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in Sets, Relations and Functions by (9.2k points)

Consider the binary operation * and 0 defined by the following tables on set S = {a, b, c, d}.


 

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We observe the following:

a * b = b * a = a a * a = a 

c * a = a * c = a b * b = b 

a * d = d * a = a c * c = d 

b * c = c * b = c d * d = c 

b * d = d * b = d 

c * d = d * c = b 

There ‘ * ’ is commutative. 

Also, a * (b * c) = a * (c) = a (From above) 

(a * b) * c = (a) * c = a (Also from above) 

Hence, ‘ * ’ is associative too. 

Therefore, to find the identity element, e for e belong to S, we need: 

a * e = e * a = a, a belong to S. 

Therefore, a * e = a 

We find that there is no unique element e which satisfies the condition. 

e = a or b or c or d for a. 

Since the identity is not unique, the inverse will also be not unique.

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