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Math Help - how to prove?

  1. #1
    Member pberardi's Avatar
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    how to prove?

    I realize that this is a quite simple proof. I can reason the answer but cannot prove it.

    problem Statement:
    Say c is a real number and has the property Q if x <= y whenever x, y in real numbers with cx <= cy. Which real numbers have property Q?
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  2. #2
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    Quote Originally Posted by pberardi View Post
    I realize that this is a quite simple proof. I can reason the answer but cannot prove it.
    problem Statement: Say c is a real number and has the property Q if x <= y whenever x, y in real numbers with cx <= cy. Which real numbers have property Q?
    I frankly donít see what there is to prove here.
    The number property Q is clearly membership in the positive real numbers.
    Are you to prove this: x \leqslant y\;\& \,c > 0\, \Rightarrow \,cx \leqslant cy ?
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  3. #3
    Member pberardi's Avatar
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    Quote Originally Posted by Plato View Post
    I frankly donít see what there is to prove here.
    The number property Q is clearly membership in the positive real numbers.
    Are you to prove this: x \leqslant y\;\& \,c > 0\, \Rightarrow \,cx \leqslant cy ?
    The problem statement is exactly as I have written it. Perhaps it is not a proof and that simply the answer would suffice. My other rationale is that we need to use a certain property that we know is true i.e. (x-y)^2 > 0 for x != y. If you Plato cannot formalize a proof for this then I am going to quit feeling bad about it. My only problem is that I would expect the prof to say "prove it".
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  4. #4
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    Quote Originally Posted by pberardi View Post
    If you Plato cannot formalize a proof for this then I am going to quit feeling bad about it. My only problem is that I would expect the prof to say "prove it".
    Once again, as I said above, I do not see what there is to formalize other than what I posted.

    BTW: I think the statement is very poorly written.
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  5. #5
    Super Member Gamma's Avatar
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    Suppose x \leq y, c\in \mathbb{R}.

    x \leq y
    0\leq (y-x)

    Case 1: c\in \mathbb{R}^+

    c0\leq c(y-x)\Rightarrow 0\leq cy-cx \Rightarrow cx \leq cy
    Which is property Q.
    Case 2: c=0

    0\cdot 0\leq 0(y-x)\Rightarrow 0\leq 0
    Which is trivially property Q.

    Case 3: c\in \mathbb{R}^-

    c0\geq c(y-x)\Rightarrow 0\geq cy-cx \Rightarrow cx \geq cy

    Which is only property Q if x=y, but Q is for all x,y \in \mathbb{R}, so case 3 does not satisfy property Q.


    Thus property Q is being non-negative.
    Last edited by Gamma; June 7th 2009 at 09:42 AM. Reason: stray = sign
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