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Prove Floor And Ceiling

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove:

    ⌊−x⌋ = −⌈x⌉


    ⌈−x⌉ = −⌊x⌋ .

    3. The attempt at a solution

    Can i use the property ⌊x⌋ = n, x = n + m where 0 <= m < 1 or do i need to incorporate the negation into this property?

    I think there would be 2 cases to this proof.

    Case 1: m = 0, the lower bound.

    Case 2: m = 1/2.

    Case 3: m = 1, the upper bound.

    I just don't know how i'd go about proving these cases.
     
  2. jcsd
  3. Nov 10, 2012 #2

    haruspex

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    Yes, that's a valid relationship regardless of the sign of x.
    That wouldn't be very satisfactory. You really need to prove it for all m in the range, not just special values.
    m < 1, so just two cases (as you wrote). But it probably isn't necessary to break it into separate cases for m at all. Separate cases for x +ve/-ve might be useful.
     
  4. Nov 10, 2012 #3
    Oh, i'm not thinking straight today.

    Looking at it again, wouldn't there be 2 cases, one where x is a whole number and one where it is a rational number? So in other words, the cases would be m = 0 and 1 > m > 0.

    My professor has thus far taught us proofs involving floor/ceiling using the property that any value being floored/ceilinged can be expressed as x = n + m, and we manipulate m to come up with the cases.
     
  5. Nov 10, 2012 #4

    haruspex

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    I hope you mean "and one where it is not a whole number"; otherwise you're leaving out the irrationals. Anyway, as I said, I doubt it's useful to break it up that way. Go ahead and try it, but do the non-integer case first; I expect you'll find it handles both.
     
  6. Nov 11, 2012 #5
    Proof for ⌊−x⌋ = −⌈x⌉

    By the definition of the floor function, x – 1 < ⌊x⌋ ≤ x

    It follows that −x + 1 > −⌊x⌋ ≥ −x

    Next, let ⌈x⌉ = n where x ≤ n < x + 1

    We note that -x ≤ −⌊x⌋ < -x + 1

    Clearly, −⌊x⌋ = ⌈-x⌉

    I found this proof online, i think it's kind of bogus since the result of the proof isn't what we set out to prove, the signs are misplaced. The steps seem correct up until the conclusion.
     
  7. Nov 11, 2012 #6

    haruspex

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    The above is correct, but the last step could be made clearer:
    By definition, y ≤ ⌈y⌉< y + 1 for all y. Set y = -x:
    -x ≤ ⌈-x⌉< -x + 1
    So −⌊x⌋ and ⌈-x⌉ are both integers in the range [-x, -x+1). Since only one integer can be in that range, they are equal.
    It is the second part of the OP.
     
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