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