Question

Prove that the relation R on set z of all integers defined by (a,b)belongs to R =(a-b) is divisible by 5 is an equivalence relation on z
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Answer 1

$\large{\underline{\rm{\orange{\bf{Question:-}}}}}$

Prove that the relation R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5 is an equivalence relation on Z.

$\large{\underline{\rm{\orange{\bf{Given:-}}}}}$

The relation R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5

$\large{\underline{\rm{\orange{\bf{To \: Find:-}}}}}$

To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.

$\large{\underline{\rm{\orange{\bf{Solution:-}}}}}$

Given that, relation R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5

Check these properties on R,

Reflexivity:

Let a be an arbitrary element of R. Then,

$\implies \sf a-a=0=0 \times 5$

$\implies \sf a-a$ is divisible by 5

$\implies \sf (a, \: a) \in R$ for all $\sf a \in Z$

So, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

$\implies \sf a-b$ is divisible by 5

$\implies \sf (b, \: a) \in R$ for all $\sf a, \: b \in Z$

So, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

$\implies \sf a-b$ is divisible by 5

$\implies \sf a-b=5p$ for some Z

Also, b − c is divisible by 5

$\implies \sf b-c=5q$ for some Z

Adding the above two equations, we get

$\sf a -b + b - c = 5p + 5q$

$\implies \sf a - c = 5 ( p + q )$

$\implies \sf a-c$ is divisible by 5

Here, p + q ∈ Z

$\implies \sf (a, \: c) \in R$ for all $\sf a, \: c \in Z$

So, R is transitive on Z.

Therefore R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.