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Predicate Logic and Proofs

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    No matter what positive real number x we choose, there exists some positive real number y
    such that yz2 > xz + 10 for every positive integer z.

    Translate the above statement to predicate logic and prove it using a direct approach.

    2. Relevant equations

    I don't believe that there are relevant equations for this problem.

    3. The attempt at a solution

    Let Q (x, y, z) = yz2 > xz + 10

    [itex]\forall[/itex]x ∈ ℝ+ [itex]\exists[/itex]y ∈ ℝ+ [itex]\forall[/itex]z ∈ [itex]Z[/itex]+ Q(x, y, z)

    Before I attempted to prove the theorem, I wanted to make sure that I got the predicate logic translation right. I don't think that the above translation is right, but I hope I'm on the right track. I've never translated into predicate logic with 3 variables. It's usually just x and y, so should it be (x, y, z)? Also, does [itex]\forall[/itex]z ∈ [itex]Z[/itex]+ come after Q (x, y, z) since it does in the statement?
     
  2. jcsd
  3. Nov 2, 2011 #2

    Mark44

    Staff: Mentor

    I would write it this way.

    [itex]\forall (x ∈ R^+, z ∈ Z^+) \exists y ∈ R^+ \ni Q(x, y, z)[/itex]

    In addition to other changes, I also replaced ℝwith R, since ℝis so tiny I can barely tell it's a version of the letter R.

    There's a nicer one that you can get with mathbb{R}, as in
    [itex]\mathbb{R}[/itex].
     
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