<s>[INST] How to prove this property involving the floor function? How does one prove that $$\lfloor x \rfloor = \lfloor x/2 \rfloor + \lfloor (x+1)/2 \rfloor $$
where $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$?
I tried to do this by using inequalities but end up with $\lfloor x \rfloor + 1/2$ on the lhs. Can someone help?
 [/INST] Let $x = a + b$ where $a \in \mathbb{N}$ and $0 \leqslant b < 1$
We need to prove that 
$$\lfloor a + b \rfloor = \left\lfloor \frac{a+b}{2} \right\rfloor + \left\lfloor \frac{a+b +1}{2} \right\rfloor $$
The L.H.S. will be just $a$
Now the R.H.S.
$$\left\lfloor \frac{a+b}{2} \right\rfloor + \left\lfloor \frac{a+b +1}{2} \right\rfloor = \left\lfloor \frac{a}{2} \right\rfloor + \left\lfloor \frac{a+1}{2} \right\rfloor $$
If $a$ is even,
$$\left\lfloor \frac{a}{2} \right\rfloor + \left\lfloor \frac{a+1}{2} \right\rfloor = \frac{a}{2} + \frac{a}{2} = a $$
If $a$ is odd, 
$$\left\lfloor \frac{a}{2} \right\rfloor + \left\lfloor \frac{a+1}{2} \right\rfloor = \frac{a-1}{2} + \frac{a+1}{2} = a $$
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