Solve this

Solution:

In order to show $R$ is an equivalence relation we need to show $R$ is Reflexive, Symmetric and Transitive.

Given that, $\forall a, b \in A, R=\{(a, b):|a-b|$ is even $\}$.

Now,

$\underline{R}$ is Reflexive if $(a, a) \in \underline{R} \underline{\forall} \underline{a} \in \underline{A}$

For any a $\in A$, we have

$|a-a|=0$, which is even.

$\Rightarrow(a, a) \in R$

Thus, $R$ is reflexive.

$\underline{R}$ is Symmetric if $(a, b) \in \underline{R} \Rightarrow \underline{(b, a)} \in \underline{R} \underline{\forall} \underline{a, b} \in \underline{A}$

$(a, b) \in R$

$\Rightarrow|\mathrm{a}-\mathrm{b}|$ is even.

$\Rightarrow|b-a|$ is even.

$\Rightarrow(b, a) \in R$

Thus, $R$ is symmetric.

$\underline{R}$ is Transitive if $(a, b) \in \underline{R}$ and $(b, c) \in \underline{R} \Rightarrow(a, c) \in \underline{R} \forall \underline{a}, b, c \in \underline{A}$

Let $(a, b) \in R$ and $(b, c) \in R \forall a, b, c \in A$

$\Rightarrow|a-b|$ is even and $|b-c|$ is even

$\Rightarrow(\mathrm{a}$ and $\mathrm{b}$ both are even or both odd) and ( $\mathrm{b}$ and $\mathrm{c}$ both are even or both odd)

Now two cases arise:

Case 1 : when b is even

Let $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow|a-b|$ is even and $|b-c|$ is even

$\Rightarrow a$ is even and $c$ is even $[\because b$ is even $]$

$\Rightarrow|a-c|$ is even [ $\because$ difference of any two even natural numbers is even]

$\Rightarrow(a, c) \in R$

Case $2:$ when b is odd

Let $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow|\mathrm{a}-\mathrm{b}|$ is even and $|\mathrm{b}-\mathrm{c}|$ is even

$\Rightarrow \mathrm{a}$ is odd and $\mathrm{c}$ is odd $[\because \mathrm{b}$ is odd $]$

$\Rightarrow|\mathrm{a}-\mathrm{c}|$ is even $[\because$ difference of any two odd

natural numbers is even]

$\Rightarrow(a, c) \in R$

Thus, $R$ is transitive on $Z$.

Since $R$ is reflexive, symmetric and transitive it is an equivalence relation on $Z$.