Question

Show that each of the relation R in the set A = { x ∈ Z : 0 ≤ x ≤ 12}, given by 
 I. R={(a,b):Ia-bI is a multiple of 4} 
 II. R={(a,b):a=b}
is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer 1

A={ x ∈ Z : 0≤ x ≤ 12}={0,1,2,3,4,5,6,7,8,9,10,11,12}
(i) R={(a,b):Ia-bI is a multiple of 4}
For any element a ∈A, we have (a, a) ∈ R as is a multiple of 4.
∴R is reflexive.

Now, let (a, b) ∈ R ⇒ is a multiple of 4.
\(\Rightarrow\)I-(a-b)I=Ib-aI is a multiple of 4.
\(\Rightarrow\) (b, a) ∈ R
∴R is symmetric.

Now, let (a, b), (b, c) ∈ R.
\(\Rightarrow\)Ia-bI is a multiple of 4 and Ib-cI is a multiple of 4.
\(\Rightarrow\)(a-b) is a multiple of 4 and (b-c) is a multiple of 4.
\(\Rightarrow\)(a-c)=(a-b)+(b-c) is a multiple of 4.
\(\Rightarrow\)Ia-cI is a multiple of 4.
\(\Rightarrow\) (a, c) ∈R
∴ R is transitive.

Hence, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9} since I1-1I=0 is a multiple of 4,
I5-1I=4 is a multiple of 4,and
I9-1I=8 is a multiple of 4.

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(ii) R = {(a, b): a = b}
For any element a ∈A, we have (a, a) ∈ R, since a = a.
∴R is reflexive.
Now, let (a, b) ∈ R.
⇒ a = b
⇒ b = a
⇒ (b, a) ∈ R
∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ a = b and b = c
⇒ a = c
⇒ (a, c) ∈ R
∴ R is transitive.

Hence, R is an equivalence relation.
The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.