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:
\(\ds \paren {-1} \circ x\) | \(=\) | \(\ds \paren {-1} + x + \paren {-1} \times x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x + \paren {-1} + x \times \paren {-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {-1}\) |
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:
\(\ds x \circ y\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x + y + x y\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds -\dfrac x {1 + x}\) |
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:
\(\ds x \circ y\) | \(=\) | \(\ds -1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x + y + x y\) | \(=\) | \(\ds -1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {-\paren {1 + x} } {1 + x}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds -1\) | |||||||||||
\(\ds \) | \(\notin\) | \(\ds \R \setminus \set {-1}\) |
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 $11$