Question #346199

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))

1
Expert's answer
2022-05-30T23:36:43-0400

In general:

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

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


So we have:

a) The negation of x(x>1)∀x(x > 1) is x(¬(x>1))=x(x1).\exist x (\lnot (x>1))=\exist x(x \leq 1).

b) The negation of x(x2)∀x(x ≤ 2) is x(¬(x2))=x(x>2).\exist x (\lnot(x ≤ 2)) = \exist x (x>2).

c) The negation of x(x4)∃x(x ≥ 4) is x(¬(x4))=x(x<4).\forall x (\lnot(x ≥ 4)) = \forall x(x<4).

d) The negation of x(x<0)∃x(x < 0) is x(¬(x<0))=x(x0).\forall x(\lnot(x < 0)) = \forall x (x ≥0).

e) The negation of x((x<1)(x>2))∀x((x < −1) ∨ (x > 2)) is x(¬((x<1)(x>2)))=x((x1)(x2))=\exist x (\lnot ((x < −1) ∨ (x > 2))) = \exists x ((x ≥ -1) \land (x \leq 2))=

x(1x2).\exist x (-1 \leq x \leq 2).

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

x(¬((x<4)(x>7)))=x((x4)(x7))=\forall x (\neg((x < 4) ∨ (x > 7))) = \forall x ((x ≥ 4) \land (x \leq 7))=

=x(4x7).=\forall x (4 \leq x \leq 7).


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