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Math Help - consider this propostiion

  1. #1
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    consider this propostiion

    For each real number a consider the proposition p(a) given by
    ∃x ∀y (square root of|x| > and = a|y|) ⇒ (|x| > and = y^2),
    where the universal set U is the set of real numbers.
    (i ) Apply the negation rule to the quantifiers to write ∼p(a) without using the implication
    sign ⇒.
    (ii ) Find all values of a for which p(a) is true. Justify your answer.

    A little confused as to what the answer is, because there is no answers in the back.... :\

    Can you show me your way and if you can also include working out thanks
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  2. #2
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    Quote Originally Posted by Khonics89 View Post
    For each real number a consider the proposition p(a) given by
    ∃x ∀y (square root of|x| > and = a|y|) ⇒ (|x| > and = y^2),
    where the universal set U is the set of real numbers.
    (i ) Apply the negation rule to the quantifiers to write ∼p(a) without using the implication
    sign ⇒.
    (ii ) Find all values of a for which p(a) is true. Justify your answer.
    Here is part (i) \left( {\forall x} \right)\left( {\exists y} \right)\left[ {\sqrt {\left| x \right|}  \geqslant ay \wedge \left| x \right| < y^2 } \right]
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  3. #3
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    Negating quantified propositions

    Hello Khonics89
    Quote Originally Posted by Khonics89 View Post
    For each real number a consider the proposition p(a) given by
    ∃x ∀y (square root of|x| > and = a|y|) ⇒ (|x| > and = y^2),
    where the universal set U is the set of real numbers.
    (i ) Apply the negation rule to the quantifiers to write ∼p(a) without using the implication
    sign ⇒.
    (ii ) Find all values of a for which p(a) is true. Justify your answer.

    A little confused as to what the answer is, because there is no answers in the back.... :\

    Can you show me your way and if you can also include working out thanks
    Two things to note first:

    • When we negate quantifiers, we flip all \forall and \exists signs and negate any propositions.


    • The proposition p \Rightarrow q, written without a \Rightarrow sign, is equivalent to \neg p \lor q.

    So \sim\Big(\exists x\, \forall y\,[p \Rightarrow q]\Big)
    \equiv\, \sim\Big(\exists x\, \forall y\,[\neg p \lor q]\Big)
    \equiv \forall x\, \exists y\, [p \land \neg q], using De Morgan's Law
    (i) So if p(a)\equiv \exists x\, \forall y\,[(\sqrt{|x|}\ge a|y|) \Rightarrow (|x| \ge y^2)], then
    \sim p(a)\equiv \forall x\, \exists y\, [(\sqrt{|x|}\ge a|y|) \land (|x| < y^2)]
    (ii) Looking first at \sim p(a), then, we have to find the values of a for which the following statement is true:
    Given an x, we can find a y such that

    • \sqrt{|x|}\ge a|y|; and

    • |x| < y^2
    So what sort of value will a have?

    First, consider a\le 0.

    a \le 0 \Rightarrow a|y| \le 0 for any value of y, and hence \sqrt{|x|} \ge a|y| will be true for any x and any y. So provided we choose y such that |x| <y^2, both statements will be true.

    Hence \sim p(a) is true for any a \le 0.


    Now let's consider a >0.

    In this case both sides of the inequality \sqrt{|x|}\ge a|y| are non-negative for all values of x and y, and hence we may square both sides to get:

    |x| \ge a^2y^2

    In addition to this, we must also have |x| < y^2. Hence we can write:

    a^2y^2 \le |x| < y^2

    \Rightarrow a^2y^2<y^2

    \Rightarrow a^2<1

    \Rightarrow |a| <1

    So for positive values of a, \sim p(a) is true whenever  a<1 .

    Therefore \sim p(a) is true for all a < 1, and hence p(a) is true for a \ge 1.


    Grandad
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  4. #4
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    Thanks Grandad your a champ
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