The distributive law












1












$begingroup$


$4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$ , but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?






algebra-precalculus arithmetic







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$endgroup$












  • $begingroup$
    It is also the distributive law or rule.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 14 '18 at 16:19










  • $begingroup$
    This is distribution done twice.
    $endgroup$
    – Randall
    Dec 14 '18 at 16:20
















1












$begingroup$


$4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$ , but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?






algebra-precalculus arithmetic







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is also the distributive law or rule.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 14 '18 at 16:19










  • $begingroup$
    This is distribution done twice.
    $endgroup$
    – Randall
    Dec 14 '18 at 16:20














1












1








1





$begingroup$


$4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$ , but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?






algebra-precalculus arithmetic







share|cite|improve this question









$endgroup$




$4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$ , but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?






algebra-precalculus arithmetic




algebra-precalculus arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 16:18









Ashraf BenmebarekAshraf Benmebarek

485




485












  • $begingroup$
    It is also the distributive law or rule.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 14 '18 at 16:19










  • $begingroup$
    This is distribution done twice.
    $endgroup$
    – Randall
    Dec 14 '18 at 16:20


















  • $begingroup$
    It is also the distributive law or rule.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 14 '18 at 16:19










  • $begingroup$
    This is distribution done twice.
    $endgroup$
    – Randall
    Dec 14 '18 at 16:20
















$begingroup$
It is also the distributive law or rule.
$endgroup$
– Dr. Sonnhard Graubner
Dec 14 '18 at 16:19




$begingroup$
It is also the distributive law or rule.
$endgroup$
– Dr. Sonnhard Graubner
Dec 14 '18 at 16:19












$begingroup$
This is distribution done twice.
$endgroup$
– Randall
Dec 14 '18 at 16:20




$begingroup$
This is distribution done twice.
$endgroup$
– Randall
Dec 14 '18 at 16:20










3 Answers
3






active

oldest

votes


















1












$begingroup$

Because:



$$left (x+y right) left (x+y right) = underbrace{(x+y)+(x+y)+...(x+y)}_{x + ytext{ times}}$$



$$=underbrace{(x+y)+(x+y)+...(x+y)}_{xtext{ times}}+underbrace{(x+y)+(x+y)+...(x+y)}_{ytext{ times}}$$



$$=underbrace{x}_{xtext{ times}}+underbrace{y}_{xtext{ times}}+underbrace{x}_{ytext{ times}}+underbrace{y}_{ytext{ times}}$$



$$=xx+xy+yx+yy$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately.
    $endgroup$
    – StackTD
    Dec 14 '18 at 16:30










  • $begingroup$
    @StackTD Excellent question! Obviously one needs to do something else in that case ... but I figured this would be in the spirit of the OP
    $endgroup$
    – Bram28
    Dec 14 '18 at 16:32










  • $begingroup$
    I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!).
    $endgroup$
    – StackTD
    Dec 14 '18 at 16:33



















0












$begingroup$


$4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$




This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.




but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?




The distributive law works even if $x$ and/or $y$ are not integers.



Keep one of the factors together in a first step, and apply distributivity twice:
$$begin{align}
(color{blue}{x}+color{red}{y})(x+y) & =color{blue}{x}(x+y)+color{red}{y}(x+y) \ & =color{blue}{x}x+color{blue}{x}y+color{red}{y}x+color{red}{y}y
end{align}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Do it in steps.



    You accept that $M(x+y) = Mx + My$



    So replace $M$ with $(x+y)$ and you get:



    $(x+y)(x+y) = M(x+y) =$



    $Mx + My = $



    $(x + y)x + (x+y)y$



    Now distribute a second time: Replace $x$ with $A$ and $y$ with $B$ to get:



    $(x+y)(x+y) = M(x+y) =$



    $Mx + My = $



    $(x + y)x + (x+y)y=$



    $(x+y)A + (x+y)B=$



    $xA + yA + xB + yB =$



    $xx + yx + xy + yy =$



    $x^2 + 2xy + y^2$.



    Of you you don't have to, and you shouldn't, do all that replacement. You should do it directly.



    $(x+y)(x+y)=$ we treat one of the $(x+y)$ as a single thing and distribute across the other $x + y$.



    $(x+y)(x+y) = (x+y)x + (x+y)y=$.



    Now we have to sums to distribute: $(x+y)x = xx + yx$ and $(x + y)y = xy + yy$. So puting them together:



    $(x+y)(x+y) = (x+y)x + (x+y)y=xx + yx + xy + yy=$.



    And then some clean up:



    $=x^2 + 2xy + y^2$






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Because:



      $$left (x+y right) left (x+y right) = underbrace{(x+y)+(x+y)+...(x+y)}_{x + ytext{ times}}$$



      $$=underbrace{(x+y)+(x+y)+...(x+y)}_{xtext{ times}}+underbrace{(x+y)+(x+y)+...(x+y)}_{ytext{ times}}$$



      $$=underbrace{x}_{xtext{ times}}+underbrace{y}_{xtext{ times}}+underbrace{x}_{ytext{ times}}+underbrace{y}_{ytext{ times}}$$



      $$=xx+xy+yx+yy$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately.
        $endgroup$
        – StackTD
        Dec 14 '18 at 16:30










      • $begingroup$
        @StackTD Excellent question! Obviously one needs to do something else in that case ... but I figured this would be in the spirit of the OP
        $endgroup$
        – Bram28
        Dec 14 '18 at 16:32










      • $begingroup$
        I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!).
        $endgroup$
        – StackTD
        Dec 14 '18 at 16:33
















      1












      $begingroup$

      Because:



      $$left (x+y right) left (x+y right) = underbrace{(x+y)+(x+y)+...(x+y)}_{x + ytext{ times}}$$



      $$=underbrace{(x+y)+(x+y)+...(x+y)}_{xtext{ times}}+underbrace{(x+y)+(x+y)+...(x+y)}_{ytext{ times}}$$



      $$=underbrace{x}_{xtext{ times}}+underbrace{y}_{xtext{ times}}+underbrace{x}_{ytext{ times}}+underbrace{y}_{ytext{ times}}$$



      $$=xx+xy+yx+yy$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately.
        $endgroup$
        – StackTD
        Dec 14 '18 at 16:30










      • $begingroup$
        @StackTD Excellent question! Obviously one needs to do something else in that case ... but I figured this would be in the spirit of the OP
        $endgroup$
        – Bram28
        Dec 14 '18 at 16:32










      • $begingroup$
        I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!).
        $endgroup$
        – StackTD
        Dec 14 '18 at 16:33














      1












      1








      1





      $begingroup$

      Because:



      $$left (x+y right) left (x+y right) = underbrace{(x+y)+(x+y)+...(x+y)}_{x + ytext{ times}}$$



      $$=underbrace{(x+y)+(x+y)+...(x+y)}_{xtext{ times}}+underbrace{(x+y)+(x+y)+...(x+y)}_{ytext{ times}}$$



      $$=underbrace{x}_{xtext{ times}}+underbrace{y}_{xtext{ times}}+underbrace{x}_{ytext{ times}}+underbrace{y}_{ytext{ times}}$$



      $$=xx+xy+yx+yy$$






      share|cite|improve this answer











      $endgroup$



      Because:



      $$left (x+y right) left (x+y right) = underbrace{(x+y)+(x+y)+...(x+y)}_{x + ytext{ times}}$$



      $$=underbrace{(x+y)+(x+y)+...(x+y)}_{xtext{ times}}+underbrace{(x+y)+(x+y)+...(x+y)}_{ytext{ times}}$$



      $$=underbrace{x}_{xtext{ times}}+underbrace{y}_{xtext{ times}}+underbrace{x}_{ytext{ times}}+underbrace{y}_{ytext{ times}}$$



      $$=xx+xy+yx+yy$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 14 '18 at 16:30

























      answered Dec 14 '18 at 16:24









      Bram28Bram28

      62.9k44793




      62.9k44793












      • $begingroup$
        If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately.
        $endgroup$
        – StackTD
        Dec 14 '18 at 16:30










      • $begingroup$
        @StackTD Excellent question! Obviously one needs to do something else in that case ... but I figured this would be in the spirit of the OP
        $endgroup$
        – Bram28
        Dec 14 '18 at 16:32










      • $begingroup$
        I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!).
        $endgroup$
        – StackTD
        Dec 14 '18 at 16:33


















      • $begingroup$
        If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately.
        $endgroup$
        – StackTD
        Dec 14 '18 at 16:30










      • $begingroup$
        @StackTD Excellent question! Obviously one needs to do something else in that case ... but I figured this would be in the spirit of the OP
        $endgroup$
        – Bram28
        Dec 14 '18 at 16:32










      • $begingroup$
        I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!).
        $endgroup$
        – StackTD
        Dec 14 '18 at 16:33
















      $begingroup$
      If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately.
      $endgroup$
      – StackTD
      Dec 14 '18 at 16:30




      $begingroup$
      If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately.
      $endgroup$
      – StackTD
      Dec 14 '18 at 16:30












      $begingroup$
      @StackTD Excellent question! Obviously one needs to do something else in that case ... but I figured this would be in the spirit of the OP
      $endgroup$
      – Bram28
      Dec 14 '18 at 16:32




      $begingroup$
      @StackTD Excellent question! Obviously one needs to do something else in that case ... but I figured this would be in the spirit of the OP
      $endgroup$
      – Bram28
      Dec 14 '18 at 16:32












      $begingroup$
      I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!).
      $endgroup$
      – StackTD
      Dec 14 '18 at 16:33




      $begingroup$
      I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!).
      $endgroup$
      – StackTD
      Dec 14 '18 at 16:33











      0












      $begingroup$


      $4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$




      This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.




      but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?




      The distributive law works even if $x$ and/or $y$ are not integers.



      Keep one of the factors together in a first step, and apply distributivity twice:
      $$begin{align}
      (color{blue}{x}+color{red}{y})(x+y) & =color{blue}{x}(x+y)+color{red}{y}(x+y) \ & =color{blue}{x}x+color{blue}{x}y+color{red}{y}x+color{red}{y}y
      end{align}$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$


        $4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$




        This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.




        but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?




        The distributive law works even if $x$ and/or $y$ are not integers.



        Keep one of the factors together in a first step, and apply distributivity twice:
        $$begin{align}
        (color{blue}{x}+color{red}{y})(x+y) & =color{blue}{x}(x+y)+color{red}{y}(x+y) \ & =color{blue}{x}x+color{blue}{x}y+color{red}{y}x+color{red}{y}y
        end{align}$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$


          $4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$




          This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.




          but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?




          The distributive law works even if $x$ and/or $y$ are not integers.



          Keep one of the factors together in a first step, and apply distributivity twice:
          $$begin{align}
          (color{blue}{x}+color{red}{y})(x+y) & =color{blue}{x}(x+y)+color{red}{y}(x+y) \ & =color{blue}{x}x+color{blue}{x}y+color{red}{y}x+color{red}{y}y
          end{align}$$






          share|cite|improve this answer









          $endgroup$




          $4left (x+y right)=4x+4y $ because $4left (x+y right) =left (x+y right) +left (x+y right) +left (x+y right) +left (x+y right)$




          This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.




          but why is $left (x+y right) left (x+y right) =xx+xy+yx+yy$?




          The distributive law works even if $x$ and/or $y$ are not integers.



          Keep one of the factors together in a first step, and apply distributivity twice:
          $$begin{align}
          (color{blue}{x}+color{red}{y})(x+y) & =color{blue}{x}(x+y)+color{red}{y}(x+y) \ & =color{blue}{x}x+color{blue}{x}y+color{red}{y}x+color{red}{y}y
          end{align}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 16:20









          StackTDStackTD

          22.9k2151




          22.9k2151























              0












              $begingroup$

              Do it in steps.



              You accept that $M(x+y) = Mx + My$



              So replace $M$ with $(x+y)$ and you get:



              $(x+y)(x+y) = M(x+y) =$



              $Mx + My = $



              $(x + y)x + (x+y)y$



              Now distribute a second time: Replace $x$ with $A$ and $y$ with $B$ to get:



              $(x+y)(x+y) = M(x+y) =$



              $Mx + My = $



              $(x + y)x + (x+y)y=$



              $(x+y)A + (x+y)B=$



              $xA + yA + xB + yB =$



              $xx + yx + xy + yy =$



              $x^2 + 2xy + y^2$.



              Of you you don't have to, and you shouldn't, do all that replacement. You should do it directly.



              $(x+y)(x+y)=$ we treat one of the $(x+y)$ as a single thing and distribute across the other $x + y$.



              $(x+y)(x+y) = (x+y)x + (x+y)y=$.



              Now we have to sums to distribute: $(x+y)x = xx + yx$ and $(x + y)y = xy + yy$. So puting them together:



              $(x+y)(x+y) = (x+y)x + (x+y)y=xx + yx + xy + yy=$.



              And then some clean up:



              $=x^2 + 2xy + y^2$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Do it in steps.



                You accept that $M(x+y) = Mx + My$



                So replace $M$ with $(x+y)$ and you get:



                $(x+y)(x+y) = M(x+y) =$



                $Mx + My = $



                $(x + y)x + (x+y)y$



                Now distribute a second time: Replace $x$ with $A$ and $y$ with $B$ to get:



                $(x+y)(x+y) = M(x+y) =$



                $Mx + My = $



                $(x + y)x + (x+y)y=$



                $(x+y)A + (x+y)B=$



                $xA + yA + xB + yB =$



                $xx + yx + xy + yy =$



                $x^2 + 2xy + y^2$.



                Of you you don't have to, and you shouldn't, do all that replacement. You should do it directly.



                $(x+y)(x+y)=$ we treat one of the $(x+y)$ as a single thing and distribute across the other $x + y$.



                $(x+y)(x+y) = (x+y)x + (x+y)y=$.



                Now we have to sums to distribute: $(x+y)x = xx + yx$ and $(x + y)y = xy + yy$. So puting them together:



                $(x+y)(x+y) = (x+y)x + (x+y)y=xx + yx + xy + yy=$.



                And then some clean up:



                $=x^2 + 2xy + y^2$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Do it in steps.



                  You accept that $M(x+y) = Mx + My$



                  So replace $M$ with $(x+y)$ and you get:



                  $(x+y)(x+y) = M(x+y) =$



                  $Mx + My = $



                  $(x + y)x + (x+y)y$



                  Now distribute a second time: Replace $x$ with $A$ and $y$ with $B$ to get:



                  $(x+y)(x+y) = M(x+y) =$



                  $Mx + My = $



                  $(x + y)x + (x+y)y=$



                  $(x+y)A + (x+y)B=$



                  $xA + yA + xB + yB =$



                  $xx + yx + xy + yy =$



                  $x^2 + 2xy + y^2$.



                  Of you you don't have to, and you shouldn't, do all that replacement. You should do it directly.



                  $(x+y)(x+y)=$ we treat one of the $(x+y)$ as a single thing and distribute across the other $x + y$.



                  $(x+y)(x+y) = (x+y)x + (x+y)y=$.



                  Now we have to sums to distribute: $(x+y)x = xx + yx$ and $(x + y)y = xy + yy$. So puting them together:



                  $(x+y)(x+y) = (x+y)x + (x+y)y=xx + yx + xy + yy=$.



                  And then some clean up:



                  $=x^2 + 2xy + y^2$






                  share|cite|improve this answer









                  $endgroup$



                  Do it in steps.



                  You accept that $M(x+y) = Mx + My$



                  So replace $M$ with $(x+y)$ and you get:



                  $(x+y)(x+y) = M(x+y) =$



                  $Mx + My = $



                  $(x + y)x + (x+y)y$



                  Now distribute a second time: Replace $x$ with $A$ and $y$ with $B$ to get:



                  $(x+y)(x+y) = M(x+y) =$



                  $Mx + My = $



                  $(x + y)x + (x+y)y=$



                  $(x+y)A + (x+y)B=$



                  $xA + yA + xB + yB =$



                  $xx + yx + xy + yy =$



                  $x^2 + 2xy + y^2$.



                  Of you you don't have to, and you shouldn't, do all that replacement. You should do it directly.



                  $(x+y)(x+y)=$ we treat one of the $(x+y)$ as a single thing and distribute across the other $x + y$.



                  $(x+y)(x+y) = (x+y)x + (x+y)y=$.



                  Now we have to sums to distribute: $(x+y)x = xx + yx$ and $(x + y)y = xy + yy$. So puting them together:



                  $(x+y)(x+y) = (x+y)x + (x+y)y=xx + yx + xy + yy=$.



                  And then some clean up:



                  $=x^2 + 2xy + y^2$







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                  answered Dec 14 '18 at 16:51









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