Floor Function (Greatest Integer Function) Identity

[tylerc1991]

Homework Statement

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

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

Homework 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}.

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!

 

[tylerc1991]

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