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Homework Help: I wish I could manipulate universal quantifiers better

  1. Apr 7, 2014 #1


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

    I am asked to prove by contradiction that:

    A real number [itex]x[/itex] that is less than every positive real cannot be positive.

    2. Relevant equations

    Axiom of Order ('A14'):

    a <= b and 0 < c → ac <= bc

    3. The attempt at a solution

    I believe I can prove it analytically as follows:

    Assume otherwise, then [itex]x[/itex] is less than every positive real and [itex]x[/itex] is positive, therefore it is the least positive number. But consider the number [itex]½ x[/itex]. Substituting into the above axiom letting a = 0, b = ½, c = x we obtain 0 < [itex]½x[/itex], demonstrating that as well as being [itex]< x, ½ x[/itex] is also positive. Contradiction since x was assumed to be the least positive real.

    But I would love to be able to throw around a few quantifiers, negate them etc and prove it far more easily. So far not so good...
    Last edited: Apr 7, 2014
  2. jcsd
  3. Apr 7, 2014 #2
    Don't be silly. Your proof is optimal. In actual substance, it is 100% efficient.
  4. Apr 8, 2014 #3


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    Thank you, I was very pleased with it even though it doesn't use the technique I set out to use and I am aware that I have not explicitly demonstrated that ½x < x by axiom.

    Even so I am very interested in being able to usefully manipulate quantifiers - not always apparent to me how to negate statements...or even what the lists of symbols I end up with actually mean.

    For example in the case of the statement I have proved above my plan was to express it as:

    [itex]\exists x\in R^{+}[/itex] | [itex] x<ε, \forall ε >0[/itex]

    making that line 2

    negate it thereby making a true statement in line 1,

    then derive a contradiction from line 2, assuring the truth of line 1.

    Any ideas?
    Last edited: Apr 8, 2014
  5. Apr 8, 2014 #4

    D H

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    You found a contradiction. Why do you want to go all formal?

    What you need here for a more formal approach is universal elimination. Assuming there exists an x>0 means x/2>0. By universal elimination, you can substitute ##x/2## for ##\varepsilon## in ##\forall \varepsilon > 0, x < \varepsilon## yielding ##x<x/2##, which is a contradiction.
  6. Apr 8, 2014 #5


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    Thanks D H I like your proof.

    The question in the back of my mind is how useful is the technique of being able to negate statements using [itex]\forall[/itex],[itex]\exists[/itex] to derrive proofs by contradiction. No doubt it will often fail, but I would like to be able to examine it as an option, however it is far from clear to me what the rules are for forming [itex]\neg[/itex]P, once P involves >2 predicates.

    Am I barking up the wrong tree?
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