Sum of Floors not greater than Floor of Sum

Theorem
Let $$\left \lfloor {x} \right \rfloor$$ be the floor function.

Then:
 * $$\left \lfloor {x} \right \rfloor + \left \lfloor {y} \right \rfloor \le \left \lfloor {x + y} \right \rfloor$$

The equality holds:
 * $$\left \lfloor {x} \right \rfloor + \left \lfloor {y} \right \rfloor = \left \lfloor {x + y} \right \rfloor$$

iff:
 * $$x \,\bmod\, 1 + y \,\bmod\, 1 < 1$$

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

Proof
From the definition of the modulo operation, we have that:
 * $$x = \left \lfloor {x}\right \rfloor + \left({x \, \bmod \, 1}\right)$$

from which we obtain:

$$ $$

Hence the inequality.

The equality holds iff:
 * $$\left \lfloor {\left({x \, \bmod \, 1}\right) + \left({y \, \bmod \, 1}\right)} \right \rfloor = 0$$

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