Question

If ‘Q’ is the set of all non – zero rational numbers, {\mathbf{R}} = \left\\{ {\left( {a,b} \right)/a = \dfrac{1}{b}} \right\\} is a relation in Q. Is R an equivalence relation?

Answer 1

We will use the definition of equivalence relation that an equivalence relation, R is a relation that is reflexive, symmetric and transitive, where, for $\forall$ a, b and c $\in$R, the reflexivity is when $a = a$, for symmetricity: if $a = b$, then $b = a$ and for transitivity, if $a = b$ and $b = c$, then $a = c$. We will check if all these properties satisfy or not. If they satisfy, then R will be an equivalence relation.

Complete step by step solution:
We are given that ‘Q’ is the set of all non – zero rational numbers and {\mathbf{R}} = \left\\{ {\left( {a,b} \right)/a = \dfrac{1}{b}} \right\\} is a relation in Q.
We are required to check if R is an equivalence relation or not.
Definition: In mathematics, an equivalence relation is the relation which is transitive, reflexive and symmetric. $\forall$ a, b and c $\in$R, the reflexive property states: $a = a$, for the symmetricity, if$a = b$, then $b = a$ and for transitivity, if $a = b$ and $b = c$, then $a = c$.
In order to check if R is an equivalence relation, we will see if {\mathbf{R}} = \left\\{ {\left( {a,b} \right)/a = \dfrac{1}{b}} \right\\} satisfies these three properties individually.
For reflexivity: We can see that $a \ne \dfrac{1}{a}$ which means that R is not reflexive and

Hence not an equivalence relation.

Note:
In this question, you need to verify the given relation R is equivalent or not only by checking the reflexive, symmetric and transitive nature. You can also verify the other two properties as:
For symmetricity: if $a = \dfrac{1}{b}$, then $b = \dfrac{1}{a}$. Hence, the relation is symmetric.
For transitivity: if $a = \dfrac{1}{b}$ and $b = \dfrac{1}{c}$, then $a = c$. Hence, the relation is transitive.