# 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.

### Isomorphic to Real Numbers

$\struct {G, \circ}$ is isomorphic to the additive group of real numbers $\struct {\R, +}$.

## Proof

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

We check the group axioms in turn:

### G1: Associativity

 $\displaystyle \paren {x \circ y} \circ z$ $=$ $\displaystyle \frac {\frac {x + y} {1 + x y} + z} {1 + \frac {x + y} {1 + xy} z}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {x + y + z + x y z} {1 + x y + x z + y z}$ $\quad$ $\quad$ $\displaystyle x \circ \paren {y \circ z}$ $=$ $\displaystyle \frac {x + \frac {y + z} {1 + y z} } {1 + x \frac {y + z} {1 + y z} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {x + y + z + x y z} {1 + x y + x z + y z}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {x \circ y} \circ z$ $\quad$ $\quad$

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

$\Box$

### G2: Identity

 $\displaystyle x$ $=$ $\displaystyle 0$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle \frac {x + y} {1 + x y}$ $=$ $\displaystyle \frac {0 + y} {1 + 0 y}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac y 1$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle y$ $\quad$ $\quad$

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

So $0$ is the identity.

$\Box$

### G3: Inverses

 $\displaystyle x \circ -x$ $=$ $\displaystyle \frac {x + \paren {-x} } {1 + x \paren {-x} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$

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

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

$\Box$

### G0: Closure

First note that:

$-1 < x, y < 1 \implies x y > -1 \implies 1 + x y > 0$

Next:

 $\displaystyle -1$ $<$ $\displaystyle x, y < 1$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle \paren {1 - x} \paren {1 - y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle 1 + x y - \paren {x + y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle \frac {1 + x y - \paren {x + y} } {1 + x y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle \frac {1 + x y} {1 + x y} - \frac {x + y} {1 + x y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle 1 - \frac {x + y} {1 + x y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {x + y} {1 + x y}$ $<$ $\displaystyle 1$ $\quad$ $\quad$

Finally:

 $\displaystyle -1$ $<$ $\displaystyle x, y < 1$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle \paren {1 + x} \paren {1 + y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle 1 + x y + \paren {x + y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle \frac {1 + x y + \paren {x + y} } {1 + x y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle \frac {1 + x y} {1 + x y} + \frac {x + y} {1 + x y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $<$ $\displaystyle 1 + \frac {x + y} {1 + x y}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle -1$ $<$ $\displaystyle \frac {x + y} {1 + x y}$ $\quad$ $\quad$

Thus:

$-1 < x, y < 1 \implies -1 < x \circ y < 1$

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

$\Box$

Thus the given set and operation form a group.

$\blacksquare$