Question

Show that each of the relation R in the set $A=\{x\in Z:0\le x\le 12\}$, given by
(i) R = {(a, b) : |a - b| is a multiple of 4}

(ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer 1

Given that $A=\{x\in Z:0\le x\le 12\}$
={0,1,2,3,4,5,6,7,8,9,10,11,12}

(i)R={(a,b):|a-b| is a multiple of 4}
For any element $a\in A$, we have $(a,a)\in R$ as |a-a|=0 is a multiple of 4. Therefore, R is reflexive.
Now, let {a,b} $\in R$
$\Rightarrow |a-b|$ is a multiple of 4, |b-a|=|-(a-b)|=|a-b|is multiple of 4.
$\Rightarrow (b,a)\in R$ Therefore, R is symmetric. Now, let (a,b),(b,c)$\in R$
So, |a-b|and |b-c| is multiple of 4.i.e a-b , b-c are multiples of 4. Then a-c=(a-b)+(b-c) is multiple of 4
$\Rightarrow$ |a-c|is also multiple of 4.
Therefore, R is transitive.
Hence, R is an equivalence relation.
$\Rightarrow$ The set of elements related to 1 is {1,5,9}, since
|1-1|=0 is multiple of 4
|5-1|=4 is a multiple of 4 and |9-1|=8 is a multiple of 4

Given that $A=\{x\in Z:0\le x\le 12\}$
={0,1,2,3,4,5,6,7,8,9,10,11,12}

R={(a,b):a=b}
For any element $a\in A$, we have $(a,a)\in R$, as a =a.
Therefore, R is reflexive.
Now, let (a,b)$\in R\Rightarrow (b,a)\in R$
$\Rightarrow a=b\Rightarrow b=a\Rightarrow (b,a)\in R$
Therefore, R is symmetric.
Now, let $(a,b)\in Rand(b,c)\in R$
$\Rightarrow a=b$ and $b=c\Rightarrow (a,c)\in R$. Therefore, R is transitive.
Hence, R is an equivalence relation.
The elements in R that are related to 1 will be those elements from set A. Which are equal to 1
Hence, the set of elements related to 1 is {1}.