How to prove that $\lfloor a\rfloor+\lfloor b\rfloor\leq \lfloor a+b\rfloor$ We have the floor function $F(x)=[x]$ such that $F(2.5)=[2.5]=2, F(-3.5)=[-3.5]=-4, F(2)=[2]=2$. 
How can I prove the following property of floor function:
$$[a]+[b] \le [a+b]$$

Consider the algebraic representation of $\mathrm{floor}(x)=\lfloor x\rfloor$ (usually the floor of $x$ is represented by $\lfloor x\rfloor$, not $[x]$):
$$
x-1<\color{blue}{\lfloor x\rfloor\leq x}.
$$
Hence, for arbitrary $a$ and $b$, we have
$$
\lfloor a\rfloor\leq a\tag{1}
$$
and
$$
\lfloor b\rfloor\leq b.\tag{2}
$$
Now simply add $(1)$ and $(2)$ together to get
$$
\lfloor a\rfloor+\lfloor b\rfloor\leq a+b.\tag{3}
$$
Finally, take the floor of both sides of $(3)$:
$$
\lfloor a+b\rfloor\geq\Bigl\lfloor\lfloor a\rfloor+\lfloor b\rfloor\Bigr\rfloor=\lfloor a\rfloor+\lfloor b\rfloor.
$$
Hence, we have that $\lfloor a\rfloor+\lfloor b\rfloor\leq \lfloor a+b\rfloor$, as desired. 
