# Floor Function (Greatest Integer Function) Identity

1. Dec 28, 2011

### tylerc1991

1. The problem statement, all variables and given/known data

Prove that, for all $x, y \in \mathbb{R}$,

$[2x] + [2y] \geq [x] + [y] + [x + y]$.

2. Relevant equations

I am using $[\cdot]$ to represent the floor function, and $\{\cdot\}$ to represent the fractional part of a real number ($\{x\} = x - [x]$ for real numbers $x$).

We may take for granted that $[x + y] \geq [x] + [y]$. (1)

We may also take for granted that $[x + n] = [x] + n$ for $n \in \mathbb{Z}$.

3. The attempt at a solution

Let $x$ and $y$ be real numbers. Using inequality (1) above, we see that $[2x] \geq 2[x]$ and $[2y] \geq 2[y]$. So I can say that

$[2x] + [2y] \geq 2[x] + 2[y]$. (a)

By definition, $x = \{x\} + [x]$, so we see that $[x + y] = [\{x\} + [x] + \{y\} + [y]] = [\{x\} + \{y\}] + [x] + [y] \leq 1 + [x] + [y]$. This is equivalent to

$-[x + y] \geq -1 - [x] - [y]$. (b)

Adding equations (a) and (b), we see that

$[2x] + [2y] + 1 \geq [x] + [y] + [x + y]$.

I am annoyingly close, and as much as I wish I could get rid of that pesky '1', I can't seem to at the moment. Could someone please give me a little direction? Thank you!

2. Dec 28, 2011

### tylerc1991

Using the fact that $[2x] = [x] + [x + 1/2]$, I figured it out!!