# Group/Examples/x+y+xy over Reals less -1

## Example of Group

Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as:

- $\forall x, y \in \R: x \circ y := x + y + x y$

Let:

- $\R' := \R \setminus \set {-1}$

that is, the set of real numbers without $-1$.

Then $\struct {\R', \circ}$ is a group whose identity is $0$.

## Proof

From Monoid Example: Operation Defined as $x + y + x y$ on Real Numbers, $\struct {\R, \circ}$ is a monoid whose identity is $0$.

Let $x \in \R$.

We have that:

\(\displaystyle \paren {-1} \circ x\) | \(=\) | \(\displaystyle \paren {-1} + x + \paren {-1} \times x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x + \paren {-1} + x \times \paren {-1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x \circ \paren {-1}\) |

demonstrating that $-1$ is a zero for $\circ$ on $\R$.

Thus for $\struct {S, \circ}$ to be a group, where $S \subseteq \R$, it is necessary for $-1 \notin S$.

It has been established that $0$ is the identity of $\struct {\R, \circ}$.

Let $y \in \R'$.

Then:

\(\displaystyle x \circ y\) | \(=\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x + y + x y\) | \(=\) | \(\displaystyle 0\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y\) | \(=\) | \(\displaystyle -\dfrac x {1 + x}\) |

Thus the inverse of $x$ is $-\dfrac x {1 + x}$, which exists for all $x$ except for $x = -1$.

Suppose $x, y \in \R \setminus \set {-1}$ such that $x \circ y = -1$.

Then:

\(\displaystyle x \circ y\) | \(=\) | \(\displaystyle -1\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x + y + x y\) | \(=\) | \(\displaystyle -1\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y\) | \(=\) | \(\displaystyle \dfrac {-\paren {1 + x} } {1 + x}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y\) | \(=\) | \(\displaystyle -1\) | ||||||||||

\(\displaystyle \) | \(\notin\) | \(\displaystyle \R \setminus \set {-1}\) |

So we have that $\struct {\R', \circ}$ is closed.

Thus all the group axioms are fulfilled, and so $\struct {\R', \circ}$ is a group.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $4$. Groups: Exercise $11$