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
(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
- When we negate quantifiers, we flip all and signs and negate any propositions.
- The proposition , written without a sign, is equivalent to .
, using De Morgan's Law(i) So if , then
(ii) Looking first at , then, we have to find the values of for which the following statement is true:
Given an , we can find a such that
- ; andSo 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 .