Group/Examples/inv x = 1 - x

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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 1

$\map {f \circ g} x = x$


Lemma 2

$\map {g \circ f} x = x$



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:

\(\displaystyle x \circ y\) \(=\) \(\displaystyle \map g {\map f x + \map f y}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1 + \map \exp {\map \log {\frac {1 - x} x} + \map \log {\frac {1 - y} y} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1 + \map \exp {\map \log {\frac {1 - x} x} } \map \exp {\map \log {\frac {1 - y} y} } }\) Exponential of Sum
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1 + \paren {\frac {1 - x} x} \paren {\frac {1 - y} y} }\) Exponential of Natural Logarithm

We see that $\circ$ is commutative.

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

\(\displaystyle \dfrac 1 2 \circ y\) \(=\) \(\displaystyle \dfrac 1 {1 + \paren {\frac {1 - y} y} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\frac y y + \frac {1 - y} y}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\frac 1 y}\)
\(\displaystyle \) \(=\) \(\displaystyle y\)

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

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

\(\displaystyle \frac 1 {1 + \paren {\frac {1 - x} x} \paren {\frac x {1 - x} } }\) \(=\) \(\displaystyle \frac 1 {1 + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2\)

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.

$\blacksquare$


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