Proving $sqrt3 + sqrt[3]{2}$ to be irrational












2












$begingroup$


In a test I tried to solve recently I came across the following question:




Prove $$sqrt3 + sqrt[3]{2}$$ is irrational




I tried proving it by saying it is equal to some rational number $$sqrt3 + sqrt[3]{2} = frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?






number-theory irrational-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
    $endgroup$
    – Watson
    Jan 29 at 9:08
















2












$begingroup$


In a test I tried to solve recently I came across the following question:




Prove $$sqrt3 + sqrt[3]{2}$$ is irrational




I tried proving it by saying it is equal to some rational number $$sqrt3 + sqrt[3]{2} = frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?






number-theory irrational-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
    $endgroup$
    – Watson
    Jan 29 at 9:08














2












2








2


1



$begingroup$


In a test I tried to solve recently I came across the following question:




Prove $$sqrt3 + sqrt[3]{2}$$ is irrational




I tried proving it by saying it is equal to some rational number $$sqrt3 + sqrt[3]{2} = frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?






number-theory irrational-numbers







share|cite|improve this question











$endgroup$




In a test I tried to solve recently I came across the following question:




Prove $$sqrt3 + sqrt[3]{2}$$ is irrational




I tried proving it by saying it is equal to some rational number $$sqrt3 + sqrt[3]{2} = frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?






number-theory irrational-numbers




number-theory irrational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 8:52









Asaf Karagila♦

307k33439770




307k33439770










asked Jan 28 at 7:38









Guysudai1Guysudai1

18011




18011








  • 1




    $begingroup$
    Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
    $endgroup$
    – Watson
    Jan 29 at 9:08














  • 1




    $begingroup$
    Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
    $endgroup$
    – Watson
    Jan 29 at 9:08








1




1




$begingroup$
Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
$endgroup$
– Watson
Jan 29 at 9:08




$begingroup$
Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
$endgroup$
– Watson
Jan 29 at 9:08










5 Answers
5






active

oldest

votes


















7












$begingroup$

It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
You can then move the $sqrt 3$ and get:



$sqrt[3] 2=frac{m}{n}-sqrt 3$



$2=(frac{m}{n}-sqrt 3)^3$



After expanding the term at the RHS, the two lemmas you proved might come in handy.






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
    $$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
    $$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
    $$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
    By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.



    This is not the easy way, of course.






    share|cite|improve this answer











    $endgroup$





















      4












      $begingroup$

      Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,



      $sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$



      then



      $sqrt[3]2 = r - sqrt 3; tag 2$



      we cube:



      $2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$



      or



      $2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$



      or



      $2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$



      or



      $sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$



      which contradicts the fact that



      $sqrt 3 notin Bbb Q; tag 7$



      therefore, (1) is false. $OEDelta$.






      share|cite|improve this answer









      $endgroup$





















        3












        $begingroup$

        Let $sqrt3+sqrt[3]2=rinmathbb Q$.



        Thus, $$2=(r-sqrt3)^3$$ or
        $$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
        $$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
        which is a contradiction.






        share|cite|improve this answer









        $endgroup$





















          2












          $begingroup$

          Squaring is simpler than cubing. ;-)



          Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
          $$
          (u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
          $$
          If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.



          Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.






          share|cite|improve this answer









          $endgroup$













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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7












            $begingroup$

            It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
            You can then move the $sqrt 3$ and get:



            $sqrt[3] 2=frac{m}{n}-sqrt 3$



            $2=(frac{m}{n}-sqrt 3)^3$



            After expanding the term at the RHS, the two lemmas you proved might come in handy.






            share|cite|improve this answer









            $endgroup$


















              7












              $begingroup$

              It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
              You can then move the $sqrt 3$ and get:



              $sqrt[3] 2=frac{m}{n}-sqrt 3$



              $2=(frac{m}{n}-sqrt 3)^3$



              After expanding the term at the RHS, the two lemmas you proved might come in handy.






              share|cite|improve this answer









              $endgroup$
















                7












                7








                7





                $begingroup$

                It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
                You can then move the $sqrt 3$ and get:



                $sqrt[3] 2=frac{m}{n}-sqrt 3$



                $2=(frac{m}{n}-sqrt 3)^3$



                After expanding the term at the RHS, the two lemmas you proved might come in handy.






                share|cite|improve this answer









                $endgroup$



                It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
                You can then move the $sqrt 3$ and get:



                $sqrt[3] 2=frac{m}{n}-sqrt 3$



                $2=(frac{m}{n}-sqrt 3)^3$



                After expanding the term at the RHS, the two lemmas you proved might come in handy.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 28 at 7:49









                GSoferGSofer

                781313




                781313























                    6












                    $begingroup$

                    It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
                    $$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
                    $$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
                    $$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
                    By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.



                    This is not the easy way, of course.






                    share|cite|improve this answer











                    $endgroup$


















                      6












                      $begingroup$

                      It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
                      $$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
                      $$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
                      $$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
                      By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.



                      This is not the easy way, of course.






                      share|cite|improve this answer











                      $endgroup$
















                        6












                        6








                        6





                        $begingroup$

                        It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
                        $$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
                        $$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
                        $$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
                        By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.



                        This is not the easy way, of course.






                        share|cite|improve this answer











                        $endgroup$



                        It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
                        $$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
                        $$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
                        $$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
                        By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.



                        This is not the easy way, of course.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Jan 28 at 8:18

























                        answered Jan 28 at 8:07









                        jmerryjmerry

                        16.5k11633




                        16.5k11633























                            4












                            $begingroup$

                            Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,



                            $sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$



                            then



                            $sqrt[3]2 = r - sqrt 3; tag 2$



                            we cube:



                            $2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$



                            or



                            $2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$



                            or



                            $2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$



                            or



                            $sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$



                            which contradicts the fact that



                            $sqrt 3 notin Bbb Q; tag 7$



                            therefore, (1) is false. $OEDelta$.






                            share|cite|improve this answer









                            $endgroup$


















                              4












                              $begingroup$

                              Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,



                              $sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$



                              then



                              $sqrt[3]2 = r - sqrt 3; tag 2$



                              we cube:



                              $2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$



                              or



                              $2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$



                              or



                              $2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$



                              or



                              $sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$



                              which contradicts the fact that



                              $sqrt 3 notin Bbb Q; tag 7$



                              therefore, (1) is false. $OEDelta$.






                              share|cite|improve this answer









                              $endgroup$
















                                4












                                4








                                4





                                $begingroup$

                                Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,



                                $sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$



                                then



                                $sqrt[3]2 = r - sqrt 3; tag 2$



                                we cube:



                                $2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$



                                or



                                $2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$



                                or



                                $2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$



                                or



                                $sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$



                                which contradicts the fact that



                                $sqrt 3 notin Bbb Q; tag 7$



                                therefore, (1) is false. $OEDelta$.






                                share|cite|improve this answer









                                $endgroup$



                                Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,



                                $sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$



                                then



                                $sqrt[3]2 = r - sqrt 3; tag 2$



                                we cube:



                                $2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$



                                or



                                $2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$



                                or



                                $2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$



                                or



                                $sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$



                                which contradicts the fact that



                                $sqrt 3 notin Bbb Q; tag 7$



                                therefore, (1) is false. $OEDelta$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 28 at 8:03









                                Robert LewisRobert Lewis

                                48.4k23167




                                48.4k23167























                                    3












                                    $begingroup$

                                    Let $sqrt3+sqrt[3]2=rinmathbb Q$.



                                    Thus, $$2=(r-sqrt3)^3$$ or
                                    $$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
                                    $$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
                                    which is a contradiction.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      3












                                      $begingroup$

                                      Let $sqrt3+sqrt[3]2=rinmathbb Q$.



                                      Thus, $$2=(r-sqrt3)^3$$ or
                                      $$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
                                      $$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
                                      which is a contradiction.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        3












                                        3








                                        3





                                        $begingroup$

                                        Let $sqrt3+sqrt[3]2=rinmathbb Q$.



                                        Thus, $$2=(r-sqrt3)^3$$ or
                                        $$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
                                        $$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
                                        which is a contradiction.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Let $sqrt3+sqrt[3]2=rinmathbb Q$.



                                        Thus, $$2=(r-sqrt3)^3$$ or
                                        $$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
                                        $$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
                                        which is a contradiction.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Jan 28 at 8:06









                                        Michael RozenbergMichael Rozenberg

                                        109k1896201




                                        109k1896201























                                            2












                                            $begingroup$

                                            Squaring is simpler than cubing. ;-)



                                            Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
                                            $$
                                            (u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
                                            $$
                                            If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.



                                            Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              2












                                              $begingroup$

                                              Squaring is simpler than cubing. ;-)



                                              Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
                                              $$
                                              (u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
                                              $$
                                              If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.



                                              Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                2












                                                2








                                                2





                                                $begingroup$

                                                Squaring is simpler than cubing. ;-)



                                                Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
                                                $$
                                                (u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
                                                $$
                                                If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.



                                                Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.






                                                share|cite|improve this answer









                                                $endgroup$



                                                Squaring is simpler than cubing. ;-)



                                                Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
                                                $$
                                                (u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
                                                $$
                                                If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.



                                                Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Jan 28 at 10:14









                                                egregegreg

                                                185k1486206




                                                185k1486206






























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