Show $(3 + sqrt{2})^{2/3}$ is irrational using RZT












0












$begingroup$


I am asked to prove that $(3+sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.



This is what I have so far:



$ x = (3+sqrt{2})^{2/3} $



$ x^3 = (3+sqrt{2})^{2} $



$ x^3 - 11 - 6sqrt{2} = 0 $



From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= frac{c}{d}$ $c,d, in mathbb{Z} $ would need to have $d= pm 1$ and $c$ divide $-11 -6sqrt{2}$.



So we have



$ -11 -6sqrt{2} = z c$, $z in mathbb{Z}$



$-6sqrt{2} = 11 +zc$



Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $sqrt{3}$ is irrational.






irrational-numbers rational-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
    $endgroup$
    – Will Jagy
    Nov 30 '18 at 23:08










  • $begingroup$
    Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
    $endgroup$
    – Daniel Schepler
    Nov 30 '18 at 23:09


















0












$begingroup$


I am asked to prove that $(3+sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.



This is what I have so far:



$ x = (3+sqrt{2})^{2/3} $



$ x^3 = (3+sqrt{2})^{2} $



$ x^3 - 11 - 6sqrt{2} = 0 $



From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= frac{c}{d}$ $c,d, in mathbb{Z} $ would need to have $d= pm 1$ and $c$ divide $-11 -6sqrt{2}$.



So we have



$ -11 -6sqrt{2} = z c$, $z in mathbb{Z}$



$-6sqrt{2} = 11 +zc$



Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $sqrt{3}$ is irrational.






irrational-numbers rational-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
    $endgroup$
    – Will Jagy
    Nov 30 '18 at 23:08










  • $begingroup$
    Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
    $endgroup$
    – Daniel Schepler
    Nov 30 '18 at 23:09
















0












0








0





$begingroup$


I am asked to prove that $(3+sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.



This is what I have so far:



$ x = (3+sqrt{2})^{2/3} $



$ x^3 = (3+sqrt{2})^{2} $



$ x^3 - 11 - 6sqrt{2} = 0 $



From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= frac{c}{d}$ $c,d, in mathbb{Z} $ would need to have $d= pm 1$ and $c$ divide $-11 -6sqrt{2}$.



So we have



$ -11 -6sqrt{2} = z c$, $z in mathbb{Z}$



$-6sqrt{2} = 11 +zc$



Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $sqrt{3}$ is irrational.






irrational-numbers rational-numbers







share|cite|improve this question











$endgroup$




I am asked to prove that $(3+sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.



This is what I have so far:



$ x = (3+sqrt{2})^{2/3} $



$ x^3 = (3+sqrt{2})^{2} $



$ x^3 - 11 - 6sqrt{2} = 0 $



From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= frac{c}{d}$ $c,d, in mathbb{Z} $ would need to have $d= pm 1$ and $c$ divide $-11 -6sqrt{2}$.



So we have



$ -11 -6sqrt{2} = z c$, $z in mathbb{Z}$



$-6sqrt{2} = 11 +zc$



Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $sqrt{3}$ is irrational.






irrational-numbers rational-numbers




irrational-numbers rational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 23:21







pmac

















asked Nov 30 '18 at 23:03









pmacpmac

6916




6916








  • 1




    $begingroup$
    $(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
    $endgroup$
    – Will Jagy
    Nov 30 '18 at 23:08










  • $begingroup$
    Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
    $endgroup$
    – Daniel Schepler
    Nov 30 '18 at 23:09
















  • 1




    $begingroup$
    $(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
    $endgroup$
    – Will Jagy
    Nov 30 '18 at 23:08










  • $begingroup$
    Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
    $endgroup$
    – Daniel Schepler
    Nov 30 '18 at 23:09










1




1




$begingroup$
$(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
$endgroup$
– Will Jagy
Nov 30 '18 at 23:08




$begingroup$
$(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
$endgroup$
– Will Jagy
Nov 30 '18 at 23:08












$begingroup$
Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
$endgroup$
– Daniel Schepler
Nov 30 '18 at 23:09






$begingroup$
Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
$endgroup$
– Daniel Schepler
Nov 30 '18 at 23:09












4 Answers
4






active

oldest

votes


















4












$begingroup$

From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    You don't need the RZT and you already did all the hard work: from



    $$x^3-11-6sqrt2=0$$



    we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Yes, but the questions asks to use RZT.
      $endgroup$
      – pmac
      Nov 30 '18 at 23:16










    • $begingroup$
      @pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
      $endgroup$
      – DonAntonio
      Nov 30 '18 at 23:16












    • $begingroup$
      But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
      $endgroup$
      – Travis
      Dec 1 '18 at 11:01










    • $begingroup$
      @Travis Yes and yes...so what? I don't understand what you're trying to tell...
      $endgroup$
      – DonAntonio
      Dec 1 '18 at 11:21



















    1












    $begingroup$

    Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.



    You got that it is a solution to



    $x^3 - (11 + 6sqrt 2) = 0$.



    So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$



    $=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.



    Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.



    ====



    D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.



    They both work and have the same result but in terms of ease in seeing and teaching... His is better.



    (Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
      $endgroup$
      – pmac
      Nov 30 '18 at 23:31



















    0












    $begingroup$

    As an alternative, we have that



    $$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$



    now suppose by contradiction that



    $$sqrt[3] x =yin mathbb{Q} iff y^3=x$$



    which is impossible.






    share|cite|improve this answer











    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.






          share|cite|improve this answer











          $endgroup$



          From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '18 at 23:09

























          answered Nov 30 '18 at 23:07









          José Carlos SantosJosé Carlos Santos

          153k22123226




          153k22123226























              1












              $begingroup$

              You don't need the RZT and you already did all the hard work: from



              $$x^3-11-6sqrt2=0$$



              we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Yes, but the questions asks to use RZT.
                $endgroup$
                – pmac
                Nov 30 '18 at 23:16










              • $begingroup$
                @pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
                $endgroup$
                – DonAntonio
                Nov 30 '18 at 23:16












              • $begingroup$
                But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
                $endgroup$
                – Travis
                Dec 1 '18 at 11:01










              • $begingroup$
                @Travis Yes and yes...so what? I don't understand what you're trying to tell...
                $endgroup$
                – DonAntonio
                Dec 1 '18 at 11:21
















              1












              $begingroup$

              You don't need the RZT and you already did all the hard work: from



              $$x^3-11-6sqrt2=0$$



              we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Yes, but the questions asks to use RZT.
                $endgroup$
                – pmac
                Nov 30 '18 at 23:16










              • $begingroup$
                @pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
                $endgroup$
                – DonAntonio
                Nov 30 '18 at 23:16












              • $begingroup$
                But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
                $endgroup$
                – Travis
                Dec 1 '18 at 11:01










              • $begingroup$
                @Travis Yes and yes...so what? I don't understand what you're trying to tell...
                $endgroup$
                – DonAntonio
                Dec 1 '18 at 11:21














              1












              1








              1





              $begingroup$

              You don't need the RZT and you already did all the hard work: from



              $$x^3-11-6sqrt2=0$$



              we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...






              share|cite|improve this answer









              $endgroup$



              You don't need the RZT and you already did all the hard work: from



              $$x^3-11-6sqrt2=0$$



              we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 30 '18 at 23:15









              DonAntonioDonAntonio

              177k1492225




              177k1492225








              • 1




                $begingroup$
                Yes, but the questions asks to use RZT.
                $endgroup$
                – pmac
                Nov 30 '18 at 23:16










              • $begingroup$
                @pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
                $endgroup$
                – DonAntonio
                Nov 30 '18 at 23:16












              • $begingroup$
                But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
                $endgroup$
                – Travis
                Dec 1 '18 at 11:01










              • $begingroup$
                @Travis Yes and yes...so what? I don't understand what you're trying to tell...
                $endgroup$
                – DonAntonio
                Dec 1 '18 at 11:21














              • 1




                $begingroup$
                Yes, but the questions asks to use RZT.
                $endgroup$
                – pmac
                Nov 30 '18 at 23:16










              • $begingroup$
                @pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
                $endgroup$
                – DonAntonio
                Nov 30 '18 at 23:16












              • $begingroup$
                But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
                $endgroup$
                – Travis
                Dec 1 '18 at 11:01










              • $begingroup$
                @Travis Yes and yes...so what? I don't understand what you're trying to tell...
                $endgroup$
                – DonAntonio
                Dec 1 '18 at 11:21








              1




              1




              $begingroup$
              Yes, but the questions asks to use RZT.
              $endgroup$
              – pmac
              Nov 30 '18 at 23:16




              $begingroup$
              Yes, but the questions asks to use RZT.
              $endgroup$
              – pmac
              Nov 30 '18 at 23:16












              $begingroup$
              @pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
              $endgroup$
              – DonAntonio
              Nov 30 '18 at 23:16






              $begingroup$
              @pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
              $endgroup$
              – DonAntonio
              Nov 30 '18 at 23:16














              $begingroup$
              But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
              $endgroup$
              – Travis
              Dec 1 '18 at 11:01




              $begingroup$
              But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
              $endgroup$
              – Travis
              Dec 1 '18 at 11:01












              $begingroup$
              @Travis Yes and yes...so what? I don't understand what you're trying to tell...
              $endgroup$
              – DonAntonio
              Dec 1 '18 at 11:21




              $begingroup$
              @Travis Yes and yes...so what? I don't understand what you're trying to tell...
              $endgroup$
              – DonAntonio
              Dec 1 '18 at 11:21











              1












              $begingroup$

              Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.



              You got that it is a solution to



              $x^3 - (11 + 6sqrt 2) = 0$.



              So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$



              $=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.



              Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.



              ====



              D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.



              They both work and have the same result but in terms of ease in seeing and teaching... His is better.



              (Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
                $endgroup$
                – pmac
                Nov 30 '18 at 23:31
















              1












              $begingroup$

              Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.



              You got that it is a solution to



              $x^3 - (11 + 6sqrt 2) = 0$.



              So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$



              $=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.



              Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.



              ====



              D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.



              They both work and have the same result but in terms of ease in seeing and teaching... His is better.



              (Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
                $endgroup$
                – pmac
                Nov 30 '18 at 23:31














              1












              1








              1





              $begingroup$

              Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.



              You got that it is a solution to



              $x^3 - (11 + 6sqrt 2) = 0$.



              So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$



              $=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.



              Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.



              ====



              D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.



              They both work and have the same result but in terms of ease in seeing and teaching... His is better.



              (Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)






              share|cite|improve this answer











              $endgroup$



              Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.



              You got that it is a solution to



              $x^3 - (11 + 6sqrt 2) = 0$.



              So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$



              $=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.



              Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.



              ====



              D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.



              They both work and have the same result but in terms of ease in seeing and teaching... His is better.



              (Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 30 '18 at 23:31

























              answered Nov 30 '18 at 23:25









              fleabloodfleablood

              68.7k22685




              68.7k22685












              • $begingroup$
                Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
                $endgroup$
                – pmac
                Nov 30 '18 at 23:31


















              • $begingroup$
                Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
                $endgroup$
                – pmac
                Nov 30 '18 at 23:31
















              $begingroup$
              Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
              $endgroup$
              – pmac
              Nov 30 '18 at 23:31




              $begingroup$
              Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
              $endgroup$
              – pmac
              Nov 30 '18 at 23:31











              0












              $begingroup$

              As an alternative, we have that



              $$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$



              now suppose by contradiction that



              $$sqrt[3] x =yin mathbb{Q} iff y^3=x$$



              which is impossible.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                As an alternative, we have that



                $$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$



                now suppose by contradiction that



                $$sqrt[3] x =yin mathbb{Q} iff y^3=x$$



                which is impossible.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  As an alternative, we have that



                  $$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$



                  now suppose by contradiction that



                  $$sqrt[3] x =yin mathbb{Q} iff y^3=x$$



                  which is impossible.






                  share|cite|improve this answer











                  $endgroup$



                  As an alternative, we have that



                  $$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$



                  now suppose by contradiction that



                  $$sqrt[3] x =yin mathbb{Q} iff y^3=x$$



                  which is impossible.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 23:31

























                  answered Nov 30 '18 at 23:08









                  gimusigimusi

                  1




                  1






























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