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Help with understanding a floor proof

  1. Feb 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that for any real number x, if x - floor(x) < 1/2, then floor(2x) = 2 floor(x)

    2. Relevant equations


    3. The attempt at a solution
    Assuming that x is a real number. Suppose that x - floor(x) < 1/2
    Multiplying both sides by 2, 2x < 2 floor(x) + 1
    from the definition, 2 floor(x) <= 2x
    2 floor(x) is an integer, and 2 floor(x) <= 2x < 2 floor(x) + 1
    Implying floor(2x) = 2 floor(x)

    My question is, how does the proof imply the result? I don't see how the logic works. Thank you for any help you can provide in advance.
     
  2. jcsd
  3. Feb 4, 2015 #2

    RUber

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    Homework Helper

    The key is that floor(x) and floor(2x) are both integers and the relation you have above is 2floor(x) < 2 floor(x) + 1. You seem to be missing the part where 2floor(x) <= floor(2x).
     
  4. Feb 5, 2015 #3
    I see... I guess I missed that part, haha, just wondering, does it come from the inequality:

    floor (2x) <= 2x < floor(2x) + 1 ?

    I can see how it relates... kind of, since

    2 floor (x) <= 2x as well, but how does

    2floor(x) <= floor (2x) ?
     
  5. Feb 6, 2015 #4

    RUber

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    2 floor(x) is always less than or equal to floor(2x).

    If you let ## x = x_{int} + x_{frac} ## where ##x_{int}\in \mathbb{Z}, 0 \leq x_{frac}<1##
    Then ##\text{floor}(x) = x_{int}=x-x_{frac}##
    ## 2\text{floor}(x) = 2x_{int}=2x-2x_{frac}##
    But floor(2x) has 2 cases:
    case 1:
    ##x_{frac}<.5 \implies 2x_{frac} < 1 \implies floor(2x) = 2 x_{int} = 2floor(x) ##
    case 2:
    ##.5 \leq x_{frac}<1 \implies 2x_{frac} \geq 1 \implies floor(2x) = 2 x_{int} + 1=2floor(x) + 1 ##
    In both cases, floor(2x) ##\geq ## 2floor(x) .
     
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