Let A = N × N. Define * on A by (a, b) * (c, d) = (a + c, b + d).
Show that
(i) A is closed for *,
(ii) * is commutative,
(iii) * is associative,
(iv) identity element does not exist in A.
(i) A is said to be closed on * if all the elements of (a, b) *(c, d) = (a+ c, b+ d) belongs to N×N for A = N×N.
Let a = 1, b = 3, c = 8, d = 2
(1, 3) * (8, 2) = (1+8, 3+2)
= (9, 5) ∈N×N
Hence A is closed for *.
(ii) For commutative,
(c, d) *(a, b) = (c+ a, d+ b)
As addition is commutative a+ c = c+ a and b+ d = d+ b, hence * is commutative binary operation.
(iii) For associative,
(a, b) *((c, d) *(e, f)) = (a, b) *(c+ e, d+ f)
= (a+ c+ e, b+ d+ f)
((a, b) *(c, d)) *(e, f) = (a+ c, b+ d) *(e, f)
= (a+ c+ e, b+ d+ f)
As (a, b) *((c, d) *(e, f)) = ((a, b) *(c, d)) *(e, f), hence * is an associative binary operation.
(iv) For identity element (e1, e2), (a, b) *(e1, e2) = (e1, e2) *(a, b) = (a, b) in a binary operation.
(a, b) *(e1, e2) = (a, b)
(a+e1, b+e2) = (a, b)
(e1, e2) = (0, 0)
As (0,0) ∉N×N, hence identity element does not exist for *.