Integer equals Floor iff between Number and One Less

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:

$\floor x = n \iff x - 1 < n \le x$

Proof

Necessary Condition

Let $x - 1 < n \le x$.

From $n \le x$, we have by Number not less than Integer iff Floor not less than Integer:

$n \le \floor x$

From $x - 1 < n$:

$x < n + 1$

Hence by Number less than Integer iff Floor less than Integer:

$\floor x < n + 1$

We have that:

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

and so:

$\floor x \le n$

Thus as:

$n \le \floor x$

and:

$\floor x \le n$

it follows that:

$\floor x = n$

$\Box$

Sufficient Condition

Let $\floor x = n$.

Then:

$\floor x \ge n$

By Number not less than Integer iff Floor not less than Integer:

$n \le x$

By definition of the floor of $x$:

$x < \floor x + 1$

and so subtracting $1$ from both sides:

$x - 1 < \floor x$

and so by hypothesis:

$x - 1 < n$

So:

$\floor x = n \implies x - 1 < n \le x$

$\Box$

Hence the result:

$\floor x = n \iff x - 1 < n \le x$

$\blacksquare$

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 $3 \ \text{(e)}$