find p(A) and n [p(A)] for the set A={x/x is a multiple of 9,0
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Answer:
as below
Step-by-step explanation:
(i)
A={x∈Z:0≤x≤12}={0,1,2,3,4,5,6,7,8,9,10,11,12}
R={(a,b):∣a−b∣is a multiple of 4}
For any element a∈A, we have (a,a)∈R as ∣a−a∣=0 is a multiple of 4.
∴R is reflexive.
Now, let (a,b)∈R⇒∣a−b∣ is a multiple of 4.
⇒∣−(a−b)∣=∣b−a∣ is a multiple of 4.
⇒(b,a)∈R
∴ is symmetric.
Now, let (a,b),(b,c)∈R
⇒∣a−b∣ is a multiple of 4 and ∣b−c∣ is a multiple of 4.
⇒(a−b) is a multiple of 4 and (b−c) is a multiple of 4.
⇒(a−c)=(a−b)+(b−c) is a multiple of 4.
⇒∣a−c∣ is a multiple of 4.
⇒(a,c)∈R.
∴R is transitive.
Hence, R is an equivalence relation.
The set of elements related to 1 is {1,5,9} since
∣1−1∣=0 is a multiple of 4,
∣5−1∣=4 is a multiple of 4, and
∣9−1∣=8 is a multiple of 4
Hence, R is an equivalence relation.
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