How to get the identity $frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)frac{1}{(b - a)}$ when $a < b$?












-1












$begingroup$


How does one arrive at the equality $frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)(b - a)$ when $a < b$? I came across this identity in a competitive programming problem, but I couldn't find out any way to get it. For example,



$$frac{1}{2 cdot 3} = left(frac{1}{2} - frac{1}{3}right)frac{1}{(3 - 2)},$$



and



$$frac{1}{3 cdot 5} = left(frac{1}{3} - frac{1}{5}right)frac{1}{(5 - 3)}.$$






algebra-precalculus proof-writing







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    They probably mean begin{eqnarray*} frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)frac{1}{(b - a)}. end{eqnarray*}
    $endgroup$
    – Donald Splutterwit
    Jan 13 at 20:46










  • $begingroup$
    Yes, I have fixed it.
    $endgroup$
    – joseph
    Jan 13 at 20:47










  • $begingroup$
    Common denominator of the difference of reciprocals.
    $endgroup$
    – Hans Engler
    Jan 13 at 20:48






  • 2




    $begingroup$
    This follows trivially from $frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$.
    $endgroup$
    – Dietrich Burde
    Jan 13 at 20:48










  • $begingroup$
    They just compute both sides and see they are the same.
    $endgroup$
    – Saucy O'Path
    Jan 13 at 20:49
















-1












$begingroup$


How does one arrive at the equality $frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)(b - a)$ when $a < b$? I came across this identity in a competitive programming problem, but I couldn't find out any way to get it. For example,



$$frac{1}{2 cdot 3} = left(frac{1}{2} - frac{1}{3}right)frac{1}{(3 - 2)},$$



and



$$frac{1}{3 cdot 5} = left(frac{1}{3} - frac{1}{5}right)frac{1}{(5 - 3)}.$$






algebra-precalculus proof-writing







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    They probably mean begin{eqnarray*} frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)frac{1}{(b - a)}. end{eqnarray*}
    $endgroup$
    – Donald Splutterwit
    Jan 13 at 20:46










  • $begingroup$
    Yes, I have fixed it.
    $endgroup$
    – joseph
    Jan 13 at 20:47










  • $begingroup$
    Common denominator of the difference of reciprocals.
    $endgroup$
    – Hans Engler
    Jan 13 at 20:48






  • 2




    $begingroup$
    This follows trivially from $frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$.
    $endgroup$
    – Dietrich Burde
    Jan 13 at 20:48










  • $begingroup$
    They just compute both sides and see they are the same.
    $endgroup$
    – Saucy O'Path
    Jan 13 at 20:49














-1












-1








-1





$begingroup$


How does one arrive at the equality $frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)(b - a)$ when $a < b$? I came across this identity in a competitive programming problem, but I couldn't find out any way to get it. For example,



$$frac{1}{2 cdot 3} = left(frac{1}{2} - frac{1}{3}right)frac{1}{(3 - 2)},$$



and



$$frac{1}{3 cdot 5} = left(frac{1}{3} - frac{1}{5}right)frac{1}{(5 - 3)}.$$






algebra-precalculus proof-writing







share|cite|improve this question











$endgroup$




How does one arrive at the equality $frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)(b - a)$ when $a < b$? I came across this identity in a competitive programming problem, but I couldn't find out any way to get it. For example,



$$frac{1}{2 cdot 3} = left(frac{1}{2} - frac{1}{3}right)frac{1}{(3 - 2)},$$



and



$$frac{1}{3 cdot 5} = left(frac{1}{3} - frac{1}{5}right)frac{1}{(5 - 3)}.$$






algebra-precalculus proof-writing




algebra-precalculus proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 20:56









Michael Rozenberg

100k1591193




100k1591193










asked Jan 13 at 20:43









josephjoseph

500111




500111








  • 1




    $begingroup$
    They probably mean begin{eqnarray*} frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)frac{1}{(b - a)}. end{eqnarray*}
    $endgroup$
    – Donald Splutterwit
    Jan 13 at 20:46










  • $begingroup$
    Yes, I have fixed it.
    $endgroup$
    – joseph
    Jan 13 at 20:47










  • $begingroup$
    Common denominator of the difference of reciprocals.
    $endgroup$
    – Hans Engler
    Jan 13 at 20:48






  • 2




    $begingroup$
    This follows trivially from $frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$.
    $endgroup$
    – Dietrich Burde
    Jan 13 at 20:48










  • $begingroup$
    They just compute both sides and see they are the same.
    $endgroup$
    – Saucy O'Path
    Jan 13 at 20:49














  • 1




    $begingroup$
    They probably mean begin{eqnarray*} frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)frac{1}{(b - a)}. end{eqnarray*}
    $endgroup$
    – Donald Splutterwit
    Jan 13 at 20:46










  • $begingroup$
    Yes, I have fixed it.
    $endgroup$
    – joseph
    Jan 13 at 20:47










  • $begingroup$
    Common denominator of the difference of reciprocals.
    $endgroup$
    – Hans Engler
    Jan 13 at 20:48






  • 2




    $begingroup$
    This follows trivially from $frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$.
    $endgroup$
    – Dietrich Burde
    Jan 13 at 20:48










  • $begingroup$
    They just compute both sides and see they are the same.
    $endgroup$
    – Saucy O'Path
    Jan 13 at 20:49








1




1




$begingroup$
They probably mean begin{eqnarray*} frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)frac{1}{(b - a)}. end{eqnarray*}
$endgroup$
– Donald Splutterwit
Jan 13 at 20:46




$begingroup$
They probably mean begin{eqnarray*} frac{1}{ab} = left(frac{1}{a} - frac{1}{b}right)frac{1}{(b - a)}. end{eqnarray*}
$endgroup$
– Donald Splutterwit
Jan 13 at 20:46












$begingroup$
Yes, I have fixed it.
$endgroup$
– joseph
Jan 13 at 20:47




$begingroup$
Yes, I have fixed it.
$endgroup$
– joseph
Jan 13 at 20:47












$begingroup$
Common denominator of the difference of reciprocals.
$endgroup$
– Hans Engler
Jan 13 at 20:48




$begingroup$
Common denominator of the difference of reciprocals.
$endgroup$
– Hans Engler
Jan 13 at 20:48




2




2




$begingroup$
This follows trivially from $frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$.
$endgroup$
– Dietrich Burde
Jan 13 at 20:48




$begingroup$
This follows trivially from $frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$.
$endgroup$
– Dietrich Burde
Jan 13 at 20:48












$begingroup$
They just compute both sides and see they are the same.
$endgroup$
– Saucy O'Path
Jan 13 at 20:49




$begingroup$
They just compute both sides and see they are the same.
$endgroup$
– Saucy O'Path
Jan 13 at 20:49










3 Answers
3






active

oldest

votes


















2












$begingroup$

For $a,bne 0$ and $ane b$, multiply both sides in $ab(b-a)$ and double-sidedly conclude what you want.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Because $$frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$$ and since $b-aneq0$, we are done.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $frac{1}{ab} = frac{b - a}{ab(b - a)} = big(frac{b-a}{ab})/(b-a) = big(frac{1}{a} - frac{1}{b}big)/(b - a)$



      if $b-a neq0$






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        For $a,bne 0$ and $ane b$, multiply both sides in $ab(b-a)$ and double-sidedly conclude what you want.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          For $a,bne 0$ and $ane b$, multiply both sides in $ab(b-a)$ and double-sidedly conclude what you want.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            For $a,bne 0$ and $ane b$, multiply both sides in $ab(b-a)$ and double-sidedly conclude what you want.






            share|cite|improve this answer









            $endgroup$



            For $a,bne 0$ and $ane b$, multiply both sides in $ab(b-a)$ and double-sidedly conclude what you want.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 13 at 20:54









            Mostafa AyazMostafa Ayaz

            15.3k3939




            15.3k3939























                3












                $begingroup$

                Because $$frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$$ and since $b-aneq0$, we are done.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Because $$frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$$ and since $b-aneq0$, we are done.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Because $$frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$$ and since $b-aneq0$, we are done.






                    share|cite|improve this answer









                    $endgroup$



                    Because $$frac{1}{a}-frac{1}{b}=frac{b-a}{ab}$$ and since $b-aneq0$, we are done.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 13 at 20:50









                    Michael RozenbergMichael Rozenberg

                    100k1591193




                    100k1591193























                        0












                        $begingroup$

                        $frac{1}{ab} = frac{b - a}{ab(b - a)} = big(frac{b-a}{ab})/(b-a) = big(frac{1}{a} - frac{1}{b}big)/(b - a)$



                        if $b-a neq0$






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          $frac{1}{ab} = frac{b - a}{ab(b - a)} = big(frac{b-a}{ab})/(b-a) = big(frac{1}{a} - frac{1}{b}big)/(b - a)$



                          if $b-a neq0$






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $frac{1}{ab} = frac{b - a}{ab(b - a)} = big(frac{b-a}{ab})/(b-a) = big(frac{1}{a} - frac{1}{b}big)/(b - a)$



                            if $b-a neq0$






                            share|cite|improve this answer











                            $endgroup$



                            $frac{1}{ab} = frac{b - a}{ab(b - a)} = big(frac{b-a}{ab})/(b-a) = big(frac{1}{a} - frac{1}{b}big)/(b - a)$



                            if $b-a neq0$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 13 at 21:03









                            J. W. Tanner

                            830112




                            830112










                            answered Jan 13 at 20:52









                            kelalakakelalaka

                            3301212




                            3301212






























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