Number not less than Integer iff Floor not less than Integer

Theorem

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

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

Let $n \in \Z$ be an integer.

Then:

$x \ge n \iff \floor x \ge n$

Proof

Necessary Condition

Let $\floor x \ge n$.

By definition of the floor of $x$:

$x \ge \floor x$

Hence:

$x \ge n$

$\Box$

Sufficient Condition

Let $x \ge n$.

Aiming for a contradiction, suppose $\floor x < n$.

We have that:

$\forall m, n \in \Z: m < n \iff m + 1 \le n$

Hence:

$\floor x + 1 \le n$

and so by hypothesis:

$\floor x + 1 \le x$

This contradicts the definition of the floor of $x$:

$\floor x + 1 > x$

Thus by Proof by Contradiction:

$\floor x \ge n$

$\Box$

Hence the result:

$\floor x \ge n \iff x \ge n$

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $3 \ \text{(b)}$