Question

Let * be a binary operation on the set Q of rational numbers as follows:
(i) a * b=a−b
(ii) a * b=a2+b2
(iii) a * b=a+ab
(iv) a * b= (a−b)2
(v) a * b= $\frac {ab} {4}$
(vi) a * b=ab2

Find which of the binary operations are commutative and which are associative.

Answer 1

(i) On Q , the operation * is defined as a * b = a − b.
It can be observed that:
$\frac {1} {2}* \frac {1} {3}= \frac {1} {2} - \frac {1} {3}= \frac {1}{6}$ and $\frac {1}{3}*\frac {1} {2}= \frac {1} {3}-\frac {1}{2}= \frac {-1} {6}$
∴ $\frac {1}{2}*\frac {1}{3}$ ≠ $\frac {1}{3}*\frac {1}{2}$ ; where $\frac {1}{2}, \frac {1} {3}$ ∈Q.
Thus, the operation * is not commutative.
It can also be observed that:
$\bigg( \frac {1}{2} * \frac{1}{3}\bigg ) * \frac {1}{4}= \bigg ( \frac {1}{2}- \frac{1} {3}\bigg )* \frac {1}{4}= \frac {1}{6}*\frac {1}{4}= \frac {1} {6}-\frac {1}{4}= -\frac{1}{12}$
$\frac {1}{2}*\bigg (\frac {1}{3}* \frac {1}{4}\bigg )= \frac {1}{2}* \bigg (\frac {1}{3}-\frac {1}{4}\bigg )=\frac {1}{2}*\frac {1}{12}=\frac {1}{2}-\frac {1}{12}=\frac {5}{12}$
∴ $\bigg (\frac {1}{2}*\frac{1}{3}\bigg)* \frac {1}{4}$≠ $\frac {1}{2}*\bigg (\frac {1}{3}*\frac {1}{4}\bigg )$;where $\frac {1}{2}, \frac {1} {3}, \frac {1}{4}$ ∈Q.
Thus, the operation * is not associative.

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(ii) On Q , the operation * is defined as a * b = a2+ b2
For a, b ∈ Q , we have:
a * b=a2+b2=b2+a2=b * a
so a * b= b * a
Thus, the operation * is commutative.
It can be observed that:
(12)3=( 12+22 ) * 3 =(1+4) * 2 = 5 * 4 = 52+42=41
1 * (2 * 3)=1 * ( 2 * 3 )=1 * (22+32 )=1 * (4+9)=113=12+132=169.
∴ (1 * 2) * 3 ≠ 1 (2 * 3) ; where 1,2,3∈Q
Thus, ,the operation * is not associative.

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(iii) On Q , the operation * is defined as a * b = a + ab.
It can be observed that: 12=1+1x2=1+2=3.
21=2+2x1=2+2=4.
therefore 12≠21;where 1,2∈Q.
Thus, the operation * is not commutative.
It can also be observed that:
(12)3=(1+1x2)3=33=3+3x3=12
1(23)=1*(2+2x3)=18=1+1x8=9.
∴ (12)3≠1(2*3); where 1,2,3∈Q
Thus, the operation * is not associative.

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(iv) On Q , the operation * is defined by a * b = (a − b)2.
For a, b ∈ Q, we have:
a * b = (a − b)2
b * a = (b − a)2
= [− (a − b)]2
= (a − b)2
∴ a * b = b * a
Thus, the operation * is commutative.
It can be observed that:
(12)3=(1-2)23=(-1)23=13=(1-3)2=(-2)2=4
1(23)=1(2-3)2=1*(-1)2=11=(1-1)2=0
∴ (12)3≠1(2*3);where 1,2,3∈Q
Thus, the operation * is not associative.

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(v) OnQ , the operation * is defined as a $*$ b= $\frac {ab}{4}$,
For a, b ∈ Q, we have:
a $*$ b= $\frac {ab}{4}=\frac {ba}{4}=b*a$
∴ a * b = b * a
Thus, the operation * is commutative.
For a, b, c ∈ Q, we have: $(a *b) * c= \frac {ab}{4}*c= \frac {\frac {ab}{4}.c} {4} = \frac{abc} {16}$
$a*(b*c)=a*\frac {bc}{4}=\frac {\frac {a.bc} {4}}{4}=\frac {abc} {16}$
∴(a * b) * c = a * (b * c)
Thus, the operation * is associative.

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(vi) On Q , the operation * is defined as a * b = ab2
It can be observed that:
$\frac {1}{2}*\frac {1}{3}= \frac {1}{2}.(\frac{1}{3})^2=\frac {1}{2}.{1}{9}=\frac {1}{18}$.
$\frac {1}{3}*\frac {1}{2}= \frac {1}{3}.(\frac {1}{2})^2=\frac {1}{3}.\frac {1}{4}=\frac{1}{12}$.
therefore $\frac {1}{2}*\frac {1}{3}$ ≠ $\frac {1}{3}*\frac {1}{2}; where \frac {1}{2}, \frac {1}{3}$ ∈Q.
Thus, the operation * is not commutative.
It can also be observed that:
$\bigg(\frac {1}{2}*\frac{1}{3}\bigg )*\frac {1}{4}= \bigg [\frac {1}{2}\Big(\frac {1}{3}\Big)^2\bigg]*\frac {1}{4}=\frac {1}{18}* \frac {1}{4}=\frac{1}{18}. (\frac {1}{4})^2=\frac {1}{18x16}$.
$\frac {1}{2}*\bigg (\frac {1}{3}*\frac{1}{4}\bigg )=\frac {1}{2}*\bigg[\frac{1}{3}.\Big(\frac {1}{4}\Big)^2\bigg]=\frac {1}{2}*\frac {1}{48}=\frac {1}{2}.(\frac {1}{48})^2=\frac {1}{2x(48)^2}$.
∴ $\bigg(\frac {1}{2}*\frac{1}{3}\bigg)*\frac {1}{4}$≠ \frac {1}{2}$$* $\bigg(\frac{1}{3}*\frac {1}{4}\bigg)$; where $\frac {1}{2}, \frac{1}{3}$ $,\frac {1}{4}$∈Q.
Thus, the operation * is not associative.

Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined
in (v) is associative.