Answer to Question #211830 in Discrete Mathematics for Jaguar

Question

Question 16

Consider the following quantified statement:

∀x ∈ Z [(x2 ≥ 0) ∨ (x2 + 2x – 8＞0)].

Which one of the alternatives provides a true statement regarding the given statement or its 

negation?

 

1. The negation ∃x ∈ Z [(x2 < 0) ∨ (x2 + 2x – 8 ≤ 0)] is not true.

2. x = – 3 would be a counterexample to prove that the negation is not true.

3. x = – 6 would be a counterexample to prove that the statement is not true.

4. The negation ∃x ∈ Z [(x2 < 0) ∧ (x2 + 2x – 8 ≤ 0)] is true.

Answer 1

Consider the following quantified statement $∀x ∈ \Z [(x^2 ≥ 0) ∨ (x^2 + 2x – 8＞0)].$ The negation is $\exists x ∈ \Z [(x^2 < 0) \land (x^2 + 2x – 8\le 0)].$

1. Since $x^2\ge 0$ for any $x\in\Z,$ we conclude that the negation $\exists x ∈ \Z [(x^2 < 0) \land (x^2 + 2x – 8\le 0)]$ is not true.
2. For $x=-3$ we have that $(-3)^2=9\ge0$ and $(-3)^2+2(-3)-8=-5<0$, and hence $[((-3)^2 < 0) \land ((-3)^2 + 2(-3) – 8 ≤ 0)]$ is not true. But this is not a counterexample because of in the prefix of the negation there is the existence quantor.
3. For $x=-6$ we have that $(-6)^2=36\ge0$ and $(-6)^2+2(-6)-8=16>0$, and hence $[((-6)^2 \ge 0) ∨ ((-6)^2 + 2(-6) – 8 > 0)]$ is true.
4. Since $x^2\ge 0$ for any $x\in\Z,$ we conclude that the negation $\exists x ∈ \Z [(x^2 < 0) \land (x^2 + 2x – 8\le 0)]$ is false.

Answer: 1