Group/Examples/inv x = 1 - x

Theorem
Let $S = \set {x \in \R: 0 < x < 1}$.

Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a group such that the inverse of $x \in S$ is $1 - x$.

Proof
Define $f: \openint 0 1 \to \R$ by:


 * $\map f x := \map \ln {\dfrac {1 - x} x}$

Let us show that $f$ is a bijection by constructing an inverse mapping $g: \R \to \openint 0 1$:


 * $\map g z := \dfrac 1 {1 + \exp z}$

Lemma 2
Thus $f$ is a bijection.

Let $\struct {\R, +}$ be the additive group on $\R$.

Now define $\circ := +_f$ to be the operation induced on $\openint 0 1$ by $f$ and $+$:


 * $x \circ y := \map {f^{-1} } {\map f x + \map f y}$

Let us determine the behaviour of $\circ$ more explicitly:

We see that $\circ$ is commutative.

Let $x = \dfrac 1 2$, $\dfrac {1 - x} x = 1$ so that:

so that $\dfrac 1 2$ is the identity element for $\circ$.

Furthermore, putting $y = 1 - x$, the following obtains:

establishing $1 - x$ to be the inverse of $x$, as desired.

That $\circ$ in fact determines a group on $\openint 0 1$ follows from Pullback of Group is Group.