For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational...












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For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational number $r$ such that $(wr+x)div(yr+z)=1$



How do I prove uniqueness? I know to show that there exists a number all I need to do is use an example.






proof-writing rational-numbers







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  • $begingroup$
    Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
    $endgroup$
    – J. W. Tanner
    Jan 29 at 19:06
















0












$begingroup$


For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational number $r$ such that $(wr+x)div(yr+z)=1$



How do I prove uniqueness? I know to show that there exists a number all I need to do is use an example.






proof-writing rational-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
    $endgroup$
    – J. W. Tanner
    Jan 29 at 19:06














0












0








0





$begingroup$


For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational number $r$ such that $(wr+x)div(yr+z)=1$



How do I prove uniqueness? I know to show that there exists a number all I need to do is use an example.






proof-writing rational-numbers







share|cite|improve this question









$endgroup$




For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational number $r$ such that $(wr+x)div(yr+z)=1$



How do I prove uniqueness? I know to show that there exists a number all I need to do is use an example.






proof-writing rational-numbers




proof-writing rational-numbers






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asked Jan 29 at 18:57









macymacy

526




526












  • $begingroup$
    Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
    $endgroup$
    – J. W. Tanner
    Jan 29 at 19:06


















  • $begingroup$
    Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
    $endgroup$
    – J. W. Tanner
    Jan 29 at 19:06
















$begingroup$
Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
$endgroup$
– J. W. Tanner
Jan 29 at 19:06




$begingroup$
Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
$endgroup$
– J. W. Tanner
Jan 29 at 19:06










2 Answers
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$begingroup$

One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.



In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).



Hope I was useful.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.






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      2 Answers
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      2 Answers
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      0












      $begingroup$

      One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.



      In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).



      Hope I was useful.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.



        In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).



        Hope I was useful.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.



          In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).



          Hope I was useful.






          share|cite|improve this answer









          $endgroup$



          One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.



          In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).



          Hope I was useful.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 29 at 19:06









          maxbpmaxbp

          1467




          1467























              0












              $begingroup$

              To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.






                  share|cite|improve this answer









                  $endgroup$



                  To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 29 at 19:13









                  J. W. TannerJ. W. Tanner

                  4,2361320




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