Real Numbers under Addition Modulo 1 form Group

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

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

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

Then $\struct {S, \circ}$ is a group.

Proof
First note that Modulo Addition is Well-Defined.

Taking the group axioms in turn:

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

Thus $\struct {S, \circ}$ is closed.

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

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

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:
 * $\floor x = 0$

So:

Hence $\struct {S, \circ}$ has $0$ as an identity element.

Let $x \in S$.

First let $x = 0$.

We have that $0$ is the identity of $\struct {S, \circ}$:
 * $0 \circ 0 = 0$

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 above, $0$ is the identity of $\struct {S, \circ}$.

Thus every element $x$ of $\struct {S, \circ}$ 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 $\struct {S, \circ}$ is a group.