Group/Examples/inv x = 1 - x

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

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 operation $\circ: S\times S\to S$ such that $x\circ y=x+y-\operatorname{floor}(x+y)$ is the one we are looking for.


 * $\circ$ is obviously closed by definition because $0\le x\circ y < 1$.


 * There is an identity element which is $0$.


 * The commutativity and associativity of $\circ$ follows from those of the sum of real numbers.


 * The inverse of $x$ is $1-x-\operatorname{floor}(1-x)$, so $inv x=1-x$ if $x>0$

Invalid Example

 * 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:


 * $x \ast \left({y \ast z}\right) = \left({2 x + y + z}\right) / 4$
 * $\left({x \ast y}\right) \ast z = \left({x + y + 2 z}\right) / 4$

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.