In: Advanced Math
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.
(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.