# 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 \( \color{green}{\textbf{conditionally true}}\).

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 \((xy)^2=(xy)(xy)=x{\color{red}{yx}} y=x{\color{red}{xy}}y=x^2 \times y^2.\ _\square\)

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/