Real Number is Floor plus Difference
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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 $t \in \hointr 0 1$:
- $x = n + t$
- $n = \floor x$
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
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 10.4 \ \text{(iii)}$: The well-ordering principle