1a) Write the negation of “if x2>4 then x>2” (your statement should be an “and” or “there exists” statement). 1b) Give all the xvalues that are described by your negation statement.

Question

I know your suppose to use universal or existential quantifier with the reverse A(for all) or E(there exist) element. However, I tried solving it a different way and would this be a possible solution or is it completely wrong.

Here is what I did.

If it isn't x^2>4, then it isn't x>2. negation

If x^2>4 then x>2 where |a|=0.

There exist x=|a|  for all a ∈ℝ

Answer 1

(b) Recall P-->Q is logically equivalent to ~PVQ (you may check this by using a truth table.).

Note that "if x^2>4, then x>2" can be written as "x^2>4-->x>2" since it is an implication. So, (x^2>4--x>2) ≡ (~(x^2>4)V(x>2)). Taking negation on both sides, we have

~(x^2>4-->x>2) ≡ ~(~(x^2>4)V(x>2))

≡~(~(x^2>4))^(~(x>2)) (DeMorgan's Law)

≡ (x^2>4) ^(~(x>2)) (Double Negation)

≡ (x^2>4) ^(x<=2) (The negation of x>2 is x<=2)

Hence, the negation of "if x^2>4, then x>2" is "x^2>4 and x<=2."

(b) By part (a), we have "x^2>4 and x<=2." It is clear that x^2>4 yields x<-2 or x>2 by basic algebra. Hence, in order to have x^2>4 and x<=2, we must have x<-2. Therefore, x<-2 is the set of numbers we are looking for.