Solution to Question 16 Consider the following quantified statement: ∀x ∈ Z [(x2 ≥ 0) ∨ (x2 … - Sikademy
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Archangel Macsika

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.

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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


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Question ID: mtid-5-stid-8-sqid-2841-qpid-1398