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 49 mins ago







      Sphygmomanometer

















      asked 51 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 42 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 42 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 37 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 37 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 5 mins 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

                                                            "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

                                                            Image
                                                            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness. ...
                                                            Read more

                                                            Alcedinidae

                                                            Image
                                                            Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout ...
                                                            Read more

                                                            Origin of the phrase “under your belt”?

                                                            Image
                                                            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 17 1 Today, I found myself discussing what students should have "under their belts" during a lecture, and I wondered to myself if there was some inappropriate undertone here that I might be unaware of. Question : What is the origin of the phrase "under your belt"? A google search revealed some webpages (e.g. [1]) that assert its origin is related to consumption of food -- once a meal has been eaten, it's under your belt (which is a relief). However, with such websites, I have no way to determine fact from "random guy on the internet making stuff up". etymology idioms ...
                                                            Read more