Sum of Floors not greater than Floor of Sum

Theorem
Let $\floor x$ denote the floor function.

Then:
 * $\floor x + \floor y \le \floor {x + y}$

The equality holds:
 * $\floor x + \floor y = \floor {x + y}$


 * $x \bmod 1 + y \bmod 1 < 1$
 * $x \bmod 1 + y \bmod 1 < 1$

where $x \bmod 1$ denotes the modulo operation.

Proof
From the definition of the modulo operation, we have that:
 * $x = \floor x + \paren {x \bmod 1}$

from which:

Hence the inequality.

The equality holds :
 * $\floor {\paren {x \bmod 1} + \paren {y \bmod 1} } = 0$

that is, :
 * $x \bmod 1 + y \bmod 1 < 1$