Question

Let * be a binary operation on the set Q of rational numbers as follows: (i) a * b = a − b (ii) a * b = a 2 + b 2 (iii) a * b = a + ab (iv) a * b = ( a − b ) 2 (v) (vi) a * b = ab 2 Find which of the binary operations are commutative and which are associative.

Answer 1

(i)

The given binary operator is *, defined as a*b=a−b.

Let, two numbers 4 and 5 belongs to set of rational numbers.

4*5=4−5 =−1

Similarly 5*4 can be calculated as,

5*4=5−4 =1

Since a*b≠b*a, the binary operation a*b=a−b is not commutative.

For binary operation to be associative, ( a*b )*c=a*( b*c ) must be true.

Consider three numbers 4, 5, 6 in the set of rational numbers and apply the binary operation.

( 4*5 )*6=4*( 5*6 ) ( 4−5 )*6=4*( 5−6 ) −1*6=4*( −1 ) −1−6=4−( −1 )

Further solve.

−1−6=4−( −1 ) −7≠5

Since, the condition for associative operation is not satisfied. Thus, the operation is not associative.

(ii)

The given binary operator is *, defined as a*b= a 2 + b 2 .

For commutative property,

b*a= b 2 + a 2 = a 2 + b 2 =a*b

Hence, the binary operation is commutative.

For binary operation to be associative ( a*b )*c=a*( b*c ) must be true.

Consider three numbers 1, 2, 3 in the set of rational numbers and apply the binary operation.

( 1*2 )*3=1*( 2*3 ) ( 1 2 + 2 2 )*3=1*( 2 2 + 3 2 ) 5*3=1*13 5 2 + 3 2 = 1 2 + 13 2

Further solve.

25+9=1+169 34≠170

Since, the condition for associative operation is not satisfied, the operation is not associative.

(iii)

The given binary operator is *, defined as a*b=a+ab.

Let, two numbers 1 and 2 belongs to set of rational numbers, then

1*2=1+1⋅2 =3

Similarly 2*1 can be calculated as,

2*1=2+2⋅1 =4

Since a*b≠b*a, the binary operation a*b=a+ab is not commutative.

For binary operation to be associative, ( a*b )*c=a*( b*c ) must be true.

Consider three numbers 1, 2, 3 in the set of rational numbers and apply the binary operation.

( 1*2 )*3=1*( 2*3 ) ( 1+1⋅2 )*3=1*( 2+2⋅3 ) ( 1+2 )*3=1*( 2+6 ) 3*3=1*8

Further solve.

3*3=1*8 3+3⋅3=1+1⋅8 12≠9

Since, the condition for associative operation is not satisfied, the operation is not associative.

(iv)

The given binary operator is *, defined as a*b= ( a−b ) 2 .

For commutative property,

b*a= ( b−a ) 2 = ( −( a−b ) ) 2 = ( a−b ) 2 =a*b

Hence, the binary operation is commutative.

For binary operation to be associative, ( a*b )*c=a*( b*c ) must be true.

Consider three numbers 1, 2, 3 in the set of rational numbers and apply the binary operation.

( 1*2 )*3=1*( 2*3 ) ( 1−2 ) 2 *3=1* ( 2−3 ) 2 ( −1 ) 2 *3=1* ( −1 ) 2 1*3=1*1

Further solve.

( 1−3 ) 2 = ( 1−1 ) 2 4≠0

Since, the condition for associative operation is not satisfied, the operation is not associative.

(v)

The given binary operator is *, defined as a*b= ab 4 .

For commutative property,

b*a= ba 4 = ab 4 =a*b

Hence, the binary operation is commutative.

For binary operation to be associative, ( a*b )*c=a*( b*c ) must be true.

Apply the binary operation on the equation.

( a*b )*c=a*( b*c ) ( ab 4 )*c=a*( bc 4 ) ( ab 4 )⋅c 4 = a⋅( bc 4 ) 4 abc 16 = abc 16

Since, the condition for associative operation is satisfied, the operation is associative.

(vi)

The given binary operator is *, defined as a*b=a b 2 .

Let, two numbers 1 and 2 belongs to set of rational numbers.

1*2=1⋅ 2 2 =4

Similarly 2*1 can be calculated as,

2*1=2⋅ 1 2 =2

Since a*b≠b*a, the binary operation a*b=a b 2 is not commutative.

For binary operation to be associative, ( a*b )*c=a*( b*c ) must be true.

Consider three numbers 1, 2, 3 in the set of rational numbers and apply the binary operation.

( 1*2 )*3=1*( 2*3 ) ( 1⋅ 2 2 )*3=1*( 2⋅ 3 2 ) 4*3=1*18 4⋅ 3 2 =1⋅ 18 2

Further solve.

4⋅9=1⋅324 36≠324

Since, the condition for associative operation is not satisfied, the operation is not associative.