true because we proved the base case, and obviously if it works for one case, we know it actually works for AT LEAST 1 k, hence we can write "some"?
– ming
Nov 19 at 8:34
You definitely can't assume that it works for all $k$ as that is what you want to show. Usually when working with induction, you want to show that if $P(n)$ holds, so does $P(n+1)$.
– Mathematician 42
Nov 19 at 9:05
@ming, That comment is true, but there's a bit more to it. "Some" in your case refers to one number $k$, not a set of them. This is important, because only if you're talking about a specific $k$ you have a "next one" ($k + 1$). However, while it denotes a single number, it is not specified which. That is, you want to be able to substitute $k$ for any number, and get a true statement.
– Todor Markov
Nov 19 at 9:07
So.. then is it saying basically "we proved that ONE k works so far, the base case. Now let's prove that it works for k+1, k+2, k+3, etc.." And that one case is just denoted by "some"??
– ming
Nov 19 at 21:13
@ming No, you refer to one general (unknown) case. Let's rephrase the whole statement. What you want to prove is that "If it works for $k$, then it works for $k + 1$." This is not much different than the statement "If $n$ is even, then $n+2$ is even." This statement refers to one specific number $n$, but it can be any number. If you replace $n$ with a number, you can get "If 4 is even, then 6 is even", or "If 5 is even, then 7 is even" (both true). But for that statement to be useful, you need to know at least one even, say 4. Then it immediately gives you 6, 8, 10, etc. are also even.
– Todor Markov
Nov 20 at 13:40
add a comment |
up vote
0
down vote
up vote
0
down vote
Consider the following situation: For every $ninmathbb{N}$ you have a statement $P(n)$, which is either true or false.
The induction principle says the following. You first explicitly show that $P(1)$ is true. Then you show that $P(n)implies P(n+1)$ for all $ngeq 1$. Now you get this nice domino effect: Since $P(1)$ is true and $P(1)implies P(2)$, $P(2)$ is true as well. Since $P(2)$ is true and $P(2)implies P(3)$, $P(3)$ is true. You can easily see that we now showed that $P(n)$ is true for all $n$.
Sometimes it is more convenient to show that $(P(1) wedge dots wedge P(n))implies P(n+1)$. In that case the left part is sometimes written as, $P(k)$ for all $kleq n$. You can show that a base step together with the above rule will also show that all $P(n)$ are true.
What you never should write is the following: $(forall nin mathbb{N}:P(n))implies P(n+1)$. This doesn't make sense for two reasons:
- You want to show $forall nin mathbb{N}:P(n)$, so why would it be your induction hypothesis? If you assume this, then obviously what you want to prove is correct, but what if it is false?
- the variable $n+1$ is not defined on the right side of the implication $(forall nin mathbb{N}:P(n))implies P(n+1)$. This is just bad mathematics.
answered Nov 19 at 8:23
Mathematician 42
8,50111438
Consider the following situation: For every $ninmathbb{N}$ you have a statement $P(n)$, which is either true or false.
The induction principle says the following. You first explicitly show that $P(1)$ is true. Then you show that $P(n)implies P(n+1)$ for all $ngeq 1$. Now you get this nice domino effect: Since $P(1)$ is true and $P(1)implies P(2)$, $P(2)$ is true as well. Since $P(2)$ is true and $P(2)implies P(3)$, $P(3)$ is true. You can easily see that we now showed that $P(n)$ is true for all $n$.
Sometimes it is more convenient to show that $(P(1) wedge dots wedge P(n))implies P(n+1)$. In that case the left part is sometimes written as, $P(k)$ for all $kleq n$. You can show that a base step together with the above rule will also show that all $P(n)$ are true.
What you never should write is the following: $(forall nin mathbb{N}:P(n))implies P(n+1)$. This doesn't make sense for two reasons:
- You want to show $forall nin mathbb{N}:P(n)$, so why would it be your induction hypothesis? If you assume this, then obviously what you want to prove is correct, but what if it is false?
- the variable $n+1$ is not defined on the right side of the implication $(forall nin mathbb{N}:P(n))implies P(n+1)$. This is just bad mathematics.
answered Nov 19 at 8:23
Mathematician 42
8,50111438
answered Nov 19 at 8:23
Mathematician 42
8,50111438
answered Nov 19 at 8:23
Mathematician 42
8,50111438
answered Nov 19 at 8:23
Mathematician 42
8,50111438
8,50111438
Ohh so the reason I can't write for all k is we can't actually assume that it works for every k, we're just saying it works for "some" k, which is true because we proved the base case, and obviously if it works for one case, we know it actually works for AT LEAST 1 k, hence we can write "some"?
– ming
Nov 19 at 8:34
You definitely can't assume that it works for all $k$ as that is what you want to show. Usually when working with induction, you want to show that if $P(n)$ holds, so does $P(n+1)$.
– Mathematician 42
Nov 19 at 9:05
@ming, That comment is true, but there's a bit more to it. "Some" in your case refers to one number $k$, not a set of them. This is important, because only if you're talking about a specific $k$ you have a "next one" ($k + 1$). However, while it denotes a single number, it is not specified which. That is, you want to be able to substitute $k$ for any number, and get a true statement.
– Todor Markov
Nov 19 at 9:07
So.. then is it saying basically "we proved that ONE k works so far, the base case. Now let's prove that it works for k+1, k+2, k+3, etc.." And that one case is just denoted by "some"??
– ming
Nov 19 at 21:13
@ming No, you refer to one general (unknown) case. Let's rephrase the whole statement. What you want to prove is that "If it works for $k$, then it works for $k + 1$." This is not much different than the statement "If $n$ is even, then $n+2$ is even." This statement refers to one specific number $n$, but it can be any number. If you replace $n$ with a number, you can get "If 4 is even, then 6 is even", or "If 5 is even, then 7 is even" (both true). But for that statement to be useful, you need to know at least one even, say 4. Then it immediately gives you 6, 8, 10, etc. are also even.
– Todor Markov
Nov 20 at 13:40
add a comment |
Ohh so the reason I can't write for all k is we can't actually assume that it works for every k, we're just saying it works for "some" k, which is true because we proved the base case, and obviously if it works for one case, we know it actually works for AT LEAST 1 k, hence we can write "some"?
– ming
Nov 19 at 8:34
You definitely can't assume that it works for all $k$ as that is what you want to show. Usually when working with induction, you want to show that if $P(n)$ holds, so does $P(n+1)$.
– Mathematician 42
Nov 19 at 9:05
@ming, That comment is true, but there's a bit more to it. "Some" in your case refers to one number $k$, not a set of them. This is important, because only if you're talking about a specific $k$ you have a "next one" ($k + 1$). However, while it denotes a single number, it is not specified which. That is, you want to be able to substitute $k$ for any number, and get a true statement.
– Todor Markov
Nov 19 at 9:07
So.. then is it saying basically "we proved that ONE k works so far, the base case. Now let's prove that it works for k+1, k+2, k+3, etc.." And that one case is just denoted by "some"??
– ming
Nov 19 at 21:13
@ming No, you refer to one general (unknown) case. Let's rephrase the whole statement. What you want to prove is that "If it works for $k$, then it works for $k + 1$." This is not much different than the statement "If $n$ is even, then $n+2$ is even." This statement refers to one specific number $n$, but it can be any number. If you replace $n$ with a number, you can get "If 4 is even, then 6 is even", or "If 5 is even, then 7 is even" (both true). But for that statement to be useful, you need to know at least one even, say 4. Then it immediately gives you 6, 8, 10, etc. are also even.
– Todor Markov
Nov 20 at 13:40
Ohh so the reason I can't write for all k is we can't actually assume that it works for every k, we're just saying it works for "some" k, which is true because we proved the base case, and obviously if it works for one case, we know it actually works for AT LEAST 1 k, hence we can write "some"?
– ming
Nov 19 at 8:34
Ohh so the reason I can't write for all k is we can't actually assume that it works for every k, we're just saying it works for "some" k, which is true because we proved the base case, and obviously if it works for one case, we know it actually works for AT LEAST 1 k, hence we can write "some"?
– ming
Nov 19 at 8:34
You definitely can't assume that it works for all $k$ as that is what you want to show. Usually when working with induction, you want to show that if $P(n)$ holds, so does $P(n+1)$.
– Mathematician 42
Nov 19 at 9:05
You definitely can't assume that it works for all $k$ as that is what you want to show. Usually when working with induction, you want to show that if $P(n)$ holds, so does $P(n+1)$.
– Mathematician 42
Nov 19 at 9:05
@ming, That comment is true, but there's a bit more to it. "Some" in your case refers to one number $k$, not a set of them. This is important, because only if you're talking about a specific $k$ you have a "next one" ($k + 1$). However, while it denotes a single number, it is not specified which. That is, you want to be able to substitute $k$ for any number, and get a true statement.
– Todor Markov
Nov 19 at 9:07
@ming, That comment is true, but there's a bit more to it. "Some" in your case refers to one number $k$, not a set of them. This is important, because only if you're talking about a specific $k$ you have a "next one" ($k + 1$). However, while it denotes a single number, it is not specified which. That is, you want to be able to substitute $k$ for any number, and get a true statement.
– Todor Markov
Nov 19 at 9:07
So.. then is it saying basically "we proved that ONE k works so far, the base case. Now let's prove that it works for k+1, k+2, k+3, etc.." And that one case is just denoted by "some"??
– ming
Nov 19 at 21:13
So.. then is it saying basically "we proved that ONE k works so far, the base case. Now let's prove that it works for k+1, k+2, k+3, etc.." And that one case is just denoted by "some"??
– ming
Nov 19 at 21:13
@ming No, you refer to one general (unknown) case. Let's rephrase the whole statement. What you want to prove is that "If it works for $k$, then it works for $k + 1$." This is not much different than the statement "If $n$ is even, then $n+2$ is even." This statement refers to one specific number $n$, but it can be any number. If you replace $n$ with a number, you can get "If 4 is even, then 6 is even", or "If 5 is even, then 7 is even" (both true). But for that statement to be useful, you need to know at least one even, say 4. Then it immediately gives you 6, 8, 10, etc. are also even.
– Todor Markov
Nov 20 at 13:40
@ming No, you refer to one general (unknown) case. Let's rephrase the whole statement. What you want to prove is that "If it works for $k$, then it works for $k + 1$." This is not much different than the statement "If $n$ is even, then $n+2$ is even." This statement refers to one specific number $n$, but it can be any number. If you replace $n$ with a number, you can get "If 4 is even, then 6 is even", or "If 5 is even, then 7 is even" (both true). But for that statement to be useful, you need to know at least one even, say 4. Then it immediately gives you 6, 8, 10, etc. are also even.
– Todor Markov
Nov 20 at 13:40
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3
Are you sure you don't mean $kinBbb N$?
– J.G.
Nov 19 at 7:58
In general, in a proof by induction we have to prove that $P(k)$ holds for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:02
The proof runs as follows : (i) Prove that $P(0)$ holds. (ii) prove that, provided that $P(k)$ holds, also $P(k+1)$ holds, i.e. that $P(k) to P(k+1)$ holds, for every $k$.
– Mauro ALLEGRANZA
Nov 19 at 8:03
In general, when you do a proof by induction, you work in a countable set, most often $mathbb{N}$, and not $mathbb{R}$. So can you provide more details? Why do you want to do induction in $mathbb{R}$?
– Todor Markov
Nov 19 at 8:17
Oh I just had to prove that a recursive sequence is non-decreasing so I just said suppose $a_k < a_{k+1} < 15$ for all $k epsilon Bbb R$ but I got marks taken off for saying "all" instead of "some"
– ming
Nov 19 at 8:32