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: 
(1*2)*3=( 1²+2² ) * 3 =(1+4) * 2 = 5 * 4 = 5²+4²=41
1 * (2 * 3)=1 * ( 2 * 3 )=1 * (2²+3² )=1 * (4+9)=1*13=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: 1*2=1+1x2=1+2=3.
2*1=2+2x1=2+2=4.
therefore 1*2≠2*1;where 1,2∈Q.
Thus, the operation * is not commutative.
It can also be observed that:
(1*2)*3=(1+1x2)*3=3*3=3+3x3=12
1*(2*3)=1*(2+2x3)=1*8=1+1x8=9.
∴ (1*2)*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:
(1*2)*3=(1-2)2*3=(-1)2*3=1*3=(1-3)²=(-2)²=4
1*(2*3)=1*(2-3)²=1*(-1)²=1*1=(1-1)²=0
∴ (1*2)*3≠1*(2*3);where 1,2,3∈Q
Thus, the operation * is not associative.

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(v) On Q, 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.