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









          $endgroup$













            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
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090585%2fproving-sqrt3-sqrt32-to-be-irrational%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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






























                                                    draft saved

                                                    draft discarded




















































                                                    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.




                                                    draft saved


                                                    draft discarded














                                                    StackExchange.ready(
                                                    function () {
                                                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090585%2fproving-sqrt3-sqrt32-to-be-irrational%23new-answer', 'question_page');
                                                    }
                                                    );

                                                    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







                                                    Popular posts from this blog

                                                    android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                                                    Image
                                                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d
                                                    Read more

                                                    SQL update select statement

                                                    Image
                                                    0 I'm trying to update an entry in my database (postgresql). I'm having a problem with updating the query: INSERT INTO prices (price, product_id) SELECT product_price, commn_id FROM products_temp as prod WHERE NOT EXISTS (SELECT * FROM prices od WHERE prod.commn_id = od.product_id) OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) Query works fine when i delete: OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) So it seems to me that it's looping through this operation. My question is, how can I fix it? sql postgresql
                                                    Read more

                                                    WPF add header to Image with URL pettitions [duplicate]

                                                    Image
                                                    up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:
                                                    Read more