Floor equals Ceiling iff Integer

Theorem

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

Let $\floor x$ denote the floor of $x$, and $\ceiling x$ denote the ceiling of $x$.

Then:

$\floor x = \begin {cases} \ceiling x & : x \in \Z \\ \ceiling x - 1 & : x \notin \Z \\ \end {cases}$

or equivalently:

$\ceiling x = \begin {cases} \floor x & : x \in \Z \\ \floor x + 1 & : x \notin \Z \\ \end {cases}$

where $\Z$ is the set of integers.

$x \in \Z \implies x = \floor x$

$x \in \Z \implies x = \ceiling x$

So:

$x \in \Z \implies \floor x = \ceiling x$

Now let $x \notin \Z$.

From the definition of the floor function:

$\floor x = \map \sup {\set {m \in \Z: m \le x} }$

From the definition of the ceiling function:

$\ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$

Thus:

$\floor x < x < \ceiling x$

Hence the result, from the definition of $\inf$ and $\sup$.

$\blacksquare$

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
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): integer part
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): integer part