Show that the relation R on N × N, defined by (a, b) R (c, d) ⇔ a

1.	Show that the relation R on N × N, defined by (a, b) R (c, d) ⇔ a + d = b + c is an equivalent relation.
Answer» In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive. Given that, R be the relation in N ×N defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in N ×N. R is Reflexive if (a, b) R (a, b) for (a, b) in N ×N Let (a,b) R (a,b) ⇒ a+b = b+a which is true since addition is commutative on N. ⇒ R is reflexive. R is Symmetric if (a,b) R (c,d) ⇒ (c,d) R (a,b) for (a, b), (c, d) in N ×N Let (a,b) R (c,d) ⇒ a+d = b+c ⇒ b+c = a+d ⇒ c+b = d+a [since addition is commutative on N] ⇒ (c,d) R (a,b) ⇒ R is symmetric. R is Transitive if (a,b) R (c,d) and (c,d) R (e,f) ⇒ (a,b) R (e,f) for (a, b), (c, d),(e,f) in N ×N Let (a,b) R (c,d) and (c,d) R (e,f) ⇒ a+d = b+c and c+f = d+e ⇒ (a+d) – (d+e) = (b+c ) – (c+f) ⇒ a-e= b-f ⇒ a+f = b+e ⇒ (a,b) R (e,f) ⇒ R is transitive. Hence, R is an equivalence relation.

Show that the relation R on N × N, defined by (a, b) R (c, d) ⇔ a + d = b + c is an equivalent relation.

Answer»

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, R be the relation in N ×N defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in N ×N.

R is Reflexive if (a, b) R (a, b) for (a, b) in N ×N

Let (a,b) R (a,b)

⇒ a+b = b+a

which is true since addition is commutative on N.

⇒ R is reflexive.

R is Symmetric if (a,b) R (c,d) ⇒ (c,d) R (a,b) for (a, b), (c, d) in N ×N

Let (a,b) R (c,d)

⇒ a+d = b+c

⇒ b+c = a+d

⇒ c+b = d+a [since addition is commutative on N]

⇒ (c,d) R (a,b)

⇒ R is symmetric.

R is Transitive if (a,b) R (c,d) and (c,d) R (e,f) ⇒ (a,b) R (e,f) for (a, b), (c, d),(e,f) in N ×N

Let (a,b) R (c,d) and (c,d) R (e,f)

⇒ a+d = b+c and c+f = d+e

⇒ (a+d) – (d+e) = (b+c ) – (c+f)

⇒ a-e= b-f

⇒ a+f = b+e

⇒ (a,b) R (e,f)

⇒ R is transitive.

Hence, R is an equivalence relation.