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Math Help - Some Simple Predicate Logic Questions

  1. #1
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    Some Simple Predicate Logic Questions

    Could someone please check my answers for these questions:

    p(x): x^2 - 7x + 10 = 0; x = {5, 2}
    q(x): x^2 - 2x - 3; x = {-1, 3}
    r(x): x < 0

    domains:
    i) Z (set of all integers)
    ii) Z+ (set of all positive integers)
    iii) integers 2 and 5

    Are these statements True or false (with counterexamples):

    <br />
\forall{x[p(x)\Rightarrow\neg}{r(x)]}<br />
    i) false x = 1
    ii) false x = 1
    iii) false

    <br />
\forall{x[q(x)\Rightarrow}{r(x)]}<br />
    i) false x = 3
    ii) false x = 3
    iii) false x = 2 ?

    <br />
\exists{x[q(x)\Rightarrow}{r(x)]}<br />
    i) true
    ii) false
    iii) false

    <br />
\exists{x[p(x)\Rightarrow}{r(x)]}<br />
    i) false
    ii) false
    iii) false

    Is it possible to find counterexamples for the existential statements? Also, in the first two problems with the universal quantifier, for these to be true, would p(x) and q(x) have to hold for every value in the domain? For example, would x^2 - 7x + 10 = 0 have to be true for every value of x in the domain, or just the solutions, 5 and 2? I understand quantifiers, but the conditional statement is confusing me a bit. Thanks!
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello aaronrj
    Quote Originally Posted by aaronrj View Post
    Could someone please check my answers for these questions:

    p(x): x^2 - 7x + 10 = 0; x = {5, 2}
    q(x): x^2 - 2x - 3; x = {-1, 3}
    r(x): x < 0

    domains:
    i) Z (set of all integers)
    ii) Z+ (set of all positive integers)
    iii) integers 2 and 5

    Are these statements True or false (with counterexamples):

    <br />
\forall{x[p(x)\Rightarrow\neg}{r(x)]}<br />
    i) false x = 1
    ii) false x = 1
    iii) false
    No - you haven't got this at all.  \forall{x[p(x)\Rightarrow\neg}{r(x)]} means that if p(x) is true, then r(x) is false. And, in all three domains, p(x) is true when x = 2 or x = 5. In both of these cases r(x) is false, since x > 0.

    So the proposition is true in all three domains.


    <br />
\forall{x[q(x)\Rightarrow}{r(x)]}<br />
    i) false x = 3
    ii) false x = 3
    iii) false x = 2 ?
    (i) and (ii) are correct.

    But (iii) is a bit sneaky, because in the domain \{2, 5\},\, q(x) is false for all x. So for every value of x that makes q(x) true, r(x) is true. In other words, there isn't a counterexample, so it's a true proposition.


    <br />
\exists{x[q(x)\Rightarrow}{r(x)]}<br />
    i) true
    ii) false
    iii) false

    <br />
\exists{x[p(x)\Rightarrow}{r(x)]}<br />
    i) false
    ii) false
    iii) false
    These are all correct.

    Is it possible to find counterexamples for the existential statements?
    No. If an existential statement is false, it simply means that no value of x exists that makes the statement true.

    Also, in the first two problems with the universal quantifier, for these to be true, would p(x) and q(x) have to hold for every value in the domain?
    No. When you form a proposition like \forall x[a(x) \Rightarrow b(x)] you are simply saying if a(x) is true, then b(x) is true. You aren't saying anything at all about a(x) actually being true at all.

    Indeed a(x) may never be true for any values of x - see for example part (iii) of the statement \forall x[q(x) \Rightarrow r(x)], where q(x) is false for all values of x in the domain \{2,5\}.

    Grandad
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