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Floor Function (Greatest Integer Function) Identity

  1. Dec 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that, for all [itex]x, y \in \mathbb{R}[/itex],

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

    2. Relevant equations

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

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

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

    3. The attempt at a solution

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

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

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

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

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

    [itex][2x] + [2y] + 1 \geq [x] + [y] + [x + y][/itex].

    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. jcsd
  3. Dec 28, 2011 #2
    Using the fact that [itex][2x] = [x] + [x + 1/2][/itex], I figured it out!!
     
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