Question -
Answer -
Given R = {(a, b): a, b ∈ Z and a + b is even} is a relation defined on R.
Also given that Z be the set of integers
To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of Z.
Then, a ∈ R
Clearly, a + a = 2a is even for all a ∈ Z.
⇒ (a, a) ∈ R for all a ∈ Z
So, R is reflexive on Z.
Symmetry:
Let (a, b) ∈ R
⇒ a + b is even
⇒ b + a is even
⇒ (b, a) ∈ R for all a, b ∈ Z
So, R is symmetric on Z.
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ a + b and b + c are even
Now, let a + b = 2x for some x ∈ Z
And b + c = 2y for some y ∈ Z
Adding the above two equations, we get
A + 2b + c = 2x + 2y
⇒ a + c = 2 (x + y − b), which is even for all x, y, b ∈ Z
Thus, (a, c) ∈ R
So, R is transitive on Z.
Therefore R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z