State/The floor sum identity

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Proposition
Proof

Proposition: (The floor sum identity) Suppose that $u$ and $v$ are rational numbers (or more generally, real numbers. Then, $$\lfloor u+v \rfloor - \left( \lfloor u \rfloor + \lfloor v \rfloor \right) \in \{ 0, 1 \}.$$

Logical Connections

This statement logically relies on the following definitions and statements: Def/Rational number

The following statements and definitions rely on the material of this page: State/Canonical decompositions of binomial coefficients

To visualize the logical connections between this statements and other items of mathematical knowledge, you can visit the following cluster(s), and click the "Visualize" tab: Clust/Basic number theory, Clust/Basic counting

State/The floor sum identity

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	Suppose that $u,v \in \RR$.
	Then, we find the following three facts: $\lfloor u+v \rfloor > u+v-1$. $\lfloor u \rfloor \leq u$. $\lfloor v \rfloor \leq v$. Hence we find that: $$\lfloor u+v \rfloor - (\lfloor u \rfloor + \lfloor v \rfloor) > u+v-1 - (u+v) = -1.$$ On the other hand, we also find the following three facts: $\lfloor u+v \rfloor \leq u+v$. $\lfloor u \rfloor > u-1$. $\lfloor v \rfloor > v-1$. Hence we find that: $$\lfloor u+v \rfloor - (\lfloor u \rfloor + \lfloor v \rfloor) < u+v - (u-1 + v-1) = 2.$$ Hence we find that: $$-1 < \lfloor u+v \rfloor - (\lfloor u \rfloor + \lfloor v \rfloor) < 2.$$ Since the middle quantity is an integer,
	we find that: $$ \lfloor u+v \rfloor - (\lfloor u \rfloor + \lfloor v \rfloor) \in \{0,1 \}.$$