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:
$$[a]+[b] \le [a+b]$$

Define $\left\{ a\right\} ,\left\{ b\right\}$ 
  the fractional part of $a$
  and $b$
  and assume $\left\{ a\right\} +\left\{ b\right\} <1$
 . Then$$\left\{ a+b\right\} =\left\{ \left[a\right]+\left[b\right]+\left\{ a\right\} +\left\{ b\right\} \right\} =\left\{ \left\{ a\right\} +\left\{ b\right\} \right\} =\left\{ a\right\} +\left\{ b\right\}.$$
 So$$\left[a+b\right]=a+b-\left\{ a+b\right\} =a+b-\left\{ a\right\} -\left\{ b\right\} =\left[a\right]+\left[b\right].$$
 Assume now $\left\{ a\right\} +\left\{ b\right\} \geq1$
  , then$$\left\{ a+b\right\} =\left\{ \left[a\right]+\left[b\right]+\left\{ a\right\} +\left\{ b\right\} \right\} =\left\{ \left\{ a\right\} +\left\{ b\right\} \right\} =\left\{ a\right\} +\left\{ b\right\} -1$$
 so$$\left[a+b\right]=a+b-\left\{ a+b\right\} =a+b-\left\{ a\right\} -\left\{ b\right\} +1>\left[a\right]+\left[b\right].$$
