Question

Prove that the relation R on Z defined by $\left( a,b \right)\in R$ $\Leftrightarrow$ a – b is divisible by 5 is an equivalence relation on Z.

Answer 1

Hint: First we are going to look at the definition of symmetric, reflexive and transitive. And after that we will check all the three conditions and if it satisfies the condition for symmetric, reflexive and transitive then we can say that it is on equivalence relation on Z.

Complete step-by-step answer:
Let’s start our solution by first writing all the definition of the terms:
Symmetric: If we have a set containing two elements ‘a’ and ‘b’, then if the relation set has (a,b) then it must have (b,a) then we can say that it is symmetric.
Reflexive: If we have a set containing two elements ‘a’ and ‘b’, then if the relation set has (a,a) and (b,b) then we can say it is reflexive.
Transitive: If we have a set containing three elements ‘a’ , ‘b’, and ‘c’ then if the relation set has (a,b) and (b,c) then it must have (a,c) for transitive.
Now we have stated all the required definitions.
Now we will check each of them,
$\left( a,a \right)\in R$ , a – a =0 which is divisible by 5, hence it is reflexive.
If $\left( a,b \right)\in R$ then a – b is divisible by 5
$\Rightarrow b-a$ is also divisible by 5
$\Rightarrow \left( b,a \right)\in R$
Hence it is symmetric.
If $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then a – b is divisible by 5 and b – c is divisible 5.
Now if we add any two numbers divisible by 5 then the result will also be divisible by 5.
Therefore,
$\left( a-b \right)+\left( b-c \right)$ is also divisible by 5.
a – c is divisible by 5
$\left( a,c \right)\in R$
Hence, it is also transitive.
The relation is called equivalence when it is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Note: All the definitions of the terms that we have used is very important, without understanding it’s meaning clearly we cannot solve this question. And what does it mean to be an equivalence relation must be known.