Group/Examples/x+y over 1+xy

Theorem
Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.

Let $\circ: G \times G \to \R$ be the binary operation defined as:
 * $\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$

The algebraic structure $\struct {G, \circ}$ is a group.

Proof
Let $-1 < x, y, z < 1$.

We check the group axioms in turn:

Thus $\circ$ has been shown to be associative.

Similarly, putting $y = 0$ we find $x \circ y = x$.

So $0$ is the identity.

Similarly, putting $x = -y$ gives us $\paren {-y} \circ y = 0$.

So each $x$ has an inverse $-x$.

First note that:
 * $-1 < x, y < 1 \implies x y > -1 \implies 1 + x y > 0$

Next:

Finally:

Thus:
 * $-1 < x, y < 1 \implies -1 < x \circ y < 1$

and we see that in this range, $\circ$ is closed.

Thus the given set and operation form a group.