Find integer solutions of $xy=-1$ (using only elementary ring theory).












1












$begingroup$


I know the answer is obvious: In $mathbb{Z}$ the only solutions of $xy=-1$ are $x=-y=1$ and $x=-y=-1$.
My problem is that I want to formally prove it and I don't know how to write it. Where do you even begin for such a trivial statement?



Edit: I would like to prove it viewing $mathbb{Z}$ as a ring. This is, just using the sum and product of integers. No order, no absolute value, etc...






ring-theory







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I know the answer is obvious: In $mathbb{Z}$ the only solutions of $xy=-1$ are $x=-y=1$ and $x=-y=-1$.
    My problem is that I want to formally prove it and I don't know how to write it. Where do you even begin for such a trivial statement?



    Edit: I would like to prove it viewing $mathbb{Z}$ as a ring. This is, just using the sum and product of integers. No order, no absolute value, etc...






    ring-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I know the answer is obvious: In $mathbb{Z}$ the only solutions of $xy=-1$ are $x=-y=1$ and $x=-y=-1$.
      My problem is that I want to formally prove it and I don't know how to write it. Where do you even begin for such a trivial statement?



      Edit: I would like to prove it viewing $mathbb{Z}$ as a ring. This is, just using the sum and product of integers. No order, no absolute value, etc...






      ring-theory







      share|cite|improve this question











      $endgroup$




      I know the answer is obvious: In $mathbb{Z}$ the only solutions of $xy=-1$ are $x=-y=1$ and $x=-y=-1$.
      My problem is that I want to formally prove it and I don't know how to write it. Where do you even begin for such a trivial statement?



      Edit: I would like to prove it viewing $mathbb{Z}$ as a ring. This is, just using the sum and product of integers. No order, no absolute value, etc...






      ring-theory




      ring-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 '18 at 4:28







      Cabex

















      asked Nov 25 '18 at 3:55









      CabexCabex

      62




      62






















          5 Answers
          5






          active

          oldest

          votes


















          3












          $begingroup$

          if $xy=-1$, then we we have $|x||y|=1$, that is we must have $|x|=1$ and $|y|=1$.



          Also, determinining $x$ would completely determine $y$.



          Hence we only need to examine what happends when $x=1$ and $x=-1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
            $endgroup$
            – Cabex
            Nov 25 '18 at 4:04










          • $begingroup$
            I think Z is defined so that 1 and -1 are the only units by definition. I think.
            $endgroup$
            – fleablood
            Nov 25 '18 at 4:21





















          2












          $begingroup$

          If xy = -1, then x = $frac{-1}{y}$. If x is an integer, y must divide -1. Therefore, y = $pm$1, which implies x $mp$1.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
            $endgroup$
            – Cabex
            Nov 25 '18 at 4:03






          • 1




            $begingroup$
            @Cabex This nearly does what you want, it can be rephrased to say that the only units in $mathbb{Z}$ are $pm 1$.
            $endgroup$
            – Michael Burr
            Nov 25 '18 at 4:06










          • $begingroup$
            @MichaelBurr I don't get it, we can't divide in $mathbb{Z}$. Isn't assuming $y$ has an inverse, or viewing the equation outside the integers?
            $endgroup$
            – Cabex
            Nov 25 '18 at 4:10










          • $begingroup$
            (xy)$^2$ = x$^2$y$^2$ = 1. Since x, y $in mathbb{Z}$, this implies that x$^2$ = y$^2$ = 1. So, x is it's own inverse. Thus x = $pm$ 1.
            $endgroup$
            – Joel Pereira
            Nov 25 '18 at 4:10










          • $begingroup$
            Right! Using idempotents... thanks a lot!
            $endgroup$
            – Cabex
            Nov 25 '18 at 4:23



















          0












          $begingroup$

          For positive integers (same can be done for negative, or just use absolute value):



          If $x >1$ and $y>1$, then $xy>1$. So either $x =1$ or $y=1$. If $x=1$, then $1 cdot y=1$, so $y=1$. If $y=1$, then $x=1$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Alternatively.



            If $x $ or $y $ is $0$, $xy=0$.



            Otherwise, $|x|ge 1$ and $|y|ge 1$. If $|x|>1$ then $|xy|=|y||x|>|y|ge 1$. So we must have $|x|=1$. And $1=|xy|=|x||y|=|y|$ so $|x|=|y|=1$.



            Of the four options only $x=y=1;x=y=-1$ work.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Oops. Sorry. I see someone else had the same idea. I'll delete when I get to an application that let's me.
              $endgroup$
              – fleablood
              Nov 25 '18 at 4:19



















            0












            $begingroup$

            Suppose that $xy=-1$, then $-xy=1$, so $x$ and $y$ are units (by definition). Since the only units in $mathbb{Z}$ are $pm 1$, we can check which combinations result in a product of $-1$.






            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









              3












              $begingroup$

              if $xy=-1$, then we we have $|x||y|=1$, that is we must have $|x|=1$ and $|y|=1$.



              Also, determinining $x$ would completely determine $y$.



              Hence we only need to examine what happends when $x=1$ and $x=-1$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:04










              • $begingroup$
                I think Z is defined so that 1 and -1 are the only units by definition. I think.
                $endgroup$
                – fleablood
                Nov 25 '18 at 4:21


















              3












              $begingroup$

              if $xy=-1$, then we we have $|x||y|=1$, that is we must have $|x|=1$ and $|y|=1$.



              Also, determinining $x$ would completely determine $y$.



              Hence we only need to examine what happends when $x=1$ and $x=-1$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:04










              • $begingroup$
                I think Z is defined so that 1 and -1 are the only units by definition. I think.
                $endgroup$
                – fleablood
                Nov 25 '18 at 4:21
















              3












              3








              3





              $begingroup$

              if $xy=-1$, then we we have $|x||y|=1$, that is we must have $|x|=1$ and $|y|=1$.



              Also, determinining $x$ would completely determine $y$.



              Hence we only need to examine what happends when $x=1$ and $x=-1$.






              share|cite|improve this answer









              $endgroup$



              if $xy=-1$, then we we have $|x||y|=1$, that is we must have $|x|=1$ and $|y|=1$.



              Also, determinining $x$ would completely determine $y$.



              Hence we only need to examine what happends when $x=1$ and $x=-1$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 25 '18 at 3:57









              Siong Thye GohSiong Thye Goh

              100k1465117




              100k1465117












              • $begingroup$
                Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:04










              • $begingroup$
                I think Z is defined so that 1 and -1 are the only units by definition. I think.
                $endgroup$
                – fleablood
                Nov 25 '18 at 4:21




















              • $begingroup$
                Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:04










              • $begingroup$
                I think Z is defined so that 1 and -1 are the only units by definition. I think.
                $endgroup$
                – fleablood
                Nov 25 '18 at 4:21


















              $begingroup$
              Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
              $endgroup$
              – Cabex
              Nov 25 '18 at 4:04




              $begingroup$
              Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
              $endgroup$
              – Cabex
              Nov 25 '18 at 4:04












              $begingroup$
              I think Z is defined so that 1 and -1 are the only units by definition. I think.
              $endgroup$
              – fleablood
              Nov 25 '18 at 4:21






              $begingroup$
              I think Z is defined so that 1 and -1 are the only units by definition. I think.
              $endgroup$
              – fleablood
              Nov 25 '18 at 4:21













              2












              $begingroup$

              If xy = -1, then x = $frac{-1}{y}$. If x is an integer, y must divide -1. Therefore, y = $pm$1, which implies x $mp$1.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:03






              • 1




                $begingroup$
                @Cabex This nearly does what you want, it can be rephrased to say that the only units in $mathbb{Z}$ are $pm 1$.
                $endgroup$
                – Michael Burr
                Nov 25 '18 at 4:06










              • $begingroup$
                @MichaelBurr I don't get it, we can't divide in $mathbb{Z}$. Isn't assuming $y$ has an inverse, or viewing the equation outside the integers?
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:10










              • $begingroup$
                (xy)$^2$ = x$^2$y$^2$ = 1. Since x, y $in mathbb{Z}$, this implies that x$^2$ = y$^2$ = 1. So, x is it's own inverse. Thus x = $pm$ 1.
                $endgroup$
                – Joel Pereira
                Nov 25 '18 at 4:10










              • $begingroup$
                Right! Using idempotents... thanks a lot!
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:23
















              2












              $begingroup$

              If xy = -1, then x = $frac{-1}{y}$. If x is an integer, y must divide -1. Therefore, y = $pm$1, which implies x $mp$1.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:03






              • 1




                $begingroup$
                @Cabex This nearly does what you want, it can be rephrased to say that the only units in $mathbb{Z}$ are $pm 1$.
                $endgroup$
                – Michael Burr
                Nov 25 '18 at 4:06










              • $begingroup$
                @MichaelBurr I don't get it, we can't divide in $mathbb{Z}$. Isn't assuming $y$ has an inverse, or viewing the equation outside the integers?
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:10










              • $begingroup$
                (xy)$^2$ = x$^2$y$^2$ = 1. Since x, y $in mathbb{Z}$, this implies that x$^2$ = y$^2$ = 1. So, x is it's own inverse. Thus x = $pm$ 1.
                $endgroup$
                – Joel Pereira
                Nov 25 '18 at 4:10










              • $begingroup$
                Right! Using idempotents... thanks a lot!
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:23














              2












              2








              2





              $begingroup$

              If xy = -1, then x = $frac{-1}{y}$. If x is an integer, y must divide -1. Therefore, y = $pm$1, which implies x $mp$1.






              share|cite|improve this answer









              $endgroup$



              If xy = -1, then x = $frac{-1}{y}$. If x is an integer, y must divide -1. Therefore, y = $pm$1, which implies x $mp$1.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 25 '18 at 4:00









              Joel PereiraJoel Pereira

              71319




              71319












              • $begingroup$
                Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:03






              • 1




                $begingroup$
                @Cabex This nearly does what you want, it can be rephrased to say that the only units in $mathbb{Z}$ are $pm 1$.
                $endgroup$
                – Michael Burr
                Nov 25 '18 at 4:06










              • $begingroup$
                @MichaelBurr I don't get it, we can't divide in $mathbb{Z}$. Isn't assuming $y$ has an inverse, or viewing the equation outside the integers?
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:10










              • $begingroup$
                (xy)$^2$ = x$^2$y$^2$ = 1. Since x, y $in mathbb{Z}$, this implies that x$^2$ = y$^2$ = 1. So, x is it's own inverse. Thus x = $pm$ 1.
                $endgroup$
                – Joel Pereira
                Nov 25 '18 at 4:10










              • $begingroup$
                Right! Using idempotents... thanks a lot!
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:23


















              • $begingroup$
                Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:03






              • 1




                $begingroup$
                @Cabex This nearly does what you want, it can be rephrased to say that the only units in $mathbb{Z}$ are $pm 1$.
                $endgroup$
                – Michael Burr
                Nov 25 '18 at 4:06










              • $begingroup$
                @MichaelBurr I don't get it, we can't divide in $mathbb{Z}$. Isn't assuming $y$ has an inverse, or viewing the equation outside the integers?
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:10










              • $begingroup$
                (xy)$^2$ = x$^2$y$^2$ = 1. Since x, y $in mathbb{Z}$, this implies that x$^2$ = y$^2$ = 1. So, x is it's own inverse. Thus x = $pm$ 1.
                $endgroup$
                – Joel Pereira
                Nov 25 '18 at 4:10










              • $begingroup$
                Right! Using idempotents... thanks a lot!
                $endgroup$
                – Cabex
                Nov 25 '18 at 4:23
















              $begingroup$
              Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
              $endgroup$
              – Cabex
              Nov 25 '18 at 4:03




              $begingroup$
              Do you think is possible to prove it using just the sum and product defined in the integers? I mean, viewing $mathbb{Z}$ as a ring.
              $endgroup$
              – Cabex
              Nov 25 '18 at 4:03




              1




              1




              $begingroup$
              @Cabex This nearly does what you want, it can be rephrased to say that the only units in $mathbb{Z}$ are $pm 1$.
              $endgroup$
              – Michael Burr
              Nov 25 '18 at 4:06




              $begingroup$
              @Cabex This nearly does what you want, it can be rephrased to say that the only units in $mathbb{Z}$ are $pm 1$.
              $endgroup$
              – Michael Burr
              Nov 25 '18 at 4:06












              $begingroup$
              @MichaelBurr I don't get it, we can't divide in $mathbb{Z}$. Isn't assuming $y$ has an inverse, or viewing the equation outside the integers?
              $endgroup$
              – Cabex
              Nov 25 '18 at 4:10




              $begingroup$
              @MichaelBurr I don't get it, we can't divide in $mathbb{Z}$. Isn't assuming $y$ has an inverse, or viewing the equation outside the integers?
              $endgroup$
              – Cabex
              Nov 25 '18 at 4:10












              $begingroup$
              (xy)$^2$ = x$^2$y$^2$ = 1. Since x, y $in mathbb{Z}$, this implies that x$^2$ = y$^2$ = 1. So, x is it's own inverse. Thus x = $pm$ 1.
              $endgroup$
              – Joel Pereira
              Nov 25 '18 at 4:10




              $begingroup$
              (xy)$^2$ = x$^2$y$^2$ = 1. Since x, y $in mathbb{Z}$, this implies that x$^2$ = y$^2$ = 1. So, x is it's own inverse. Thus x = $pm$ 1.
              $endgroup$
              – Joel Pereira
              Nov 25 '18 at 4:10












              $begingroup$
              Right! Using idempotents... thanks a lot!
              $endgroup$
              – Cabex
              Nov 25 '18 at 4:23




              $begingroup$
              Right! Using idempotents... thanks a lot!
              $endgroup$
              – Cabex
              Nov 25 '18 at 4:23











              0












              $begingroup$

              For positive integers (same can be done for negative, or just use absolute value):



              If $x >1$ and $y>1$, then $xy>1$. So either $x =1$ or $y=1$. If $x=1$, then $1 cdot y=1$, so $y=1$. If $y=1$, then $x=1$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                For positive integers (same can be done for negative, or just use absolute value):



                If $x >1$ and $y>1$, then $xy>1$. So either $x =1$ or $y=1$. If $x=1$, then $1 cdot y=1$, so $y=1$. If $y=1$, then $x=1$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  For positive integers (same can be done for negative, or just use absolute value):



                  If $x >1$ and $y>1$, then $xy>1$. So either $x =1$ or $y=1$. If $x=1$, then $1 cdot y=1$, so $y=1$. If $y=1$, then $x=1$.






                  share|cite|improve this answer









                  $endgroup$



                  For positive integers (same can be done for negative, or just use absolute value):



                  If $x >1$ and $y>1$, then $xy>1$. So either $x =1$ or $y=1$. If $x=1$, then $1 cdot y=1$, so $y=1$. If $y=1$, then $x=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 '18 at 4:05









                  OviOvi

                  12.4k1038111




                  12.4k1038111























                      0












                      $begingroup$

                      Alternatively.



                      If $x $ or $y $ is $0$, $xy=0$.



                      Otherwise, $|x|ge 1$ and $|y|ge 1$. If $|x|>1$ then $|xy|=|y||x|>|y|ge 1$. So we must have $|x|=1$. And $1=|xy|=|x||y|=|y|$ so $|x|=|y|=1$.



                      Of the four options only $x=y=1;x=y=-1$ work.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Oops. Sorry. I see someone else had the same idea. I'll delete when I get to an application that let's me.
                        $endgroup$
                        – fleablood
                        Nov 25 '18 at 4:19
















                      0












                      $begingroup$

                      Alternatively.



                      If $x $ or $y $ is $0$, $xy=0$.



                      Otherwise, $|x|ge 1$ and $|y|ge 1$. If $|x|>1$ then $|xy|=|y||x|>|y|ge 1$. So we must have $|x|=1$. And $1=|xy|=|x||y|=|y|$ so $|x|=|y|=1$.



                      Of the four options only $x=y=1;x=y=-1$ work.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Oops. Sorry. I see someone else had the same idea. I'll delete when I get to an application that let's me.
                        $endgroup$
                        – fleablood
                        Nov 25 '18 at 4:19














                      0












                      0








                      0





                      $begingroup$

                      Alternatively.



                      If $x $ or $y $ is $0$, $xy=0$.



                      Otherwise, $|x|ge 1$ and $|y|ge 1$. If $|x|>1$ then $|xy|=|y||x|>|y|ge 1$. So we must have $|x|=1$. And $1=|xy|=|x||y|=|y|$ so $|x|=|y|=1$.



                      Of the four options only $x=y=1;x=y=-1$ work.






                      share|cite|improve this answer









                      $endgroup$



                      Alternatively.



                      If $x $ or $y $ is $0$, $xy=0$.



                      Otherwise, $|x|ge 1$ and $|y|ge 1$. If $|x|>1$ then $|xy|=|y||x|>|y|ge 1$. So we must have $|x|=1$. And $1=|xy|=|x||y|=|y|$ so $|x|=|y|=1$.



                      Of the four options only $x=y=1;x=y=-1$ work.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 25 '18 at 4:17









                      fleabloodfleablood

                      68.8k22685




                      68.8k22685












                      • $begingroup$
                        Oops. Sorry. I see someone else had the same idea. I'll delete when I get to an application that let's me.
                        $endgroup$
                        – fleablood
                        Nov 25 '18 at 4:19


















                      • $begingroup$
                        Oops. Sorry. I see someone else had the same idea. I'll delete when I get to an application that let's me.
                        $endgroup$
                        – fleablood
                        Nov 25 '18 at 4:19
















                      $begingroup$
                      Oops. Sorry. I see someone else had the same idea. I'll delete when I get to an application that let's me.
                      $endgroup$
                      – fleablood
                      Nov 25 '18 at 4:19




                      $begingroup$
                      Oops. Sorry. I see someone else had the same idea. I'll delete when I get to an application that let's me.
                      $endgroup$
                      – fleablood
                      Nov 25 '18 at 4:19











                      0












                      $begingroup$

                      Suppose that $xy=-1$, then $-xy=1$, so $x$ and $y$ are units (by definition). Since the only units in $mathbb{Z}$ are $pm 1$, we can check which combinations result in a product of $-1$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Suppose that $xy=-1$, then $-xy=1$, so $x$ and $y$ are units (by definition). Since the only units in $mathbb{Z}$ are $pm 1$, we can check which combinations result in a product of $-1$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Suppose that $xy=-1$, then $-xy=1$, so $x$ and $y$ are units (by definition). Since the only units in $mathbb{Z}$ are $pm 1$, we can check which combinations result in a product of $-1$.






                          share|cite|improve this answer









                          $endgroup$



                          Suppose that $xy=-1$, then $-xy=1$, so $x$ and $y$ are units (by definition). Since the only units in $mathbb{Z}$ are $pm 1$, we can check which combinations result in a product of $-1$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 25 '18 at 15:01









                          Michael BurrMichael Burr

                          26.7k23262




                          26.7k23262






























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