Sure, let's break down the solution step by step to find the set [tex]\( A = \{ x \in \mathbb{N} \mid x < 80 \text{ and } x \text{ is a multiple of 9} \} \)[/tex].
1. Definition of [tex]\( \mathbb{N} \)[/tex]:
- [tex]\( \mathbb{N} \)[/tex] is the set of all natural numbers: [tex]\( \{0, 1, 2, 3, \ldots \} \)[/tex].
2. Condition 1: [tex]\( x < 80 \)[/tex]:
- We are looking for numbers [tex]\( x \)[/tex] that are less than 80.
3. Condition 2: [tex]\( x \text{ is a multiple of 9} \)[/tex]:
- A number [tex]\( x \)[/tex] is a multiple of 9 if there exists an integer [tex]\( k \)[/tex] such that [tex]\( x = 9k \)[/tex].
4. Finding all multiples of 9 that are less than 80:
- We start with the smallest multiple of 9, which is 0 (since [tex]\( 9 \times 0 = 0 \)[/tex]).
- We then continue by adding 9 to the previous multiple until we reach a number that is 80 or more.
Here are the multiples of 9 less than 80:
- [tex]\( 9 \times 0 = 0 \)[/tex]
- [tex]\( 9 \times 1 = 9 \)[/tex]
- [tex]\( 9 \times 2 = 18 \)[/tex]
- [tex]\( 9 \times 3 = 27 \)[/tex]
- [tex]\( 9 \times 4 = 36 \)[/tex]
- [tex]\( 9 \times 5 = 45 \)[/tex]
- [tex]\( 9 \times 6 = 54 \)[/tex]
- [tex]\( 9 \times 7 = 63 \)[/tex]
- [tex]\( 9 \times 8 = 72 \)[/tex]
- [tex]\( 9 \times 9 = 81 \)[/tex] (81 is not less than 80, so we stop here)
So, the set [tex]\( A \)[/tex] of all natural numbers less than 80 and multiples of 9 is:
[tex]\[ A = \{0, 9, 18, 27, 36, 45, 54, 63, 72\} \][/tex]
Therefore, the final answer is:
[tex]\[ \boxed{\{0, 9, 18, 27, 36, 45, 54, 63, 72\}} \][/tex]
