The $n$-th root of $2$ is irrational for $n>1$, and $lim_{ntoinfty}sqrt[n]{2}=1$. Is the latter fact a...
So I was looking at a theorem that popped up into my head:
The $n$-th root of $2$ always being irrational if $n$ is greater than $1$.
I also noticed that
The limit of the $n$-th root of $2$, as $n$ approaches infinity, is $1$.
Is there a connection between this "$n$-th root of $2$ Theorem" and the limit I mentioned?
Can I consider that limit a corollary to the theorem?
limits elementary-number-theory
edited Nov 29 at 4:55
Jyrki Lahtonen
108k12166366
asked Nov 28 at 23:33
Xavier Stanton
330211
add a comment |
So I was looking at a theorem that popped up into my head:
The $n$-th root of $2$ always being irrational if $n$ is greater than $1$.
I also noticed that
The limit of the $n$-th root of $2$, as $n$ approaches infinity, is $1$.
Is there a connection between this "$n$-th root of $2$ Theorem" and the limit I mentioned?
Can I consider that limit a corollary to the theorem?
limits elementary-number-theory
edited Nov 29 at 4:55
Jyrki Lahtonen
108k12166366
asked Nov 28 at 23:33
Xavier Stanton
330211
3
Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
– T. Bongers
Nov 28 at 23:35
If $n$ is irrational then the nth root of $2$ can be rational.
– kingW3
Nov 28 at 23:40
Just so I know, how do I keep the revisions? It sounds stupid, but it wouldn't let me keep the revisions that were made.
– Xavier Stanton
Nov 29 at 0:18
Also, how do you make the formulas?
– Xavier Stanton
Nov 29 at 0:19
We use MathJax to typeset formulas. A tutorial is here. To see the edit history you can click on the "edited xxx ago" in the center bottom of the post.
– Ross Millikan
Nov 29 at 0:39
add a comment |
So I was looking at a theorem that popped up into my head:
The $n$-th root of $2$ always being irrational if $n$ is greater than $1$.
I also noticed that
The limit of the $n$-th root of $2$, as $n$ approaches infinity, is $1$.
Is there a connection between this "$n$-th root of $2$ Theorem" and the limit I mentioned?
Can I consider that limit a corollary to the theorem?
limits elementary-number-theory
edited Nov 29 at 4:55
Jyrki Lahtonen
108k12166366
asked Nov 28 at 23:33
Xavier Stanton
330211
So I was looking at a theorem that popped up into my head:
The $n$-th root of $2$ always being irrational if $n$ is greater than $1$.
I also noticed that
The limit of the $n$-th root of $2$, as $n$ approaches infinity, is $1$.
Is there a connection between this "$n$-th root of $2$ Theorem" and the limit I mentioned?
Can I consider that limit a corollary to the theorem?
limits elementary-number-theory
limits elementary-number-theory
edited Nov 29 at 4:55
Jyrki Lahtonen
108k12166366
asked Nov 28 at 23:33
Xavier Stanton
330211
edited Nov 29 at 4:55
Jyrki Lahtonen
108k12166366
asked Nov 28 at 23:33
Xavier Stanton
330211
edited Nov 29 at 4:55
Jyrki Lahtonen
108k12166366
edited Nov 29 at 4:55
Jyrki Lahtonen
108k12166366
edited Nov 29 at 4:55
Jyrki Lahtonen
108k12166366
108k12166366
asked Nov 28 at 23:33
Xavier Stanton
330211
asked Nov 28 at 23:33
Xavier Stanton
330211
asked Nov 28 at 23:33
Xavier Stanton
330211
330211
3
Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
– T. Bongers
Nov 28 at 23:35
If $n$ is irrational then the nth root of $2$ can be rational.
– kingW3
Nov 28 at 23:40
Just so I know, how do I keep the revisions? It sounds stupid, but it wouldn't let me keep the revisions that were made.
– Xavier Stanton
Nov 29 at 0:18
Also, how do you make the formulas?
– Xavier Stanton
Nov 29 at 0:19
We use MathJax to typeset formulas. A tutorial is here. To see the edit history you can click on the "edited xxx ago" in the center bottom of the post.
– Ross Millikan
Nov 29 at 0:39
add a comment |
3
Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
– T. Bongers
Nov 28 at 23:35
If $n$ is irrational then the nth root of $2$ can be rational.
– kingW3
Nov 28 at 23:40
Just so I know, how do I keep the revisions? It sounds stupid, but it wouldn't let me keep the revisions that were made.
– Xavier Stanton
Nov 29 at 0:18
Also, how do you make the formulas?
– Xavier Stanton
Nov 29 at 0:19
We use MathJax to typeset formulas. A tutorial is here. To see the edit history you can click on the "edited xxx ago" in the center bottom of the post.
– Ross Millikan
Nov 29 at 0:39
3
3
Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
– T. Bongers
Nov 28 at 23:35
Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
– T. Bongers
Nov 28 at 23:35
If $n$ is irrational then the nth root of $2$ can be rational.
– kingW3
Nov 28 at 23:40
If $n$ is irrational then the nth root of $2$ can be rational.
– kingW3
Nov 28 at 23:40
Just so I know, how do I keep the revisions? It sounds stupid, but it wouldn't let me keep the revisions that were made.
– Xavier Stanton
Nov 29 at 0:18
Just so I know, how do I keep the revisions? It sounds stupid, but it wouldn't let me keep the revisions that were made.
– Xavier Stanton
Nov 29 at 0:18
Also, how do you make the formulas?
– Xavier Stanton
Nov 29 at 0:19
Also, how do you make the formulas?
– Xavier Stanton
Nov 29 at 0:19
We use MathJax to typeset formulas. A tutorial is here. To see the edit history you can click on the "edited xxx ago" in the center bottom of the post.
– Ross Millikan
Nov 29 at 0:39
We use MathJax to typeset formulas. A tutorial is here. To see the edit history you can click on the "edited xxx ago" in the center bottom of the post.
– Ross Millikan
Nov 29 at 0:39
add a comment |
3 Answers
3
active
oldest
votes
As far as I understand your question, no. The sequence
$$2sqrt{2},2root3of2,2root4of2,cdots$$
(multiplying every term of your nth-root-of-$2$ sequence by $2$) contains only irrational terms and tends to $2$, while the sequence
$$1+1,1+frac{1}{2},1+frac{1}{3},cdots$$
contains only rational terms and tends to $1$. The two properties (containing irrational terms and tending to 1) have nothing to do with one another.
answered Nov 28 at 23:37
Carl Schildkraut
11.1k11441
add a comment |
The two things are not related, for the limit indeed we have that
$$sqrt[n] 2=e^{frac{log 2}n}to 1$$
since $frac{log 2}nto 0$, and the result is not related to any hypotesis about the irrationality of $sqrt[n] 2$.
answered Nov 28 at 23:37
gimusi
1
add a comment |
The $n^{th}$ root of any positive number tends toward $1$ as $n to infty$. This is regardless of whether the roots are rational or irrational. For any given positive number there will only be a finite number of rational roots, but it can be arbitrarily large. Find some $k$ with many factors, like one of the highly composite numbers. Now form $1.01^k$. That will have as many rational roots as $k$ has factors.
answered Nov 28 at 23:41
Ross Millikan
291k23196370
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
As far as I understand your question, no. The sequence
$$2sqrt{2},2root3of2,2root4of2,cdots$$
(multiplying every term of your nth-root-of-$2$ sequence by $2$) contains only irrational terms and tends to $2$, while the sequence
$$1+1,1+frac{1}{2},1+frac{1}{3},cdots$$
contains only rational terms and tends to $1$. The two properties (containing irrational terms and tending to 1) have nothing to do with one another.
answered Nov 28 at 23:37
Carl Schildkraut
11.1k11441
add a comment |
As far as I understand your question, no. The sequence
$$2sqrt{2},2root3of2,2root4of2,cdots$$
(multiplying every term of your nth-root-of-$2$ sequence by $2$) contains only irrational terms and tends to $2$, while the sequence
$$1+1,1+frac{1}{2},1+frac{1}{3},cdots$$
contains only rational terms and tends to $1$. The two properties (containing irrational terms and tending to 1) have nothing to do with one another.
answered Nov 28 at 23:37
Carl Schildkraut
11.1k11441
add a comment |
As far as I understand your question, no. The sequence
$$2sqrt{2},2root3of2,2root4of2,cdots$$
(multiplying every term of your nth-root-of-$2$ sequence by $2$) contains only irrational terms and tends to $2$, while the sequence
$$1+1,1+frac{1}{2},1+frac{1}{3},cdots$$
contains only rational terms and tends to $1$. The two properties (containing irrational terms and tending to 1) have nothing to do with one another.
answered Nov 28 at 23:37
Carl Schildkraut
11.1k11441
As far as I understand your question, no. The sequence
$$2sqrt{2},2root3of2,2root4of2,cdots$$
(multiplying every term of your nth-root-of-$2$ sequence by $2$) contains only irrational terms and tends to $2$, while the sequence
$$1+1,1+frac{1}{2},1+frac{1}{3},cdots$$
contains only rational terms and tends to $1$. The two properties (containing irrational terms and tending to 1) have nothing to do with one another.
answered Nov 28 at 23:37
Carl Schildkraut
11.1k11441
answered Nov 28 at 23:37
Carl Schildkraut
11.1k11441
answered Nov 28 at 23:37
Carl Schildkraut
11.1k11441
answered Nov 28 at 23:37
Carl Schildkraut
11.1k11441
11.1k11441
add a comment |
add a comment |
The two things are not related, for the limit indeed we have that
$$sqrt[n] 2=e^{frac{log 2}n}to 1$$
since $frac{log 2}nto 0$, and the result is not related to any hypotesis about the irrationality of $sqrt[n] 2$.
answered Nov 28 at 23:37
gimusi
1
add a comment |
The two things are not related, for the limit indeed we have that
$$sqrt[n] 2=e^{frac{log 2}n}to 1$$
since $frac{log 2}nto 0$, and the result is not related to any hypotesis about the irrationality of $sqrt[n] 2$.
answered Nov 28 at 23:37
gimusi
1
add a comment |
The two things are not related, for the limit indeed we have that
$$sqrt[n] 2=e^{frac{log 2}n}to 1$$
since $frac{log 2}nto 0$, and the result is not related to any hypotesis about the irrationality of $sqrt[n] 2$.
answered Nov 28 at 23:37
gimusi
1
The two things are not related, for the limit indeed we have that
$$sqrt[n] 2=e^{frac{log 2}n}to 1$$
since $frac{log 2}nto 0$, and the result is not related to any hypotesis about the irrationality of $sqrt[n] 2$.
answered Nov 28 at 23:37
gimusi
1
answered Nov 28 at 23:37
gimusi
1
answered Nov 28 at 23:37
gimusi
1
answered Nov 28 at 23:37
gimusi
1
1
add a comment |
add a comment |
The $n^{th}$ root of any positive number tends toward $1$ as $n to infty$. This is regardless of whether the roots are rational or irrational. For any given positive number there will only be a finite number of rational roots, but it can be arbitrarily large. Find some $k$ with many factors, like one of the highly composite numbers. Now form $1.01^k$. That will have as many rational roots as $k$ has factors.
answered Nov 28 at 23:41
Ross Millikan
291k23196370
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
add a comment |
The $n^{th}$ root of any positive number tends toward $1$ as $n to infty$. This is regardless of whether the roots are rational or irrational. For any given positive number there will only be a finite number of rational roots, but it can be arbitrarily large. Find some $k$ with many factors, like one of the highly composite numbers. Now form $1.01^k$. That will have as many rational roots as $k$ has factors.
answered Nov 28 at 23:41
Ross Millikan
291k23196370
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
add a comment |
The $n^{th}$ root of any positive number tends toward $1$ as $n to infty$. This is regardless of whether the roots are rational or irrational. For any given positive number there will only be a finite number of rational roots, but it can be arbitrarily large. Find some $k$ with many factors, like one of the highly composite numbers. Now form $1.01^k$. That will have as many rational roots as $k$ has factors.
answered Nov 28 at 23:41
Ross Millikan
291k23196370
The $n^{th}$ root of any positive number tends toward $1$ as $n to infty$. This is regardless of whether the roots are rational or irrational. For any given positive number there will only be a finite number of rational roots, but it can be arbitrarily large. Find some $k$ with many factors, like one of the highly composite numbers. Now form $1.01^k$. That will have as many rational roots as $k$ has factors.
answered Nov 28 at 23:41
Ross Millikan
291k23196370
answered Nov 28 at 23:41
Ross Millikan
291k23196370
answered Nov 28 at 23:41
Ross Millikan
291k23196370
answered Nov 28 at 23:41
Ross Millikan
291k23196370
291k23196370
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
add a comment |
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
add a comment |
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3
Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
– T. Bongers
Nov 28 at 23:35
If $n$ is irrational then the nth root of $2$ can be rational.
– kingW3
Nov 28 at 23:40
Just so I know, how do I keep the revisions? It sounds stupid, but it wouldn't let me keep the revisions that were made.
– Xavier Stanton
Nov 29 at 0:18
Also, how do you make the formulas?
– Xavier Stanton
Nov 29 at 0:19
We use MathJax to typeset formulas. A tutorial is here. To see the edit history you can click on the "edited xxx ago" in the center bottom of the post.
– Ross Millikan
Nov 29 at 0:39