1. ## 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

2. Originally Posted by Khonics89
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) $\left( {\forall x} \right)\left( {\exists y} \right)\left[ {\sqrt {\left| x \right|} \geqslant ay \wedge \left| x \right| < y^2 } \right]$

3. ## Negating quantified propositions

Hello Khonics89
Originally Posted by Khonics89
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| , 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

$\Rightarrow a^2<1$

$\Rightarrow |a| <1$

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 $a \ge 1$.