BINARY OPERATIONS 2

Definition 3 (Commutative operation)

A binary operation * on a set S is commutative if a*b = b*a for all a, b in S.

Example 1 :

On Q, define a binary operation * by a*b = ab + 1. Show that the binary operation * is commutative on Q.
Let a, b element of Q,
(To check whether a*b = b*a)

LHS : a*b
= ab+1
= ba+1 since multiplication is commutative on Q.
= b*a = RHS
Therefore, * is commutative on Q.

Example 2 :
On Z, define a*b = a-b. Determine whether * is commutative on Z.

Let a, b element of Z
To check whether a*b = b*a

Note that,
by using counter example
let 1, 2 in Z.
LHS : 1 - 2 = -1
RHS : 2 - 1 = 1
but, since LHS is not equal to RHS, then * is not commutative on Z.



Definition 4 (Associative Operation)

A binary operation * on a set S is associative if (a*b)*c = a*(b*c) for all a, b, c in S.

Example 1 :

On Q, define * by a*b = (ab)/2. Determine whether * is associative on Q.
Let a, b, c element of Q.

(To show (a*b)*c = a*(b*c))
LHS : (a*b)*c
= (ab)/2 * c let x = (ab)/2
= x*c = xc/2
= (ab)c/4
= a(bc)/4 since multiplication is commutative on Q.
= a*(bc/2)
= a*(b*c) : RHS
Therefore, * is commutative on Q.

Note :
For infinite set, a binary operation on a set can be defined by means of a table.

* is commutative if and only if the entries in the table are symmetric with respect to the diagonal.

Since the entries are symmetric with respect to the diagonal, then * is commutative.

Example 2 :


a*c = b
c*a = c
a*c is not equal to c*a.
Therefore, * is not commutative on S.

Example 3 :
Complete the following table so that * is a commutative binary operation on the set
S = {a, b, c, d}