R = {(a, b) | a/b ∈ Q, a,b ∈ R - {0}}
Reflexivity :→ Let a ∈ R - {0}
Then a/a = 1 ∈ Q.
∴ (a, a) ∈ R \(\forall\) a ∈ R - {0}.
Which implies that relation R is a reflexive relation.
Symmetricity :→ Let a, b ∈ R - {0} such that
(a, b) ∈ R.
(i.e.) a/b ∈ Q
Now since a ≠ 0 and a ∈ R and b ∈ R - {0}.
Then b/a ∈ Q
⇒ (b, a) ∈ R.
This implies if (a,b) ∈ R then (b, a) ∈ R
\(\forall\) a, b ∈ R - {0}.
i.e., relation R is a symmetric relation.
Transitivity :→ Let a, b, c ∈ R - {0} such that (a, b) ∈ R and (b, c) ∈ R.
i.e., a/b ∈ Q and b/c ∈ Q
Then a/b x b/c ∈ Q (∵ product of two rational number is rational)
⇒ a/c ∈ Q
⇒ (a/c) ∈ R \(\forall\) a, b, c ∈ R - {0}.
Which implies that relation R is transitive relation. Since, relation R is reflexive, symmetric and transitive.
Therefore, relation R is an equivalence relation.