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
By definition of $S$:


 * $0 \in S$

Let $x \in S$.

We have that:
 * $0 \le x < 1$

and so by definition of floor function:
 * $\left\lfloor{x}\right\rfloor = 0$

So:

Hence $\left({S, \circ}\right)$ has $0$ as an identity element.

G3: Inverses
Let $x \in S$.

First let $x = 0$.

We have that $0$ is the identity of $\left({S, \circ}\right)$:
 * $0 \circ 0 = 0$

and so from Inverse of Identity Element is Itself, $0$ is its own inverse.

Now let $x \ne 0$.

By definition of $S$:
 * $0 \le x < 1$

Hence:
 * $1 - x \in S$

(Note that because $1 - 0 \notin S$, the above is not true for $x = 0$, which is why it has been treated as a special case.)

Therefore:

From G3: Identity, $0$ is the identity of $\left({S, \circ}\right)$.

Thus every element $x$ of $\left({S, \circ}\right)$ has an inverse:
 * $\begin{cases} 1 - x & : x \ne 0 \\ 0 & : x = 0 \end{cases}$

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