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Math Help - Discrete Math

  1. #1
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    Discrete Math

    Apparently this question is hard

    Let Q(a,b) be a property which involves a and b. Let P(c) be a property which involves c. For this problem, the domain will be the reals. For each of the following below, find properties P(c) and Q(a,b) such that:

    a.) (\forall x)(\forall z)[Q(x,z) \implies Q(z,x)] is true.

    b.) (\forall x)(\forall z)[Q(x,z) \implies Q(z,x)] is false.

    c.) (\forall x)[P(x) \implies (\exists z) Q(x,z)] is true.

    d.) (\forall x)[P(x) \implies (\exists z) Q(x,z)] is false.
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  2. #2
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    a) x=z
    b) x<z
    c) x is rational then for some y, xy is irrational.
    d) x is an even integer then for some integer xy is odd.
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  3. #3
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    Quote Originally Posted by Plato View Post
    a) x=z
    b) x<z
    c) x is rational then for some y, xy is irrational.
    d) x is an even integer then for some integer xy is odd.
    Two questions:

    where does y come from, and for c do you mean "if x is rational, then for some z, xz is rational" ?
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  4. #4
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    Quote Originally Posted by DiscreteW View Post
    Two questions:

    where does y come from, and for c do you mean "if x is rational, then for some z, xz is rational" ?
    NO!
    P(x)="x is rational"
    Q(x,y)="xy is not rational"
    These predicates work for both (c)

    For (d) Q(x,y)="x+y is not rational"
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  5. #5
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    Quote Originally Posted by Plato View Post
    NO!
    P(x)="x is rational"
    Q(x,y)="xy is not rational"
    These predicates work for both (c)

    For (d) Q(x,y)="x+y is not rational"
    Why y? Why not z?

    Oh I see I guess it doesn't matter. So, my answer can be:

    a) x=z
    b) x<z
    c.) If P(x) = x \in \mathbb{Q}, then Q(x,z) = xz \notin \mathbb{Q}
    d.) If P(x) = x \in \mathbb{Q}, then Q(x,z) = x + z \notin \mathbb{Q}

    Right? Although not sure how c and d are both not in the rationals (that is irrational), since one is true and the other is false, and we know that rational + rational = rational and similarly for rational*rational.

    c is true, but isn't xz \notin \mathbb{Q} false?
    Last edited by DiscreteW; February 14th 2008 at 04:26 PM.
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