Show that the relation 'a R b if and only if a−b is an even integer defined on the Z of integers is an equivalence relation.

Solutions

Expert Solution

(i) Since a−a=0 and 0 is an even integer

(a,a)∈R

∴ R is reflexive.

(ii) If (a−b) is even, then (b−a) is also even. then, if (a−b)∈R,(b,a)∈R

∴ The relation is symmetric.

(iii) If (a,b)∈R,(b,c)∈R, then (a−b) is even, (b−c) is even, then $$(a-b

+b-c)=a-c$$ is even.

∴ If (a,b)∈R,(b,c)∈R implies (a,c)∈R

∴ R is transitive.

Since R is reflexive, symmetric and transitive, it is an equivalence relation.

(i) Since a−a=0 and 0 is an even integer

(a,a)∈R

∴ R is reflexive.

(ii) If (a−b) is even, then (b−a) is also even. then, if (a−b)∈R,(b,a)∈R

∴ The relation is symmetric.

(iii) If (a,b)∈R,(b,c)∈R, then (a−b) is even, (b−c) is even, then $$(a-b

+b-c)=a-c$$ is even.

∴ If (a,b)∈R,(b,c)∈R implies (a,c)∈R

∴ R is transitive.

Since R is reflexive, symmetric and transitive, it is an equivalence relation.

Nipun Maity answered 2 years ago

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