The $n$-th root of $2$ is irrational for $n>1$, and $lim_{ntoinfty}sqrt[n]{2}=1$. Is the latter fact a...












-1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
    $endgroup$
    – T. Bongers
    Nov 28 '18 at 23:35










  • $begingroup$
    If $n$ is irrational then the nth root of $2$ can be rational.
    $endgroup$
    – kingW3
    Nov 28 '18 at 23:40












  • $begingroup$
    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.
    $endgroup$
    – Xavier Stanton
    Nov 29 '18 at 0:18










  • $begingroup$
    Also, how do you make the formulas?
    $endgroup$
    – Xavier Stanton
    Nov 29 '18 at 0:19










  • $begingroup$
    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.
    $endgroup$
    – Ross Millikan
    Nov 29 '18 at 0:39
















-1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
    $endgroup$
    – T. Bongers
    Nov 28 '18 at 23:35










  • $begingroup$
    If $n$ is irrational then the nth root of $2$ can be rational.
    $endgroup$
    – kingW3
    Nov 28 '18 at 23:40












  • $begingroup$
    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.
    $endgroup$
    – Xavier Stanton
    Nov 29 '18 at 0:18










  • $begingroup$
    Also, how do you make the formulas?
    $endgroup$
    – Xavier Stanton
    Nov 29 '18 at 0:19










  • $begingroup$
    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.
    $endgroup$
    – Ross Millikan
    Nov 29 '18 at 0:39














-1












-1








-1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 4:55









Jyrki Lahtonen

109k13169372




109k13169372










asked Nov 28 '18 at 23:33









Xavier StantonXavier Stanton

311211




311211








  • 3




    $begingroup$
    Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
    $endgroup$
    – T. Bongers
    Nov 28 '18 at 23:35










  • $begingroup$
    If $n$ is irrational then the nth root of $2$ can be rational.
    $endgroup$
    – kingW3
    Nov 28 '18 at 23:40












  • $begingroup$
    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.
    $endgroup$
    – Xavier Stanton
    Nov 29 '18 at 0:18










  • $begingroup$
    Also, how do you make the formulas?
    $endgroup$
    – Xavier Stanton
    Nov 29 '18 at 0:19










  • $begingroup$
    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.
    $endgroup$
    – Ross Millikan
    Nov 29 '18 at 0:39














  • 3




    $begingroup$
    Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
    $endgroup$
    – T. Bongers
    Nov 28 '18 at 23:35










  • $begingroup$
    If $n$ is irrational then the nth root of $2$ can be rational.
    $endgroup$
    – kingW3
    Nov 28 '18 at 23:40












  • $begingroup$
    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.
    $endgroup$
    – Xavier Stanton
    Nov 29 '18 at 0:18










  • $begingroup$
    Also, how do you make the formulas?
    $endgroup$
    – Xavier Stanton
    Nov 29 '18 at 0:19










  • $begingroup$
    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.
    $endgroup$
    – Ross Millikan
    Nov 29 '18 at 0:39








3




3




$begingroup$
Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
$endgroup$
– T. Bongers
Nov 28 '18 at 23:35




$begingroup$
Not really, no. Irrationality doesn't tend to interact well with limits; perhaps there' something more specific you have in mind.
$endgroup$
– T. Bongers
Nov 28 '18 at 23:35












$begingroup$
If $n$ is irrational then the nth root of $2$ can be rational.
$endgroup$
– kingW3
Nov 28 '18 at 23:40






$begingroup$
If $n$ is irrational then the nth root of $2$ can be rational.
$endgroup$
– kingW3
Nov 28 '18 at 23:40














$begingroup$
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.
$endgroup$
– Xavier Stanton
Nov 29 '18 at 0:18




$begingroup$
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.
$endgroup$
– Xavier Stanton
Nov 29 '18 at 0:18












$begingroup$
Also, how do you make the formulas?
$endgroup$
– Xavier Stanton
Nov 29 '18 at 0:19




$begingroup$
Also, how do you make the formulas?
$endgroup$
– Xavier Stanton
Nov 29 '18 at 0:19












$begingroup$
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.
$endgroup$
– Ross Millikan
Nov 29 '18 at 0:39




$begingroup$
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.
$endgroup$
– Ross Millikan
Nov 29 '18 at 0:39










3 Answers
3






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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 $sqrt[n] 2$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks, that's interesting.
        $endgroup$
        – Xavier Stanton
        Nov 28 '18 at 23:47











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '18 at 23:37









          Carl SchildkrautCarl Schildkraut

          11.3k11441




          11.3k11441























              0












              $begingroup$

              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 $sqrt[n] 2$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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 $sqrt[n] 2$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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 $sqrt[n] 2$.






                  share|cite|improve this answer









                  $endgroup$



                  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 $sqrt[n] 2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 '18 at 23:37









                  gimusigimusi

                  92.8k84494




                  92.8k84494























                      0












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Thanks, that's interesting.
                        $endgroup$
                        – Xavier Stanton
                        Nov 28 '18 at 23:47
















                      0












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Thanks, that's interesting.
                        $endgroup$
                        – Xavier Stanton
                        Nov 28 '18 at 23:47














                      0












                      0








                      0





                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$



                      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.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 28 '18 at 23:41









                      Ross MillikanRoss Millikan

                      295k23198371




                      295k23198371












                      • $begingroup$
                        Thanks, that's interesting.
                        $endgroup$
                        – Xavier Stanton
                        Nov 28 '18 at 23:47


















                      • $begingroup$
                        Thanks, that's interesting.
                        $endgroup$
                        – Xavier Stanton
                        Nov 28 '18 at 23:47
















                      $begingroup$
                      Thanks, that's interesting.
                      $endgroup$
                      – Xavier Stanton
                      Nov 28 '18 at 23:47




                      $begingroup$
                      Thanks, that's interesting.
                      $endgroup$
                      – Xavier Stanton
                      Nov 28 '18 at 23:47


















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