How do I prove this function is monotonic?

Question

Let $f:\mathbb R\to \mathbb R$ be a function such that $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for every $x,y\in \mathbb R$ and $f(1)=1$. In order to prove this function is 1-1, I just need to prove this function is monotonic.

Anyone has some ideas how to proceed?

Thanks

Answer 1

We'll show that $f$ is monotone increasing.

Notice that if $x\geq 0$ then $f(x)=f(\sqrt{x})^2\geq 0$.

Thus if $x\geq y$, then $x-y \geq 0$, so $f(x)-f(y) = f(x-y) \geq 0$, so that $f(x) \geq f(y)$.

Answer 2

It is not necessary, as data of this question, the condition f(1) = 1 which is easily deduced as well as f(0) = 0. This is important to get f(-x) = -f(x) and f(1/x) = 1/f(x) and justify the steps f(x-y) = f(x) - f(y) and f (x/y) = f(x)/f(y)