Question

Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a − b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

Answer 1

Given that A ={1,2,3,4,5}, R ={(a,b): |a-b| is even}

Let $a\in A\Rightarrow |a-a|=0$ (which is even), $\forall$ a
So, R is reflexive.
Let $(a,b)\in R\Rightarrow$ |a-b|is even $\Rightarrow |-(b-a)|=|b-a|$ is also even
$\Rightarrow (b,a)\in R$. So, 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 $\Rightarrow (a,c)\in R$
So, 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 set are odd. So, 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 set are even. So, the modulus of the difference between any two elements will be even.
Also, no element of the {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. So, the modulus of the difference between the two elements (one from each of these two subsets)will not be even.