Given the sets $A = \{1, 2, ..., 16\}$, $B = \{x: 0 < x < 16, x \text{ is an odd integer}\}$, and $C = \{P: P < 16, P \text{ is prime}\}$, we want to find the elements in $A \cap B'$ and $(A \cup B \cup C)'$.

1. Problem Description

Given the sets $A = \{1, 2, ..., 16\}$, $B = \{x: 0 < x < 16, x \text{ is an odd integer}\}$, and $C = \{P: P < 16, P \text{ is prime}\}$, we want to find the elements in $A \cap B'$ and $(A \cup B \cup C)'$.

2. Solution Steps

First, we need to find the elements in sets $B$ and $C$.

$B = \{1, 3, 5, 7, 9, 11, 13, 15\}$

$C = \{2, 3, 5, 7, 11, 13\}$

i. Find $A \cap B'$:

$B'$ is the complement of $B$, which means all elements in the universal set $U$ (real numbers) that are not in $B$.

Since $A$ contains only integers from 1 to 16, $A \cap B'$ contains elements in $A$ that are not in $B$.

$A \cap B' = \{x: x \in A \text{ and } x \notin B\} = \{2, 4, 6, 8, 10, 12, 14, 16\}$

ii. Find $(A \cup B \cup C)'$:

First, find $A \cup B \cup C$.

$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16\}$

$B = \{1, 3, 5, 7, 9, 11, 13, 15\}$

$C = \{2, 3, 5, 7, 11, 13\}$

$A \cup B \cup C$ is the set of all elements in $A$, $B$, or $C$. Since $B$ is a subset of $A$, and $C$ contains elements also found in $A$,

$A \cup B \cup C = A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16\}$

So $A \cup B \cup C = A$.

Then, $(A \cup B \cup C)' = A'$

Since $A$ contains integers from 1 to 16, $A'$ will be all real numbers that are not integers from 1 to

1

6. Since the problem asks for a listing of all elements, and $A'$ consists of an infinite amount of numbers, we must interpret the universal set as $A$.

If the universal set is $A$, then $A' = \emptyset$, which is the empty set.

Therefore, $(A \cup B \cup C)' = \emptyset = \{\}$.

3. Final Answer

i. $A \cap B' = \{2, 4, 6, 8, 10, 12, 14, 16\}$

ii. $(A \cup B \cup C)' = \{\}$