Question

Question 17

Consider the following proposition:

For any predicates P(x) and Q(x) over a domain D, the negation of the statement 

∃x ∈ D, P(x) ∧ Q(x) 

is the statement

∀x ∈ D, P(x) → ¬Q(x).

We can use this truth to write the negation of the following statement:

“There exist integers a and d such that a and d are negative and a/d = 1 + d/a.”

Which one of the alternatives provides the negation of this statement?

1. There exist integers a and d such that a and d are positive and a/d = 1 + d/a.

2. For all integers a and d, if a and d are positive then a/d  1 + d/a.

3. For all integers a and d, if a and d are negative then a/d  1 + d/a.

4. For all integers a and d, a and d are positive and a/d  1 + d/a.

Answer 1

Let $P(x)$ be the statement “ a and d are negative" and $Q(x)$ be the statement “$\frac{a}{d} = 1 + \frac{d}{a}$ ”. Then the statement “There exist integers a and d such that a and d are negative and $\frac{a}{d} = 1 + \frac{d}{a}$ .” is $∃x ∈ D, P(x) ∧ Q(x)$ and its negation is the statement $∀x ∈ D, P(x) → ¬Q(x),$ that is

the statement "For all integers a and d, if a and d are negative then $\frac{a}{d} \ne 1 + \frac{d}{a}$ ."

Answer: 3