Group/Examples/inv x = 1 - x

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

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

Proof

 * The first thing we can do is find the identity.

From Identities all Self-Inverse, the identity must satisfy $$x = \left({1-x}\right)$$, which means the identity must be $$1/2$$.


 * Now we investigate an equation $$x \circ \left({1 - x}\right) = 1/2$$.

Symmetry of the set about the element $$x = 1/2$$ suggests we might want to average. So try it out:

Let $$\ast$$ be defined as $$x \ast y = \left({x + y}\right) / 2$$.


 * Next, note that $$\left({x + \left({1 - x}\right)}\right) / 2 = 1/2$$ as required.


 * It is straightforward to check that $$\left({S, \ast}\right)$$ is closed.


 * Note, however, that:


 * $$x \ast \left({y \ast z}\right) = \left({2 x + y + z} / 2\right)$$
 * $$\left({x \ast y}\right) \ast z = \left({x + y + 2 z} / 2\right)$$

Thus $$x \ast \left({y \ast z}\right) \not \equiv \left({x \ast y}\right) \ast z$$ and $$\ast$$ is not associative.

So $$\ast$$ can not be the operation we are after.