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






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      share|cite|improve this question













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      edited 36 mins ago







      Sphygmomanometer

















      asked 38 mins ago









      SphygmomanometerSphygmomanometer

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          4 Answers
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          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









                $endgroup$













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                  4 Answers
                  4






                  active

                  oldest

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                  4 Answers
                  4






                  active

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                  active

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                  active

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                  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 29 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 30 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 25 mins ago









                                      AdityaAditya

                                      112




                                      112




                                      New contributor




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                                      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 25 mins ago









                                              Rhys HughesRhys Hughes

                                              6,7101530




                                              6,7101530






























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