Prove that $x * y = \frac{x+y}{1+xy}$ is a stable part of $G=(-1, 1)$

I have to prove that the result of $x * y \in G$ so $\frac{x+y}{1+xy} \in (-1, 1)$. So $x > -1$ and $y > -1$ at the same time $x < 1$ and $y < 1$.

If I multiply the first 2 expressions I obtain $xy < 1$ which is true, and if I sum up the last 2 I get $x + y < 2$. At this point I am not sure if I am doing well. How should I continue this problem?

Solution 1:

If $x,y\in (-1,1)$ then we can find $\alpha,\beta$ such that $x = \tanh(\alpha)$ and $y=\tanh(\beta)$ and it follows that

$$x*y = \frac{x+y}{1+xy} = \tanh(\alpha+\beta) \in (-1,1)$$

As a sidenote: the operation $*$ is what defines velocity addition in special relativity (when working in units of $c=1$).

Solution 2:

Hint: Consider the product $(1-x)(1-y)$ or $(x-1)(y-1)$ and think about its sign based on the fact that $x,y \in (-1,1)$.