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Floor function proof

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Note: [x] denotes the floor of x.

    Prove that [3x] = [x] + [x + 1/3] + [x + 2/3]


    2. Relevant equations



    3. The attempt at a solution

    Let x = n + E, where n is an integer and 0≤ E < 1. We have three cases:

    Case 1: 0 ≤ E < 1/3
    3x = 3n + 3E and [3x] = 3n, since 0 ≤ 3E < 1.
    [x + 1/3} = n, since x + 1/3 = n + (1/3 + E) and 0 ≤ 1/3 + E < 1.
    [x + 2/3} = n, since x + 2/3 = n + (2/3 + E) and 0 ≤ 1/3 + E < 1.
    Thus, [3x] = 3n, and [x] + [x + 1/3] + [x + 2/3] = n + n + n = 3n.

    Case 2: 1/3 ≤ E < 2/3
    3x = 3n = 3E = (3n + 1) + (3E -1) and [3x] = 3n + 1, since 0 ≤ 3E - 1 < 1.
    [x + 1/3] = n, since x + 1/3 = n + (1/3 + E) and 0 ≤ 1/3 + E < 1.
    [x + 2/3] = [n + (2/3 + E)] = [n + 1 + (E - 1/3)] = n + 1, since 0 ≤ E - 1/3 < 1.
    Thus, [3x] = 3n + 1, and [x] + [x + 1/3] + [x + 2/3] = n + n + (n + 1) = 3n + 1.

    Case 3: 2/3 ≤ E < 1
    3x = 3n = 3E = (3n + 2) + (3E -2) and [3x] = 3n + 2, since 0 ≤ 3E - 2 < 1.
    [x + 1/3] = [n + (1/3 + E)] = [n + 1 + (E - 2/3)] = n + 1, since 0 ≤ E - 2/3 < 1.
    [x + 2/3] = [n + (2/3 + E)] = [n + 1 + (E - 1/3)] = n + 1, since 0 ≤ E - 1/3 < 1.
    Thus, [3x] = 3n + 2, and [x] + [x + 1/3] + [x + 2/3] = n + (n + 1) + (n + 1) = 3n + 2.

    How does this look? I just want to make sure I'm not over looking something.

    Thanks.
     
  2. jcsd
  3. Nov 19, 2012 #2
    Any comments?
     
  4. Nov 19, 2012 #3

    Dick

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    Science Advisor
    Homework Helper

    Doesn't look like you missed anything to me. I might have missed a typo in a line or two, but it's certainly the right idea.
     
  5. Nov 19, 2012 #4
    Yeah, there's a pretty good chance there's at least one typo. Thanks for looking it over I know it's tedious--I spent more time writing it than I care to admit.
     
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