How is rearranging 56 x 100 ÷ 8 into 56 ÷8 x100 allowed by the commutative property?












1












$begingroup$


So according to the commutative property for multiplication:



$a * b = b * a$



However this does not hold for division



$a ÷ b$ $!= b ÷a$



Why is it that in the following case:



$56 * 100 ÷ 8 = 56 ÷ 8* 100$



It seems like division is breaking the rule. There is something I am misunderstanding here.



Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?



If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?






arithmetic







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    So according to the commutative property for multiplication:



    $a * b = b * a$



    However this does not hold for division



    $a ÷ b$ $!= b ÷a$



    Why is it that in the following case:



    $56 * 100 ÷ 8 = 56 ÷ 8* 100$



    It seems like division is breaking the rule. There is something I am misunderstanding here.



    Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?



    If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?






    arithmetic







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So according to the commutative property for multiplication:



      $a * b = b * a$



      However this does not hold for division



      $a ÷ b$ $!= b ÷a$



      Why is it that in the following case:



      $56 * 100 ÷ 8 = 56 ÷ 8* 100$



      It seems like division is breaking the rule. There is something I am misunderstanding here.



      Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?



      If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?






      arithmetic







      share|cite|improve this question











      $endgroup$




      So according to the commutative property for multiplication:



      $a * b = b * a$



      However this does not hold for division



      $a ÷ b$ $!= b ÷a$



      Why is it that in the following case:



      $56 * 100 ÷ 8 = 56 ÷ 8* 100$



      It seems like division is breaking the rule. There is something I am misunderstanding here.



      Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?



      If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?






      arithmetic




      arithmetic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 44 mins ago







      Sphygmomanometer

















      asked 46 mins ago









      SphygmomanometerSphygmomanometer

      717




      717






















          5 Answers
          5






          active

          oldest

          votes


















          1












          $begingroup$

          Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
          $$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.



            So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              Hope this makes sense.



              $$ atimes b ÷ c $$



              $$=atimesdfrac{b}{c}$$



              $$=atimes btimesdfrac{1}{c}$$



              $$=(atimesdfrac{1}{c})times b$$



              $$=dfrac{a}{c}times b$$



              $$=a÷ctimes b$$






              share|cite|improve this answer








              New contributor




              Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$





















                1












                $begingroup$

                Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$



                So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
                Which is obvious.






                share|cite|improve this answer









                $endgroup$





















                  0












                  $begingroup$

                  Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.



                  So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.



                  Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.



                  So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.






                  share|cite









                  $endgroup$













                    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%2f3117260%2fhow-is-rearranging-56-x-100-%25c3%25b7-8-into-56-%25c3%25b78-x100-allowed-by-the-commutative-prope%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown

























                    5 Answers
                    5






                    active

                    oldest

                    votes








                    5 Answers
                    5






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes









                    1












                    $begingroup$

                    Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
                    $$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$






                    share|cite|improve this answer









                    $endgroup$


















                      1












                      $begingroup$

                      Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
                      $$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$






                      share|cite|improve this answer









                      $endgroup$
















                        1












                        1








                        1





                        $begingroup$

                        Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
                        $$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$






                        share|cite|improve this answer









                        $endgroup$



                        Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
                        $$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 37 mins ago









                        AndreiAndrei

                        12.1k21127




                        12.1k21127























                            2












                            $begingroup$

                            The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.



                            So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$






                            share|cite|improve this answer









                            $endgroup$


















                              2












                              $begingroup$

                              The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.



                              So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$






                              share|cite|improve this answer









                              $endgroup$
















                                2












                                2








                                2





                                $begingroup$

                                The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.



                                So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$






                                share|cite|improve this answer









                                $endgroup$



                                The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.



                                So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 37 mins ago









                                Angela RichardsonAngela Richardson

                                5,28911733




                                5,28911733























                                    1












                                    $begingroup$

                                    Hope this makes sense.



                                    $$ atimes b ÷ c $$



                                    $$=atimesdfrac{b}{c}$$



                                    $$=atimes btimesdfrac{1}{c}$$



                                    $$=(atimesdfrac{1}{c})times b$$



                                    $$=dfrac{a}{c}times b$$



                                    $$=a÷ctimes b$$






                                    share|cite|improve this answer








                                    New contributor




                                    Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    $endgroup$


















                                      1












                                      $begingroup$

                                      Hope this makes sense.



                                      $$ atimes b ÷ c $$



                                      $$=atimesdfrac{b}{c}$$



                                      $$=atimes btimesdfrac{1}{c}$$



                                      $$=(atimesdfrac{1}{c})times b$$



                                      $$=dfrac{a}{c}times b$$



                                      $$=a÷ctimes b$$






                                      share|cite|improve this answer








                                      New contributor




                                      Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






                                      $endgroup$
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        Hope this makes sense.



                                        $$ atimes b ÷ c $$



                                        $$=atimesdfrac{b}{c}$$



                                        $$=atimes btimesdfrac{1}{c}$$



                                        $$=(atimesdfrac{1}{c})times b$$



                                        $$=dfrac{a}{c}times b$$



                                        $$=a÷ctimes b$$






                                        share|cite|improve this answer








                                        New contributor




                                        Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        $endgroup$



                                        Hope this makes sense.



                                        $$ atimes b ÷ c $$



                                        $$=atimesdfrac{b}{c}$$



                                        $$=atimes btimesdfrac{1}{c}$$



                                        $$=(atimesdfrac{1}{c})times b$$



                                        $$=dfrac{a}{c}times b$$



                                        $$=a÷ctimes b$$







                                        share|cite|improve this answer








                                        New contributor




                                        Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.









                                        share|cite|improve this answer



                                        share|cite|improve this answer






                                        New contributor




                                        Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.









                                        answered 32 mins ago









                                        AdityaAditya

                                        112




                                        112




                                        New contributor




                                        Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.





                                        New contributor





                                        Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.























                                            1












                                            $begingroup$

                                            Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$



                                            So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
                                            Which is obvious.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              1












                                              $begingroup$

                                              Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$



                                              So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
                                              Which is obvious.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                1












                                                1








                                                1





                                                $begingroup$

                                                Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$



                                                So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
                                                Which is obvious.






                                                share|cite|improve this answer









                                                $endgroup$



                                                Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$



                                                So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
                                                Which is obvious.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 32 mins ago









                                                Rhys HughesRhys Hughes

                                                6,7101530




                                                6,7101530























                                                    0












                                                    $begingroup$

                                                    Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.



                                                    So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.



                                                    Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.



                                                    So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.






                                                    share|cite









                                                    $endgroup$


















                                                      0












                                                      $begingroup$

                                                      Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.



                                                      So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.



                                                      Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.



                                                      So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.






                                                      share|cite









                                                      $endgroup$
















                                                        0












                                                        0








                                                        0





                                                        $begingroup$

                                                        Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.



                                                        So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.



                                                        Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.



                                                        So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.






                                                        share|cite









                                                        $endgroup$



                                                        Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.



                                                        So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.



                                                        Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.



                                                        So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.







                                                        share|cite












                                                        share|cite



                                                        share|cite










                                                        answered 57 secs ago









                                                        Mark BennetMark Bennet

                                                        81.3k983180




                                                        81.3k983180






























                                                            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%2f3117260%2fhow-is-rearranging-56-x-100-%25c3%25b7-8-into-56-%25c3%25b78-x100-allowed-by-the-commutative-prope%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

                                                            Loup dans la culture

                                                            Image
                                                            Loups sur une enluminure du Bestiaire d'Aberdeen Le loup est l'animal le plus emblématique de l'histoire de l'Eurasie, il était à l'honneur durant l'Antiquité chez la totalité des anciens peuples européens [ 1 ] . Sommaire 1 Le loup dans la mythologie 2 Les loups dans le folklore 2.1 L'évolution des mentalités 2.2 Le loup dans la psychanalyse 3 Le loup dans les œuvres culturelles 3.1 Le loup dans la littérature 3.1.1 Fables 3.1.2 Poésie 3.1.3 Romans 3.1.4 Littérature jeunesse 3.1.4.1 Quelques titres d'albums contemporains 3.1.5 Bande dessinée 3.2 Le loup à l'écran 3.2.1 Le cinéma et la télévision 3.2.2 Documentaires 3.2.3 Dessins animés 3.2.4 Jeux vidéo 3.3 Le loup en peinture et en sculpture 3.4 Le loup dans les œuvres musicales 4 Le loup et les noms propres 5 Bibliographie 6 Notes et références 7 Voir aussi 7.1 Articles connexes ...
                                                            Read more

                                                            How to solve the problem of ntp “Unable to contact time server” from KDE?

                                                            Image
                                                            1 0 I want to set date and time automatically but this error appears when I try to .. I'm using openSUSE, I tried different servers but it didn't work either, help please! It doesn't seem to be a network problem: $ ping pool.ntp.org 64 bytes from 41.78.128.17: icmp_seq=1 ttl=41 time=260 ms To the question of whether ntpd is installed and running: $ ps -C ntpd PID TTY TIME CMD i.e. there's no ntpd process running. $ rpm -qa | grep ntp yast2-ntp-client-3.1.12-1.7.noarch ntp-4.2.6p5-25.2.1.i586 Further information from the comments: $ sudo ntpdate pool.ntp.org 25 Sep 19:41:51 ntpdate[3162]: no server suitable for synchronization found networking kde opensuse ntp ntpd sh...
                                                            Read more

                                                            ASUS Zenbook UX433/UX333 — Configure Touchpad-embedded numpad on Linux

                                                            Image
                                                            0 1 I have an ASUS Zenbook UX433,on which ASUS decided to innovate by embedding a Numpad inside the touchpad (see picture 1). Toggling between classic touchpad mode and numpad (also turns the touchpad LEDs on) can be done through the golden button in the top-right corner of the touchpad. However, on my Manjaro arch Linux config, this behavior does not work by default and I can only use the touchpad, no numpad. Has anyone been able to configure the numpad to work on Linux, or have any idea how this could be achieved? EDIT1: I have had a look at the official Windows drivers for the UX433FA TouchPad, but was not able to make anything of it. I tried installing them on Linux using tools like NDISWrapper, but they do not detect the TouchPad/Numpad hardware, and my understanding of drivers is too limited. ...
                                                            Read more