Real Numbers under Addition Modulo 1 form Group

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

Let $\circ: S \times S \to S$ be the operation defined as:
 * $x \circ y = x + y - \left \lfloor {x + y} \right \rfloor$

That is, $\circ$ is defined as addition modulo $1$.

Then $\left({S, \circ}\right)$ is a group.

Proof
First note that Modulo Addition is Well-Defined.

Taking the group axioms in turn:

G0: Closure
In Real Number Minus Floor it is demonstrated that:
 * $\forall x, y \in S: x \circ y \in S$

Thus $\left({S, \circ}\right)$ is closed.

G1: Associativity
The associativity of $\circ$ follows from that of the sum of real numbers.

Thus $\circ$ is associative on $S$.

G2: Identity
There is an identity element of $\circ$ which is $0$.

Thus ... has an identity element.

G3: Inverses
The inverse of $x$ is $1 - x - \left\lfloor{1 - x}\right\rfloor$, so $x^{-1} = 1 - x$ if $x > 0$

Thus every element of $\left({S, \circ}\right)$ has an inverse.

All the group axioms are thus seen to be fulfilled, and so $\left({S, \circ}\right)$ is a group.