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:
Group Axiom $\text G 0$: Closure
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.
$\Box$
Group Axiom $\text G 1$: Associativity
The associativity of $\circ$ follows from that of the sum of real numbers.
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Thus $\circ$ is associative on $S$.
$\Box$
Group Axiom $\text G 2$: Existence of Identity Element
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:
| \(\ds x \circ 0\) | \(=\) | \(\ds x + 0 - \floor {x + 0}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x - \floor x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x - 0\) | from above | |||||||||||
| \(\ds \) | \(=\) | \(\ds x\) |
Hence $\struct {S, \circ}$ has $0$ as an identity element.
$\Box$
Group Axiom $\text G 3$: Existence of Inverse 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:
| \(\ds x \circ \paren {1 - x}\) | \(=\) | \(\ds x + 1 - x - \floor {x + 1 - x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - \floor 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - 1\) | Real Number is Integer iff equals Floor | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
From above, $0$ is the identity of $\struct {S, \circ}$.
Thus every element $x$ of $\struct {S, \circ}$ has an inverse:
- $\begin {cases} 1 - x & : x \ne 0 \\ 0 & : x = 0 \end {cases}$
$\Box$
All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.
$\blacksquare$