Question

if the set u = {all positive integers} and set a = {x|x ∈ u and x is an odd positive integer}, which describes the complement of set a, a^c?
a^c = {x|x ∈ u and is a negative integer}
a^c = {x|x ∈ u and is an even positive integer}
a^c = {x|x ∈ u and is not an integer}
a^c = {x|x ∈ u and is zero}

Explanation:

Step1: Recall Complement Definition

The complement of a set \( A \) (denoted \( A^c \)) in a universal set \( U \) is the set of all elements in \( U \) that are not in \( A \). Here, \( U = \{ \text{all positive integers} \} \) and \( A = \{ x | x \in U \text{ and } x \text{ is an odd positive integer} \} \).

Step2: Analyze Elements in \( U \) and \( A \)

Positive integers are either odd or even (by the definition of parity). So, elements in \( U \) not in \( A \) (i.e., \( A^c \)) must be positive integers that are not odd, which means they are even positive integers.

Step3: Evaluate Each Option

• Option 1: \( A^c = \{ x | x \in U \text{ and is a negative integer} \} \) – \( U \) has positive integers, so negative integers are not in \( U \). Eliminate.
• Option 2: \( A^c = \{ x | x \in U \text{ and is an even positive integer} \} \) – Matches the complement definition (even positive integers are in \( U \) and not odd).
• Option 3: \( A^c = \{ x | x \in U \text{ and is not an integer} \} \) – \( U \) is all positive integers, so all elements of \( U \) are integers. Eliminate.
• Option 4: \( A^c = \{ x | x \in U \text{ and is zero} \} \) – Zero is not a positive integer, so not in \( U \). Eliminate.

Answer:

\( \boldsymbol{A^c = \{x|x \in U \text{ and is an even positive integer}\}} \) (the second option)