Apostol's Calulus: Prove that $[x+y] = [x]+[y]$ or $[x]+[y]+1$, where $[·]$ is the floor function. 
Prove that $[x+y] = [x]+[y]$ or $[x]+[y]+1$, where $[·]$ is the floor
  function

I'm Having a little bit of trouble with the last part of this proof.
First, I will use the definition of floor function:
$[x] = m ≡ m ≤ x < m+1$ 
and 
$[y] = n ≡ n ≤ y < n+1$
so, $[x]+[y] = m+n ≡ m+n ≤ x+y < m+n+2$
This is where I get stuck; I have that $[x+y] = t ≡ t ≤ x < t+1 $, so putting 
$m+n ≤ x+y < m+n+2$ in that form, seems impossible, let alone the one that corresponds to $[x]+[y]+1$.
Could you help me with this last part? 
Thanks in advance.

I didn't read the question exactly, but you have floor(x)=n iff. $n<x<n+1$, hence you get $...<m+n+2$ instead of $\leq$.
