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:
$$[a]+[b] \le [a+b]$$
 A: 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. 
A: By the definition of $[x]$ as the greatest integer less than or equal to $x$, we have
$$[a]\le a\quad\text{and}\quad [b]\le b$$
and thus
$$[a]+[b]\le a+b$$
Now $[a]+[b]$ is certainly an integer (because it's the sum of two integers).  Therefore, by the definition of $[x]$ as the greatest integer less than or equal to $x$, we have
$$[a]+[b]\le [a+b]$$
A: 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$$\left[a+b\right]=a+b-\left\{ a+b\right\} =a+b-\left\{ a\right\} -\left\{ b\right\} +1>\left[a\right]+\left[b\right].$$
A: First, note that we always have $[x] \le x < [x] + 1$.  Thus, $0 \le x - [x] < 1$ for all real $x$. To show your inequality, let $u = a - [a]$ and $v = b - [b]$.  My claim is that, if $u + v < 1$, then $[a] + [b] = [a + b]$, and if $u + v \ge 1$, then $[a] + [b] < [a + b]$.  Can you show that?
A: Working along similar lines to an answer I gave to another question, note that $f:\mathbb R\rightarrow\mathbb Z$ given by $f(x)=\lfloor x\rfloor$ is an increasing function, that is the identity on the integers. Also note that $f(x)\leq x$. Thus
$$
f(a)+f(b)\leq a+b
$$
and applying the increasing function $f$ on both sides, noting that the LHS is an integer, we have
$$
f(f(a)+f(b))=f(a)+f(b)\leq f(a+b)
$$
A: Let $a,b \in \Bbb{R}$ such that $a = n+x$ and $b = m+y$ where $x,y \in [0,1)$  and $n,m \in \Bbb{Z}$. Then $$\lfloor a\rfloor+\lfloor b\rfloor = \lfloor n+x\rfloor+\lfloor m+y\rfloor \\ = n+m$$ 
Case 1: $x+y\geq 1$. Then $x+y = 1+z$ for $z \in [0,1)$ and $\lfloor a+b\rfloor = n+m+1$, demonstrating that $$\lfloor a\rfloor+\lfloor b\rfloor < \lfloor a+b\rfloor $$
Case 2: $x+y<1$. Then $\lfloor a+b\rfloor = n+m$ so $$\lfloor a\rfloor+\lfloor b\rfloor = \lfloor a+b\rfloor $$ The combination of cases has shown that $$\lfloor a\rfloor+\lfloor b\rfloor \leq \lfloor a+b\rfloor $$ for all $a,b \in \Bbb{R}$.
