Find the truth set of each predicate. (If your answer is an interval, enter it using interval notation; otherwise enter it using set-roster notation.) 8 (a) Predicate: is an integer, domain: Z d {1,2,4,8) (b) Predicate: {1} 80 is an integer, domain: Z+ (c) Predicate: 1 = x² = 4, domain: R [-2,-1] U 1,2 Predicate: 1 = x² = 4, domain: Z {-1}
The correct answer and explanation is:
Here are the solutions and explanations for the truth sets of the given predicates:
(a) Predicate: x is an integerx \text{ is an integer}, Domain: Z∖{1,2,4,8}\mathbb{Z} \setminus \{1, 2, 4, 8\}
Solution:
The predicate specifies that xx is an integer and must not belong to the set {1,2,4,8}\{1, 2, 4, 8\}. Since Z\mathbb{Z} is the set of all integers and we exclude {1,2,4,8}\{1, 2, 4, 8\}, the truth set is: Z∖{1,2,4,8}={x∈Z∣x≠1,x≠2,x≠4,x≠8}.\mathbb{Z} \setminus \{1, 2, 4, 8\} = \{x \in \mathbb{Z} \mid x \neq 1, x \neq 2, x \neq 4, x \neq 8\}.
Explanation:
The domain excludes specific values {1,2,4,8}\{1, 2, 4, 8\}, while the predicate simply requires xx to be an integer. Therefore, the truth set includes all integers except those explicitly removed.
(b) Predicate: x=80 and x is an integerx = 80 \text{ and } x \text{ is an integer}, Domain: Z+\mathbb{Z}^+ (Positive integers)
Solution:
Since the predicate specifies x=80x = 80, the truth set depends on whether 8080 belongs to the domain Z+\mathbb{Z}^+. Since Z+\mathbb{Z}^+ represents all positive integers and 8080 is a positive integer, the truth set is: {80}.\{80\}.
Explanation:
The predicate narrows the condition to a single value x=80x = 80, and 8080 satisfies both the conditions of being a positive integer and an integer.
(c) Predicate: 1≤x2≤41 \leq x^2 \leq 4, Domain: R\mathbb{R}
Solution:
The inequality 1≤x2≤41 \leq x^2 \leq 4 implies: x2∈[1,4].x^2 \in [1, 4].
Taking the square root, the solution for xx is: x∈[−2,−1]∪[1,2].x \in [-2, -1] \cup [1, 2].
Explanation:
For real numbers, x2≥1x^2 \geq 1 corresponds to values outside the interval (−1,1)(-1, 1), and x2≤4x^2 \leq 4 ensures the values remain within [−2,2][-2, 2]. Combining these gives x∈[−2,−1]∪[1,2]x \in [-2, -1] \cup [1, 2].
(d) Predicate: 1≤x2≤41 \leq x^2 \leq 4, Domain: Z∖{−1}\mathbb{Z} \setminus \{-1\}
Solution:
For x2∈[1,4]x^2 \in [1, 4] within the integers, xx must satisfy: x∈{−2,−1,1,2}.x \in \{-2, -1, 1, 2\}.
Excluding −1-1 from the domain, the truth set becomes: {−2,1,2}.\{-2, 1, 2\}.
Explanation:
The condition x2∈[1,4]x^2 \in [1, 4] provides the integer solutions {−2,−1,1,2}\{-2, -1, 1, 2\}. However, −1-1 is excluded, leaving {−2,1,2}\{-2, 1, 2\} as the truth set.