Number less than Integer iff Floor 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:

$\floor x < n \iff x < n$

Proof

Necessary Condition

Let $x < n$.

By definition of the floor of $x$:

$\floor x \le x$

Hence:

$\floor x < n$

$\Box$

Sufficient Condition

Let $\floor x < n$.

We have that:

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

and so:

$(1): \quad \floor x + 1 \le n$

Then:

\(\ds x\)

\(<\)

\(\ds \floor x + 1\)

Definition of Floor Function

\(\ds \)

\(\le\)

\(\ds n\)

from $(1)$

$\Box$

Hence the result:

$\floor x < n \iff x < n$

$\blacksquare$

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 $3 \ \text{(a)}$