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Logic: difference between very similar statements

  1. Jul 27, 2012 #1
    What is the difference between the following two questions:

    (a) For every positive real number x, there is a positive real number y less
    than x with the property that for all positive real numbers z, yz ≥ z.

    (b) For every positive real number x, there is a positive real number y with
    the property that if y < z, then for all positive real numbers z, yz ≥ z.


    (b) I understand as

    [itex](\forall x\inℝ\stackrel{+}{})(\exists y\inℝ\stackrel{+}{})[(y<x)\Rightarrow(\forall z\inℝ\stackrel{+}{})(yz≥z)][/itex]

    I am unsure of how to understand (a) but this is my interpretation:

    [itex](\forall x\inℝ\stackrel{+}{})(\exists y\inℝ\stackrel{+}{})[y<x\wedge(\forall z\inℝ\stackrel{+}{})(yz≥z)][/itex]

    Other than the fact that (b) has an implication and (a) does not, I do not see any difference between (a) and (b) and they both seem false to me because if you choose x=1 and 0<y<1 and z=1, then it is not the case that yz>=z. However, according to the back of my book, it says that x=1 is a counterexample to (a), not (b). It also says that (b) is actually a true statement...please help explain?

    edit: I think I see why (b) is true..is it because for all x, if you choose y>x, then y<x is false, and so false=>false and false=>true are both true ?
    So then x=1 would just be a counterexample to (a). But am I expressing (a) correctly?
     
    Last edited: Jul 27, 2012
  2. jcsd
  3. Jul 28, 2012 #2

    Stephen Tashi

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    I don't know the details of your books notation, but I think you expressed the statements correctly and gave a correct analysis of why (b) is true in the case of x = 1.
     
  4. Jul 28, 2012 #3

    haruspex

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    Doesn't look like that to me. The y < z has become y < x. If y < z is the correct version then x doesn't seem to have any role. If it should be y < x then the wording is strange, and the obvious way to straighten it makes it the same as (a).
     
  5. Jul 28, 2012 #4

    mfb

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    In terms of the analysis, a is wrong.

    For every positive real number x? Take x=1/2.
    yz ≥ z for positive real z is equivalent to y ≥ 1, and there is not positive real number y < 1/2 which satisfies y ≥ 1.

    b (with the fix "y<x")... well, I would not use such a statement, as it is a bit ill to analyze, but your analysis looks correct.
     
  6. Jul 28, 2012 #5

    Stephen Tashi

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    I hope you mean "a is false", which is in agreement with the original post.
     
  7. Jul 29, 2012 #6

    mfb

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    Right. I hope the counterexample to (a) was clear enough to see that.
     
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