Negation of the statement p: for every real number, either x ≥ or x

Question

Negation of the statement p: for every real number, either x ≥ or x < 1 is (A) There exist a real number x such that neither x > 1 nor x < 1 (B) There exist a real number x such that 0 < x < 1 . (C) There exist a real number x such that neither x ≥ 1 , nor x ≤ 1 (D) none of these

Tardigrade Admin · Accepted Answer

Negation of given statement p ∨ q is ∼ (p ∨ q) = ∼ p ∧ ∼ q ∴ .

Q. Negation of the statement $p$ : for every real number, either $x \geq$ or $x < 1$ is

There exist a real number x such that neither $x > 1$ nor $x < 1$

There exist a real number x such that $0 < x < 1$ .

There exist a real number x such that neither $x \geq 1$ , nor $x \leq 1$

none of these

Negation of given statement $p \lor q$ is $\sim (p \lor q) =\sim p \land \sim q ∴$ .

Q. Negation of the statement p : for every real number, either x≥ or x<1 is

There exist a real number x such that neither x>1 nor x<1

There exist a real number x such that 0<x<1 .

There exist a real number x such that neither x≥1 , nor x≤1