Question

Let D = E = {−2, −1, 0, 1, 2} Write negations for each of the following statements and determine which is true, the given statement or its negation. a. ∀x ∈ D, ∃y ∈ E such that x + y = 1. b ∃x ∈ D such that ∀y ∈ E, x + y = y. c. ∀x ∈ D, ∃y ∈ E such that xy ≥ y. d. ∃x ∈ D such that ∀y ∈ E, x ≤ y.

Answer 1

Answer:

See explanation below.

Step-by-step explanation:

We use laws of quantifiers to write the negations. The negation of ∀w∃zP(w,z) is ∃w∀z¬P(w,z). The negation of ∃w∀zP(w,z) is ∀w∃z¬P(w,z), for a predicate P depending on w and z.  

a) Negation: ∃x ∈ D, such that ∀y ∈ E, x + y ≠ 1. In this case the negation is true, hence the original statement is false. Take x=-2∈D, and note that x+y≤0 for all y ∈ E. Therefore x+y is never 1.

b) Negation: ∀x ∈ D, ∃y ∈ E such that x + y ≠ y. Here, the original statement is true. Take x=0∈D, then x+y=0+y=y for all y∈E.

c) Negation: ∃x ∈ D, such that ∀y ∈ E, xy < y. In this case, the original statement is true. Given x∈D, take y=x∈E. Hence xy=x²≥x for all x∈D (this inequality is true for integers, and we can check it case by case in D).

d) Negation: ∀x ∈ D, ∃y ∈ E such that x > y. In this case, the original statement is true, take x=-2∈D. Then -2≤y for all y∈E.