Let R=\{(a, b): a, b \in Z and (a+b) is even \}. Show that R is an equivalence relation on Z.
Let R=\{(a, b): a, b \in Z and (a+b) is even \}. Show that R is an equivalence relation on Z.

Solution:

\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{Z} and (\mathrm{a}+\mathrm{b}) is even \} (As given)
If R is Reflexive, Symmetric and Transitive, then R is an equivalence relation.

Reflexivity:
Suppose a be an arbitrary element of Z
\mathrm{a}+\mathrm{a}=2 \mathrm{a}
As 2 \mathrm{a} is even
\Rightarrow(\mathrm{a}, \mathrm{a}) \in \mathrm{R}
Therefore, \mathrm{R} is reflexive.

Symmetric:
Suppose a and \mathrm{b} \in \mathrm{Z}, such that (\mathrm{a}, \mathrm{b}) \in \mathrm{R}
\Rightarrow \mathrm{a}+\mathrm{b}= even.
\begin{array}{l} \Rightarrow \mathrm{b}+\mathrm{a}=\text { even } \\ \Rightarrow(\mathrm{b}, \mathrm{a}) \in \mathrm{R} \end{array}
Therefore, \mathrm{R} is symmetric.

Transitivity:
Suppose \mathrm{a}, \mathrm{b} and c \in \mathrm{Z}, such that (\mathrm{a}, \mathrm{b}) \in \mathrm{R} and (\mathrm{b}, \mathrm{c}) \in \mathrm{R} \Rightarrow a+b=2 k (which is an even)
and b+c=21 (which is an even)
On adding both the above equations, we obtain
\begin{array}{l} \Rightarrow \mathrm{a}+\mathrm{c}+2 \mathrm{~b}=2(\mathrm{k}+1) \\ \Rightarrow \mathrm{a}+\mathrm{c}=2(\mathrm{k}+1)-2 \mathrm{~b} \end{array}
\Rightarrow \mathrm{a}+\mathrm{c} is an even number
\Rightarrow(a, c) \in R
Therefore, R is transitive.
As a result, R is an equivalence relation.