Is it valid to simplify $(x-y)^2 > 0$ to $x-y > 0$

Question

Is it valid to simplify $(x-y)^2 > 0$ to $x-y > 0$ by taking the square root of both sides ? and if not, why?

Answer 1

We have that

$$(x-y)^2 > 0 \iff (x-y)(x-y) > 0 \iff (x-y)>0 \lor (x-y)<0$$

that is

$$(x-y)^2 > 0 \iff (x-y)\neq 0$$

Answer 2

No, it is not valid, because $\sqrt{(x-y)^2}=\lvert x-y\rvert$, not $x-y$. Besides, if it was valid, since we always have $(x-y)^2>0$ unless $x\neq y$, it would always be true that $x-y>0$, that is, it would always be true that $x>y$ (again, unless $x\neq y$). Doesn't that strike you as a bit odd?

Answer 3

The inequation $(x-y)^2>0$ just says that $x\ne y$ because a square is always non-negative.

This can also be expressed as

$$x<y\lor x>y.$$