Let * be a binary operation defined on NXN, by (a,b)*(c,d) = (ac, bd). Show that * is commutative and associative. Also find the identity element for * on NxN.

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Matthew Fonda | eNotes Employee

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First, let's recall what it means for a binary operation * on a set S to be commutative and associative. Furthermore, we'll recall what it means for an element to be an identity.

We say * is commutative if for all x, y in S, x*y = y*x

We say * is associative if for all x, y, z in S, (x * y) * z = x * (y * z)

We say an element e in S is an identity of for all x in S, x * e = x and e * x = x.



Let x = (a,b), y = (c,d) be elements of NxN.

Then x*y = (ac, bd) and y*x = (ca, db).

But ac=ca and bd=db, therefore x*y = y*x. Thus * is commutative.


Let x = (a, b), y = (c, d), z = (e, f) be elements of NxN.

Then, (x * y) * z = (ac, bd) * (e, f) = (ace, bdf) and x * (y * z) = (a, b) * (ce, df) = (ace, bdf). Thus * is associative.


Let x = (a, b). We are looking for an element e = (c, d) such that x * e = x.

x * e = (a, b) * (c, d) = (ac, bd)

Therefore for e to be an identify: ac = a, bd = b.

=> c = a/a = 1, d = b/b = 1

Therefore e = (1, 1) is an identity for (NxN, *).

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