Question

(i) Is the binary operation * , defined on set N, given by \(a*b =\large \frac{a+b}{2}\) for all \( a,b \in\: N\), commutative?
(ii) Is the above binary operation * associative?

Answer 1

Answer:

(i) yes it is commutative

(ii) no it is not associative

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Step-by-step explanation:

(i)

b * a = ( b + a ) / 2    [ by definition of * ]

= ( a + b ) / 2            [ by commutativity of + ]

= a * b

Since b * a = a * b for all a, b ∈ N, the operation * is commutative.

(ii)

( a * b ) * c

= ( (a*b) + c ) / 2      [ by definition of * ]

= ( (a+b)/2 + c ) / 2   [ by defnition of * ]

= ( a + b + 2c ) / 4

Compare this with

a * ( b * c )

= ( a + (b*c) ) / 2    [ by definition of * ]

= ( a + (b+c)/2 ) / 2  [ by definition of * ]

= ( 2a + b + c ) / 4

Are these two expressions equal for all a, b, c ∈ N.  No!

For example, if a = 1, b = 2, c = 3, then we have

( a * b ) * c = ( a + b + 2c ) / 4 = ( 1 + 2 + 6 ) / 4 = 9 / 4

while

a * ( b * c ) = ( 2a + b + c ) / 4 = ( 2 + 2 + 3 ) / 4 = 7 / 4

Since it is not true that

( a * b ) * c = a * ( b * c ) for all a, b, c ∈ N,

the operation * is not associative.