I am asked to prove that the following function is one-to-one.

$\displaystyle f(x)=\sqrt(x), x \geq 0$

I understand that the proof template for this is:

Suppose f(x) = f(y) ... then x = y.

So, is this sufficient?

Suppose $\displaystyle f(x) = f(y)$. Then $\displaystyle \sqrt(x) = \sqrt(y)$. Square both sides, and we have that $\displaystyle x = y$. Therefore, $\displaystyle f(x)$ is one-to-one.