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Thread: 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) $\displaystyle \left( {\forall x} \right)\left( {\exists y} \right)\left[ {\sqrt {\left| x \right|} \geqslant ay \wedge \left| x \right| < y^2 } \right]$
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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 $\displaystyle \forall$ and $\displaystyle \exists$ signs and negate any propositions.


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

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

    • $\displaystyle \sqrt{|x|}\ge a|y|$; and

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

    First, consider $\displaystyle a\le 0$.

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

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


    Now let's consider $\displaystyle a >0$.

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

    $\displaystyle |x| \ge a^2y^2$

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

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

    $\displaystyle \Rightarrow a^2y^2<y^2$

    $\displaystyle \Rightarrow a^2<1$

    $\displaystyle \Rightarrow |a| <1$

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

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


    Grandad
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  4. #4
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    Thanks Grandad your a champ
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