# The distributive law

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

• It is also the distributive law or rule. – Dr. Sonnhard Graubner Dec 14 '18 at 16:19
• This is distribution done twice. – Randall Dec 14 '18 at 16:20

Because:

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

$$=\underbrace{(x+y)+(x+y)+...(x+y)}_{x\text{ times}}+\underbrace{(x+y)+(x+y)+...(x+y)}_{y\text{ times}}$$

$$=\underbrace{x}_{x\text{ times}}+\underbrace{y}_{x\text{ times}}+\underbrace{x}_{y\text{ times}}+\underbrace{y}_{y\text{ times}}$$

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

• If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately. – StackTD Dec 14 '18 at 16:30
• @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 – Bram28 Dec 14 '18 at 16:32
• 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!). – StackTD Dec 14 '18 at 16:33

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

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}

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:

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