Question

Let n be a fixed positive integer. Define a relation R on Z as follows:
$\left( a,b \right)\in R\Leftrightarrow n\text{ divides }a-b$.
Show that R is an equivalence relation on Z.

Answer 1

Hint:For solving this problem, first we have to prove that the relation is reflexive, symmetric and transitive by using appropriate conditions. Once these three relations are established, then R becomes an equivalence relation on Z.

Complete step-by-step answer:
According to the problem statement, we are given a relation R on Z such that $\left( a,b \right)\in R\Leftrightarrow n\text{ divides }a-b$. This implies that aRb = a – b is divisible by n for $a,b\in Z\text{ and }n\in {{Z}^{+}}$
First, we have to prove that this relation is reflexive. So, for a relation to be reflexive, $\left( a,a \right)\in R$.
Putting it in the relation, we get
a – a = 0 is divisible by n. So, R is a reflexive relation.
Now, we have to prove that this relation is symmetric. So, for a relation to be symmetric, $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$.
Now, from the relation statement, a – b is divisible by n. Therefore, a – b = pn where $p\in Z$. Taking negative common from both sides we get, b – a = - pn where $-p\in Z$. So, we proved $\left( b,a \right)\in R\text{ by using }\left( a,b \right)\in R$. Hence, this relation is symmetric.
Now, we have to prove that this relation is transitive. So, for a relation to be transitive, $\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
Now, from the relation statement, a – b is divisible by n. Therefore, a – b = pn where $p\in Z\ldots (1)$.
Also, for (b, c), b – c is divisible by n. Therefore, b – c = qn where $q\in Z\ldots (2)$. Now, adding equation (1) and equation (2), we get
$\begin{aligned} & a-b+b-c=pn+qn \\\ & a-c=\left( p+q \right)n\text{ where }p+q\in Z \\\ \end{aligned}$
Therefore, a – c is divisible by n. So, we proved $\left( a,c \right)\in R\text{ by using }\left( a,b \right)\in R,\left( b,c \right)\in R$. Hence, it is transitive also.
So, we proved that relation R is reflexive, symmetric and transitive. Hence, relation R is an equivalence relation.

Note: The knowledge of equivalence of a relation is must for solving this problem. Students must remember all the necessary conditions for proving a set is reflexive, symmetric and transitive. All these relations together yield an equivalence relation.