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The integer part is (? distributive ?)

  1. Jan 26, 2009 #1
    the integer part is ... (?? distributive ??)

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

    Define the floor of a real number k where [k] is the least smallest integer from k.

    I want to show that [a - b] = [a] -

    2. Relevant equations
    n/a


    3. The attempt at a solution
    [1.2 - 5.7] = [-3.8] = -4
    [1.2] - [5.7] = 1 - 5 = -4

    I am not sure how to go about generalizing the observation above to all numbers.
     
  2. jcsd
  3. Jan 26, 2009 #2

    CompuChip

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    Re: the integer part is ... (?? distributive ??)

    You don't:
    [5.2 - 2.3] = [2.9] = 2
    [5.2] - [2.3] = [5] - [2] = 3

    Sorry :smile:
     
  4. Jan 27, 2009 #3
    Re: the integer part is ... (?? distributive ??)

    Yes, thank you CompuChip. However I stated the problem wrong. What I am thinking is this:

    Let a > 0. Then
    [a - 0] = [a] - 0 and
    [a - 1] = [a] - 1.

    I only care about the values being subtracted from a of 0 and 1, nothing else. Then if b = 0 or 1, [a-b] = [a] - b. This should hold, correct? Demonstrating it works for 0 is easy. I don't know how to show it for 1, but here is my attempt:

    a - 1 < a
    Suppose a is an integer. Then [a - 1] = a - 1 = [a] - 1.
    Suppose a is not an integer. I am not sure. It seems to work fine with all the tests I can give (a>0). I don't know how to write this.
     
  5. Jan 27, 2009 #4

    CompuChip

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    Re: the integer part is ... (?? distributive ??)

    Let x be any (positive, although it probably works for any) number. Then write x as n + f, with n an integer and f the fractional part (0 <= f < 1).
     
  6. Jan 27, 2009 #5
    Re: the integer part is ... (?? distributive ??)

    Yes, thank you. Work backwards, of course.
     
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