Let R = {(a, b): a, b ∈ Z and (a + b) is even}. Show that R is an eq

1.	Let R = {(a, b): a, b ∈ Z and (a + b) is even}. Show that R is an equivalence relation on Z.
Answer» It is given that R = {(a, b): a, b ∈ Z and (a + b) is even} Reflexive: If a ∈ Z then a + a = 2a, which is even So we get (a, a) ∈ R Therefore, it is reflexive. Symmetric: Consider (a, b) ∈ Z then a + b is even The same way b + a is also even. So we know that (b, a) also belongs to R. We get (a, b) ∈ R and (b, a) ∈ R Therefore, R is symmetric. Transitive: Consider (a, b) and (b, c) ∈ R then a + b = 2k and b + c = 2r is even By adding them a + 2b + c = 2k + r We get a + c = 2 (k + r – b) So a + c is even (a, c) ∈ R Therefore, R is transitive. We know that R is reflexive, symmetric and transitive. Therefore, R is equivalence.

Let R = {(a, b): a, b ∈ Z and (a + b) is even}. Show that R is an equivalence relation on Z.

Answer»

It is given that

R = {(a, b): a, b ∈ Z and (a + b) is even}

Reflexive:

If a ∈ Z then a + a = 2a, which is even

So we get (a, a) ∈ R

Therefore, it is reflexive.

Symmetric:

Consider (a, b) ∈ Z then a + b is even

The same way b + a is also even.

So we know that (b, a) also belongs to R.

We get (a, b) ∈ R and (b, a) ∈ R

Therefore, R is symmetric.

Transitive:

Consider (a, b) and (b, c) ∈ R then a + b = 2k and b + c = 2r is even

By adding them

a + 2b + c = 2k + r

We get

a + c = 2 (k + r – b)

So a + c is even (a, c) ∈ R

Therefore, R is transitive.

We know that R is reflexive, symmetric and transitive.

Therefore, R is equivalence.

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