Find two irrational numbers $x,y$ such that $x+y$ and $xy$ are both rational.












3












$begingroup$


I know how to satisfy one of the statements but never both together.



$(a+b)*(a-b)=a^2-b^2$ so taking $a=sqrt k_1$ and $b=sqrt k_2$, $k_1,k_2inmathbb{Q}$ such that $a,b notinmathbb{Q}$ would suffice for the second one. Also, for every $xinmathbb{R} - mathbb{Q}$, taking $y=x^{-1}$ would also work for the product.



For the sum one could do something boring like given $xinmathbb{R}-mathbb{Q}$, take $-xinmathbb{R}-mathbb{Q}$ and $x-x=0inmathbb{R}$.



Ideally only "easy-to-define functions" (like square root, sum, subtraction...) should be used since stuff like $log,e,cos$... are not formally defined until later stages of my real analysis book (this is a question from the real numbers chapter).






real-analysis irrational-numbers rational-numbers







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I know how to satisfy one of the statements but never both together.



    $(a+b)*(a-b)=a^2-b^2$ so taking $a=sqrt k_1$ and $b=sqrt k_2$, $k_1,k_2inmathbb{Q}$ such that $a,b notinmathbb{Q}$ would suffice for the second one. Also, for every $xinmathbb{R} - mathbb{Q}$, taking $y=x^{-1}$ would also work for the product.



    For the sum one could do something boring like given $xinmathbb{R}-mathbb{Q}$, take $-xinmathbb{R}-mathbb{Q}$ and $x-x=0inmathbb{R}$.



    Ideally only "easy-to-define functions" (like square root, sum, subtraction...) should be used since stuff like $log,e,cos$... are not formally defined until later stages of my real analysis book (this is a question from the real numbers chapter).






    real-analysis irrational-numbers rational-numbers







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I know how to satisfy one of the statements but never both together.



      $(a+b)*(a-b)=a^2-b^2$ so taking $a=sqrt k_1$ and $b=sqrt k_2$, $k_1,k_2inmathbb{Q}$ such that $a,b notinmathbb{Q}$ would suffice for the second one. Also, for every $xinmathbb{R} - mathbb{Q}$, taking $y=x^{-1}$ would also work for the product.



      For the sum one could do something boring like given $xinmathbb{R}-mathbb{Q}$, take $-xinmathbb{R}-mathbb{Q}$ and $x-x=0inmathbb{R}$.



      Ideally only "easy-to-define functions" (like square root, sum, subtraction...) should be used since stuff like $log,e,cos$... are not formally defined until later stages of my real analysis book (this is a question from the real numbers chapter).






      real-analysis irrational-numbers rational-numbers







      share|cite|improve this question









      $endgroup$




      I know how to satisfy one of the statements but never both together.



      $(a+b)*(a-b)=a^2-b^2$ so taking $a=sqrt k_1$ and $b=sqrt k_2$, $k_1,k_2inmathbb{Q}$ such that $a,b notinmathbb{Q}$ would suffice for the second one. Also, for every $xinmathbb{R} - mathbb{Q}$, taking $y=x^{-1}$ would also work for the product.



      For the sum one could do something boring like given $xinmathbb{R}-mathbb{Q}$, take $-xinmathbb{R}-mathbb{Q}$ and $x-x=0inmathbb{R}$.



      Ideally only "easy-to-define functions" (like square root, sum, subtraction...) should be used since stuff like $log,e,cos$... are not formally defined until later stages of my real analysis book (this is a question from the real numbers chapter).






      real-analysis irrational-numbers rational-numbers




      real-analysis irrational-numbers rational-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 13 at 17:06









      BroccoliFinancialsBroccoliFinancials

      404




      404






















          5 Answers
          5






          active

          oldest

          votes


















          5












          $begingroup$

          What about $a=sqrt k$, $b=-sqrt k$?






          share|cite|improve this answer









          $endgroup$









          • 5




            $begingroup$
            where $k$ is not a perfect square
            $endgroup$
            – J. W. Tanner
            Jan 13 at 17:08



















          2












          $begingroup$

          $-sqrt a+b, sqrt a+b$ are examples






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            where $a$ is not a perfect square
            $endgroup$
            – J. W. Tanner
            Jan 13 at 18:34



















          2












          $begingroup$

          It is not difficult to show what is the general solution here.



          If $a+b=2r$ and $ab=q$ we find (eg by substitution, or vieta) that $a$ and $b$ are the roots of the quadratic equation $$x^2-2rx+q=0$$ which has the roots $$x=rpm sqrt{r^2-q}$$giving the two values of $a$ and $b$. If $r^2-q$ is not a rational square then the values of $a$ and $b$ will be irrational.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            $$x=text{Golden Ratio}={1+sqrt 5over 2}\y={1-sqrt 5over 2}$$More generally, take the roots of $x^2+ax+b=0$ when $a,b$ are rational and the equation has two irrational roots. For example, my $x,y$ are roots of $$x^2-x-1=0$$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Take any polynomial $$f(x)=x^2+ax+1$$ where $ainBbb Z$ and $a^2>4$. Let $alphainBbb R$ such that $f(alpha)=0$. By the rational root test $alpha=1$, but this is impossible since $f(1)=a+2neq 0$. As the roots comes in pairs there is another root $baralpha$, and $alphabaralpha=1$, $-alpha-baralpha=a$.






              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









                5












                $begingroup$

                What about $a=sqrt k$, $b=-sqrt k$?






                share|cite|improve this answer









                $endgroup$









                • 5




                  $begingroup$
                  where $k$ is not a perfect square
                  $endgroup$
                  – J. W. Tanner
                  Jan 13 at 17:08
















                5












                $begingroup$

                What about $a=sqrt k$, $b=-sqrt k$?






                share|cite|improve this answer









                $endgroup$









                • 5




                  $begingroup$
                  where $k$ is not a perfect square
                  $endgroup$
                  – J. W. Tanner
                  Jan 13 at 17:08














                5












                5








                5





                $begingroup$

                What about $a=sqrt k$, $b=-sqrt k$?






                share|cite|improve this answer









                $endgroup$



                What about $a=sqrt k$, $b=-sqrt k$?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 13 at 17:07









                Matt SamuelMatt Samuel

                38.8k63769




                38.8k63769








                • 5




                  $begingroup$
                  where $k$ is not a perfect square
                  $endgroup$
                  – J. W. Tanner
                  Jan 13 at 17:08














                • 5




                  $begingroup$
                  where $k$ is not a perfect square
                  $endgroup$
                  – J. W. Tanner
                  Jan 13 at 17:08








                5




                5




                $begingroup$
                where $k$ is not a perfect square
                $endgroup$
                – J. W. Tanner
                Jan 13 at 17:08




                $begingroup$
                where $k$ is not a perfect square
                $endgroup$
                – J. W. Tanner
                Jan 13 at 17:08











                2












                $begingroup$

                $-sqrt a+b, sqrt a+b$ are examples






                share|cite|improve this answer









                $endgroup$









                • 1




                  $begingroup$
                  where $a$ is not a perfect square
                  $endgroup$
                  – J. W. Tanner
                  Jan 13 at 18:34
















                2












                $begingroup$

                $-sqrt a+b, sqrt a+b$ are examples






                share|cite|improve this answer









                $endgroup$









                • 1




                  $begingroup$
                  where $a$ is not a perfect square
                  $endgroup$
                  – J. W. Tanner
                  Jan 13 at 18:34














                2












                2








                2





                $begingroup$

                $-sqrt a+b, sqrt a+b$ are examples






                share|cite|improve this answer









                $endgroup$



                $-sqrt a+b, sqrt a+b$ are examples







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 13 at 17:08









                Tsemo AristideTsemo Aristide

                59.6k11446




                59.6k11446








                • 1




                  $begingroup$
                  where $a$ is not a perfect square
                  $endgroup$
                  – J. W. Tanner
                  Jan 13 at 18:34














                • 1




                  $begingroup$
                  where $a$ is not a perfect square
                  $endgroup$
                  – J. W. Tanner
                  Jan 13 at 18:34








                1




                1




                $begingroup$
                where $a$ is not a perfect square
                $endgroup$
                – J. W. Tanner
                Jan 13 at 18:34




                $begingroup$
                where $a$ is not a perfect square
                $endgroup$
                – J. W. Tanner
                Jan 13 at 18:34











                2












                $begingroup$

                It is not difficult to show what is the general solution here.



                If $a+b=2r$ and $ab=q$ we find (eg by substitution, or vieta) that $a$ and $b$ are the roots of the quadratic equation $$x^2-2rx+q=0$$ which has the roots $$x=rpm sqrt{r^2-q}$$giving the two values of $a$ and $b$. If $r^2-q$ is not a rational square then the values of $a$ and $b$ will be irrational.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  It is not difficult to show what is the general solution here.



                  If $a+b=2r$ and $ab=q$ we find (eg by substitution, or vieta) that $a$ and $b$ are the roots of the quadratic equation $$x^2-2rx+q=0$$ which has the roots $$x=rpm sqrt{r^2-q}$$giving the two values of $a$ and $b$. If $r^2-q$ is not a rational square then the values of $a$ and $b$ will be irrational.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    It is not difficult to show what is the general solution here.



                    If $a+b=2r$ and $ab=q$ we find (eg by substitution, or vieta) that $a$ and $b$ are the roots of the quadratic equation $$x^2-2rx+q=0$$ which has the roots $$x=rpm sqrt{r^2-q}$$giving the two values of $a$ and $b$. If $r^2-q$ is not a rational square then the values of $a$ and $b$ will be irrational.






                    share|cite|improve this answer









                    $endgroup$



                    It is not difficult to show what is the general solution here.



                    If $a+b=2r$ and $ab=q$ we find (eg by substitution, or vieta) that $a$ and $b$ are the roots of the quadratic equation $$x^2-2rx+q=0$$ which has the roots $$x=rpm sqrt{r^2-q}$$giving the two values of $a$ and $b$. If $r^2-q$ is not a rational square then the values of $a$ and $b$ will be irrational.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 13 at 17:12









                    Mark BennetMark Bennet

                    81.6k984181




                    81.6k984181























                        1












                        $begingroup$

                        $$x=text{Golden Ratio}={1+sqrt 5over 2}\y={1-sqrt 5over 2}$$More generally, take the roots of $x^2+ax+b=0$ when $a,b$ are rational and the equation has two irrational roots. For example, my $x,y$ are roots of $$x^2-x-1=0$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          $$x=text{Golden Ratio}={1+sqrt 5over 2}\y={1-sqrt 5over 2}$$More generally, take the roots of $x^2+ax+b=0$ when $a,b$ are rational and the equation has two irrational roots. For example, my $x,y$ are roots of $$x^2-x-1=0$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            $$x=text{Golden Ratio}={1+sqrt 5over 2}\y={1-sqrt 5over 2}$$More generally, take the roots of $x^2+ax+b=0$ when $a,b$ are rational and the equation has two irrational roots. For example, my $x,y$ are roots of $$x^2-x-1=0$$






                            share|cite|improve this answer









                            $endgroup$



                            $$x=text{Golden Ratio}={1+sqrt 5over 2}\y={1-sqrt 5over 2}$$More generally, take the roots of $x^2+ax+b=0$ when $a,b$ are rational and the equation has two irrational roots. For example, my $x,y$ are roots of $$x^2-x-1=0$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 13 at 17:08









                            Mostafa AyazMostafa Ayaz

                            16.6k3939




                            16.6k3939























                                0












                                $begingroup$

                                Take any polynomial $$f(x)=x^2+ax+1$$ where $ainBbb Z$ and $a^2>4$. Let $alphainBbb R$ such that $f(alpha)=0$. By the rational root test $alpha=1$, but this is impossible since $f(1)=a+2neq 0$. As the roots comes in pairs there is another root $baralpha$, and $alphabaralpha=1$, $-alpha-baralpha=a$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Take any polynomial $$f(x)=x^2+ax+1$$ where $ainBbb Z$ and $a^2>4$. Let $alphainBbb R$ such that $f(alpha)=0$. By the rational root test $alpha=1$, but this is impossible since $f(1)=a+2neq 0$. As the roots comes in pairs there is another root $baralpha$, and $alphabaralpha=1$, $-alpha-baralpha=a$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Take any polynomial $$f(x)=x^2+ax+1$$ where $ainBbb Z$ and $a^2>4$. Let $alphainBbb R$ such that $f(alpha)=0$. By the rational root test $alpha=1$, but this is impossible since $f(1)=a+2neq 0$. As the roots comes in pairs there is another root $baralpha$, and $alphabaralpha=1$, $-alpha-baralpha=a$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Take any polynomial $$f(x)=x^2+ax+1$$ where $ainBbb Z$ and $a^2>4$. Let $alphainBbb R$ such that $f(alpha)=0$. By the rational root test $alpha=1$, but this is impossible since $f(1)=a+2neq 0$. As the roots comes in pairs there is another root $baralpha$, and $alphabaralpha=1$, $-alpha-baralpha=a$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 13 at 17:21









                                    cansomeonehelpmeoutcansomeonehelpmeout

                                    7,1473935




                                    7,1473935






























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