The $n$-th root of $2$ is irrational for $n>1$, and $lim_{ntoinfty}sqrt[n]{2}=1$. Is the latter fact a...
up vote
-1
down vote
favorite
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
107k12166365
asked Nov 28 at 23:33
Xavier Stanton
330211
add a comment |
up vote
-1
down vote
favorite
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
107k12166365
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 |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
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
107k12166365
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
107k12166365
asked Nov 28 at 23:33
Xavier Stanton
330211
edited Nov 29 at 4:55
Jyrki Lahtonen
107k12166365
asked Nov 28 at 23:33
Xavier Stanton
330211
edited Nov 29 at 4:55
Jyrki Lahtonen
107k12166365
edited Nov 29 at 4:55
Jyrki Lahtonen
107k12166365
edited Nov 29 at 4:55
Jyrki Lahtonen
107k12166365
107k12166365
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
up vote
1
down vote
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 |
up vote
0
down vote
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 |
up vote
0
down vote
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
290k23196369
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017910%2fthe-n-th-root-of-2-is-irrational-for-n1-and-lim-n-to-infty-sqrtn2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
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 |
up vote
1
down vote
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 |
up vote
1
down vote
up vote
1
down vote
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 |
up vote
0
down vote
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 |
up vote
0
down vote
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 |
up vote
0
down vote
up vote
0
down vote
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 |
up vote
0
down vote
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
290k23196369
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
add a comment |
up vote
0
down vote
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
290k23196369
Thanks, that's interesting.
– Xavier Stanton
Nov 28 at 23:47
add a comment |
up vote
0
down vote
up vote
0
down vote
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
290k23196369
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
290k23196369
answered Nov 28 at 23:41
Ross Millikan
290k23196369
answered Nov 28 at 23:41
Ross Millikan
290k23196369
answered Nov 28 at 23:41
Ross Millikan
290k23196369
290k23196369
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017910%2fthe-n-th-root-of-2-is-irrational-for-n1-and-lim-n-to-infty-sqrtn2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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