Show that the relation R in the set A = {1, 2, 3, 4, 5} given by

Solution:

A = {1, 2, 3, 4, 5}

$\mathrm{R}=\{(a, b):|a-b|$ is even $\}$

It is clear that for any element $a \in A$, we have $|a-a|=0$ (which is even).

∴R is reflexive.

Let $(a, b) \in R$.

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

$\Rightarrow|-(a-b)|=|b-a|$ is also even.

$\Rightarrow(b, a) \in \mathrm{R}$

∴R is symmetric.

Now, let $(a, b) \in R$ and $(b, c) \in R$.

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

$\Rightarrow(a-b)$ is even and $(b-c)$ is even.

$\Rightarrow(a-c)=(a-b)+(b-c)$ is even.      [Sum of two even integers is even]

$\Rightarrow|a-c|$ is even.     

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

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.