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

108k1895200




108k1895200










asked Jan 13 at 20:43









josephjoseph

496111




496111








  • 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











      $endgroup$













        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%2f3072489%2fhow-to-get-the-identity-frac1ab-left-frac1a-frac1b-right-f%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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

            17k3939




            17k3939























                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

                    108k1895200




                    108k1895200























                        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

                            3,4851320




                            3,4851320










                            answered Jan 13 at 20:52









                            kelalakakelalaka

                            3351314




                            3351314






























                                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%2f3072489%2fhow-to-get-the-identity-frac1ab-left-frac1a-frac1b-right-f%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

                                Human spaceflight

                                Image
                                Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
                                Read more

                                Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                Image
                                0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
                                Read more

                                張江高科駅

                                Image
                                張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
                                Read more