If [tex]\( U \)[/tex] is the set of all multiples of 3 less than 20, and [tex]\( M = \{3, 12, 18\} \)[/tex], find [tex]\( M' \)[/tex] (the complement of [tex]\( M \)[/tex]). Express your answer as a set in roster form (i.e., [tex]\(\{a, b, c, \ldots\}\)[/tex]).

Provide your answer below:



Answer :

To solve this problem, we need to find the complement of set [tex]\( M \)[/tex] in set [tex]\( U \)[/tex]. Let's break down the step-by-step solution:

1. Define the Universal Set [tex]\( U \)[/tex]:
The universal set [tex]\( U \)[/tex] is given as the set of all multiples of 3 that are less than 20. Here are these multiples:
[tex]\[ U = \{3, 6, 9, 12, 15, 18\} \][/tex]

2. Define the Set [tex]\( M \)[/tex]:
The set [tex]\( M \)[/tex] is given as:
[tex]\[ M = \{3, 12, 18\} \][/tex]

3. Find the Complement of [tex]\( M \)[/tex] in [tex]\( U \)[/tex]:
The complement of [tex]\( M \)[/tex] in [tex]\( U \)[/tex], denoted as [tex]\( M' \)[/tex] (or [tex]\( M^c \)[/tex]), is the set of elements that are in [tex]\( U \)[/tex] but not in [tex]\( M \)[/tex]. To find these elements, we list all elements of [tex]\( U \)[/tex] and then remove the elements that are in [tex]\( M \)[/tex]:

- Elements in [tex]\( U \)[/tex]: [tex]\( \{3, 6, 9, 12, 15, 18\} \)[/tex]
- Elements in [tex]\( M \)[/tex]: [tex]\( \{3, 12, 18\} \)[/tex]

Removing the elements of [tex]\( M \)[/tex] from [tex]\( U \)[/tex], we get:
[tex]\[ M' = \{6, 9, 15\} \][/tex]

4. Express the Complement of [tex]\( M \)[/tex] in Roster Form:
The complement of [tex]\( M \)[/tex] in [tex]\( U \)[/tex], expressed in roster form, is:
[tex]\[ M' = \{9, 6, 15\} \][/tex]

Therefore, the set representing the complement of [tex]\( M \)[/tex] in the universal set [tex]\( U \)[/tex] is:
[tex]\[ M' = \{9, 6, 15\} \][/tex]