Does $pi$ contain every real?
2
$begingroup$
One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
$pi = 3.141592653589793238462643383279502884197169399...$
for instance, for $m=3$, $n=5$, this will be
3.141592653589793238462643383279502884197169399...
giving
$0.469363019...$
I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.
So I guess this method can't create any real. But why?
elementary-set-theory
edited Dec 12 '18 at 18:14
verret
3,3161923
asked Dec 12 '18 at 8:52
MrMartinMrMartin
757
$endgroup$
add a comment |
2
$begingroup$
One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
$pi = 3.141592653589793238462643383279502884197169399...$
for instance, for $m=3$, $n=5$, this will be
3.141592653589793238462643383279502884197169399...
giving
$0.469363019...$
I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.
So I guess this method can't create any real. But why?
elementary-set-theory
edited Dec 12 '18 at 18:14
verret
3,3161923
asked Dec 12 '18 at 8:52
MrMartinMrMartin
757
$endgroup$
3
$begingroup$
You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
$endgroup$
– mathworker21
Dec 12 '18 at 8:57
1
$begingroup$
The number of pairs $n, : k$ is countable!
$endgroup$
– R.C.Cowsik
Dec 12 '18 at 8:58
4
$begingroup$
I think the more interesting question is : give an example of a real number that cannot be produced like this.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 9:07
$begingroup$
My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:36
$begingroup$
@SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
$endgroup$
– Holo
Dec 12 '18 at 9:53
add a comment |
2
2
2
$begingroup$
One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
$pi = 3.141592653589793238462643383279502884197169399...$
for instance, for $m=3$, $n=5$, this will be
3.141592653589793238462643383279502884197169399...
giving
$0.469363019...$
I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.
So I guess this method can't create any real. But why?
elementary-set-theory
edited Dec 12 '18 at 18:14
verret
3,3161923
asked Dec 12 '18 at 8:52
MrMartinMrMartin
757
$endgroup$
One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
$pi = 3.141592653589793238462643383279502884197169399...$
for instance, for $m=3$, $n=5$, this will be
3.141592653589793238462643383279502884197169399...
giving
$0.469363019...$
I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.
So I guess this method can't create any real. But why?
elementary-set-theory
elementary-set-theory
edited Dec 12 '18 at 18:14
verret
3,3161923
asked Dec 12 '18 at 8:52
MrMartinMrMartin
757
edited Dec 12 '18 at 18:14
verret
3,3161923
asked Dec 12 '18 at 8:52
MrMartinMrMartin
757
edited Dec 12 '18 at 18:14
verret
3,3161923
edited Dec 12 '18 at 18:14
verret
3,3161923
edited Dec 12 '18 at 18:14
verret
3,3161923
3,3161923
asked Dec 12 '18 at 8:52
MrMartinMrMartin
757
asked Dec 12 '18 at 8:52
MrMartinMrMartin
757
asked Dec 12 '18 at 8:52
MrMartinMrMartin
757
757
3
$begingroup$
You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
$endgroup$
– mathworker21
Dec 12 '18 at 8:57
1
$begingroup$
The number of pairs $n, : k$ is countable!
$endgroup$
– R.C.Cowsik
Dec 12 '18 at 8:58
4
$begingroup$
I think the more interesting question is : give an example of a real number that cannot be produced like this.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 9:07
$begingroup$
My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:36
$begingroup$
@SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
$endgroup$
– Holo
Dec 12 '18 at 9:53
add a comment |
3
$begingroup$
You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
$endgroup$
– mathworker21
Dec 12 '18 at 8:57
1
$begingroup$
The number of pairs $n, : k$ is countable!
$endgroup$
– R.C.Cowsik
Dec 12 '18 at 8:58
4
$begingroup$
I think the more interesting question is : give an example of a real number that cannot be produced like this.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 9:07
$begingroup$
My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:36
$begingroup$
@SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
$endgroup$
– Holo
Dec 12 '18 at 9:53
3
3
$begingroup$
You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
$endgroup$
– mathworker21
Dec 12 '18 at 8:57
$begingroup$
You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
$endgroup$
– mathworker21
Dec 12 '18 at 8:57
1
1
$begingroup$
The number of pairs $n, : k$ is countable!
$endgroup$
– R.C.Cowsik
Dec 12 '18 at 8:58
$begingroup$
The number of pairs $n, : k$ is countable!
$endgroup$
– R.C.Cowsik
Dec 12 '18 at 8:58
4
4
$begingroup$
I think the more interesting question is : give an example of a real number that cannot be produced like this.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 9:07
$begingroup$
I think the more interesting question is : give an example of a real number that cannot be produced like this.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 9:07
$begingroup$
My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:36
$begingroup$
My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:36
$begingroup$
@SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
$endgroup$
– Holo
Dec 12 '18 at 9:53
$begingroup$
@SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
$endgroup$
– Holo
Dec 12 '18 at 9:53
add a comment |
1 Answer
1
active
oldest
votes
5
$begingroup$
You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.
In particular, the argument would go like so:
- Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
- Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.
- Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$
- Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$
- Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.
- From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.
- Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.
answered Dec 12 '18 at 9:05
5xum5xum
91.8k394161
$endgroup$
$begingroup$
Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
$endgroup$
– MrMartin
Dec 12 '18 at 9:20
$begingroup$
@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
$endgroup$
– 5xum
Dec 12 '18 at 9:24
$begingroup$
@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
$endgroup$
– Yves Daoust
Dec 12 '18 at 9:30
$begingroup$
Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
$endgroup$
– MrMartin
Dec 12 '18 at 9:32
3
$begingroup$
@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
$endgroup$
– 5xum
Dec 12 '18 at 9:40
|
show 2 more comments
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.
In particular, the argument would go like so:
- Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
- Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.
- Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$
- Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$
- Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.
- From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.
- Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.
answered Dec 12 '18 at 9:05
5xum5xum
91.8k394161
$endgroup$
$begingroup$
Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
$endgroup$
– MrMartin
Dec 12 '18 at 9:20
$begingroup$
@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
$endgroup$
– 5xum
Dec 12 '18 at 9:24
$begingroup$
@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
$endgroup$
– Yves Daoust
Dec 12 '18 at 9:30
$begingroup$
Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
$endgroup$
– MrMartin
Dec 12 '18 at 9:32
3
$begingroup$
@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
$endgroup$
– 5xum
Dec 12 '18 at 9:40
|
show 2 more comments
5
$begingroup$
You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.
In particular, the argument would go like so:
- Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
- Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.
- Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$
- Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$
- Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.
- From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.
- Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.
answered Dec 12 '18 at 9:05
5xum5xum
91.8k394161
$endgroup$
$begingroup$
Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
$endgroup$
– MrMartin
Dec 12 '18 at 9:20
$begingroup$
@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
$endgroup$
– 5xum
Dec 12 '18 at 9:24
$begingroup$
@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
$endgroup$
– Yves Daoust
Dec 12 '18 at 9:30
$begingroup$
Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
$endgroup$
– MrMartin
Dec 12 '18 at 9:32
3
$begingroup$
@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
$endgroup$
– 5xum
Dec 12 '18 at 9:40
|
show 2 more comments
5
5
5
$begingroup$
You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.
In particular, the argument would go like so:
- Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
- Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.
- Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$
- Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$
- Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.
- From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.
- Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.
answered Dec 12 '18 at 9:05
5xum5xum
91.8k394161
$endgroup$
You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.
In particular, the argument would go like so:
- Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.
- Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.
- Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$
- Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$
- Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.
- From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.
- Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.
answered Dec 12 '18 at 9:05
5xum5xum
91.8k394161
edited Dec 12 '18 at 9:37
answered Dec 12 '18 at 9:05
5xum5xum
91.8k394161
answered Dec 12 '18 at 9:05
5xum5xum
91.8k394161
answered Dec 12 '18 at 9:05
5xum5xum
91.8k394161
91.8k394161
$begingroup$
Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
$endgroup$
– MrMartin
Dec 12 '18 at 9:20
$begingroup$
@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
$endgroup$
– 5xum
Dec 12 '18 at 9:24
$begingroup$
@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
$endgroup$
– Yves Daoust
Dec 12 '18 at 9:30
$begingroup$
Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
$endgroup$
– MrMartin
Dec 12 '18 at 9:32
3
$begingroup$
@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
$endgroup$
– 5xum
Dec 12 '18 at 9:40
|
show 2 more comments
$begingroup$
Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
$endgroup$
– MrMartin
Dec 12 '18 at 9:20
$begingroup$
@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
$endgroup$
– 5xum
Dec 12 '18 at 9:24
$begingroup$
@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
$endgroup$
– Yves Daoust
Dec 12 '18 at 9:30
$begingroup$
Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
$endgroup$
– MrMartin
Dec 12 '18 at 9:32
3
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@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
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– 5xum
Dec 12 '18 at 9:40
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Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
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– MrMartin
Dec 12 '18 at 9:20
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Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
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– MrMartin
Dec 12 '18 at 9:20
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@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
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– 5xum
Dec 12 '18 at 9:24
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@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
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– 5xum
Dec 12 '18 at 9:24
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@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
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– Yves Daoust
Dec 12 '18 at 9:30
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@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
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– Yves Daoust
Dec 12 '18 at 9:30
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Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
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– MrMartin
Dec 12 '18 at 9:32
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Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
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– MrMartin
Dec 12 '18 at 9:32
3
3
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@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
$endgroup$
– 5xum
Dec 12 '18 at 9:40
$begingroup$
@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
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– 5xum
Dec 12 '18 at 9:40
|
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3
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You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
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– mathworker21
Dec 12 '18 at 8:57
1
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The number of pairs $n, : k$ is countable!
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– R.C.Cowsik
Dec 12 '18 at 8:58
4
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I think the more interesting question is : give an example of a real number that cannot be produced like this.
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– астон вілла олоф мэллбэрг
Dec 12 '18 at 9:07
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My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
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– SmileyCraft
Dec 12 '18 at 9:36
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@SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
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– Holo
Dec 12 '18 at 9:53