Real Number is Floor plus Difference

Theorem

Let $x \in \R$ be a real number.

Let $\floor x$ be the floor of $x$.

Then:

There exists an integer $n \in \Z$ such that for some $t \in \hointr 0 1$:

$x = n + t$

if and only if:

$n = \floor x$

where:

$\hointr 0 1$ is the real interval half open on the right from $0$ to $1$

$\floor x$ is the floor of $x$

Proof

Sufficient Condition

Let there exist $n \in \Z$ such that $x = n + t$, where $t \in \hointr 0 1$.

We have that $1 - t > 0$.

Thus:

$0 \le x - n < 1$

Thus:

$n \le x < n + 1$

That is, $n$ is the floor of $x$.

$\Box$

Necessary Condition

Let $n = \floor x$.

Let $t = x - \floor x$.

Then $x = n + t$.

From Real Number minus Floor:

$t = x - \floor x \in \hointr 0 1$

and so:

$x = n + t: t \in \hointr 0 1$

$\blacksquare$

Also see

• Real Number is Ceiling minus Difference
• Definition:Fractional Part

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 10.4 \ \text{(iii)}$: The well-ordering principle