# Does $(xy)^2 = x^2 \times y^2?$

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Is this true or false?

$(xy)^2=x^2 \times y^2$

**Why some people say it's true:** It's easy: $(xy)^2=(xy)(xy)=xxyy=x^2 \times y^2.$

**Why some people say it's false:** It's not that easy... there must be something I have missed...

The statement is $\color{#20A900}{\textbf{conditionally true}}$.

The statement is true if the operator $\times$ is commutative $($that is, $x\times y=y\times x \; \forall x,y)$ and associative $($that is, $x(yz)=(xy)z\;\forall x,y,z).$ This is always true with real numbers, but not always for imaginary numbers.

We have $(xy)^2=(xy)(xy)=x{\color{#D61F06}{yx}} y=x{\color{#D61F06}{xy}}y=x^2 \times y^2.\ _\square$

For non-commutative operators under some algebraic structure, it is not always true:

Let $\mathbb Q$ be the set of quaternions, and let $x=i,y=j\in\mathbb Q$.

Then $(xy)^2=(ij)^2=k^2=-1.$

However, $x^2\times y^2=i^2\times j^2=(-1) \times (-1)=1$.

This is because $ij \neq ji$.

In fact, $ij=-ji$.

Let $S_3$ be the symmetry group of $3$ elements, and let $x=(1\;2),y=(2\;3)$ in cycle notation:

- $(xy)^2=\big((1\;2)(2\;3)\big)^2=(1\;3\;2)^2=(1\;3\;2)(1\;3\;2)=(1\;2\;3)$
- $x^2\times y^2=(1\;2)^2(1\;3)^2=()()=(),$
which are clearly unequal.

**Cite as:**Does $(xy)^2 = x^2 \times y^2?$.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/is-a-plus-b-squared-equal-a-squared-plus-b-squared/