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- Oct 3rd 2009, 02:15 AM #1

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- Apr 2009
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- 108

## 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

- Oct 3rd 2009, 04:00 AM #2

- Oct 6th 2009, 11:39 AM #3
## Negating quantified propositions

Hello Khonics89Two things to note first:

- When we negate quantifiers, we flip all and signs and negate any propositions.

- The proposition , written without a sign, is equivalent to .

So

, using De Morgan's Law(i) So if , then

Given an , we can find a such that

- ; and

So what sort of value will have?

First, consider .

for any value of , and hence will be true for any and any . So provided we choose such that , both statements will be true.

Hence is true for any .

Now let's consider .

In this case both sides of the inequality are non-negative for all values of and , and hence we may square both sides to get:

In addition to this, we must also have . Hence we can write:

So for positive values of , is true whenever .

Therefore is true for all , and hence is true for .

Grandad

- Oct 8th 2009, 05:46 PM #4

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- Apr 2009
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