Answer to Question #346199 in Discrete Mathematics for Aish

Question

Express the negation of each of these statements in terms

of quantifiers without using the negation symbol.

a) ∀x(x > 1)

b) ∀x(x ≤ 2)

c) ∃x(x ≥ 4)

d) ∃x(x < 0)

e) ∀x((x < −1) ∨ (x > 2))

f ) ∃x((x < 4) ∨ (x > 7))

Answer 1

In general:

1) The negation of $\forall x (P(x))$ is $\exist x(\lnot P(x)).$

2) The negation of $\exist x (P(x))$ is $\forall x (\lnot P(x)).$

So we have:

a) The negation of $∀x(x > 1)$ is $\exist x (\lnot (x>1))=\exist x(x \leq 1).$

b) The negation of $∀x(x ≤ 2)$ is $\exist x (\lnot(x ≤ 2)) = \exist x (x>2).$

c) The negation of $∃x(x ≥ 4)$ is $\forall x (\lnot(x ≥ 4)) = \forall x(x<4).$

d) The negation of $∃x(x < 0)$ is $\forall x(\lnot(x < 0)) = \forall x (x ≥0).$

e) The negation of $∀x((x < −1) ∨ (x > 2))$ is $\exist x (\lnot ((x < −1) ∨ (x > 2))) = \exists x ((x ≥ -1) \land (x \leq 2))=$

$\exist x (-1 \leq x \leq 2).$

f) The negation of $∃x((x < 4) ∨ (x > 7))$ is

$\forall x (\neg((x < 4) ∨ (x > 7))) = \forall x ((x ≥ 4) \land (x \leq 7))=$

$=\forall x (4 \leq x \leq 7).$