On Q, the set of all rational numbers, * is defined by a ^* b= a-b 2 , shown that * is no associative.

The binary operator * defined as, a ^* b= a-b 2 , for all a, b . Now, Associativity: Let a, b, c . Then, (a ^* b) ^* c= a-b 2 ^* c= a-b 2 -c 2 = a-b-2c 4 ....(i) and, a ^* (b ^* c)=a ^* b-c 2 = a- b-c 2 2 = 2a-b+c 4 .....(ii) From (i) and (ii),(a ^* b) ^* c ^* (b ^* c) Hence, '*' is not associative on Q.

On Q, the set of all rational numbers, * is defined by a ^* b= a-…

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On Q, the set of all rational numbers, * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}-\text{b}}{2},$ shown that * is no associative.

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The binary operator * defined as, $\text{a}\ ^*\ \text{b}=\frac{\text{a}-\text{b}}{2},$ for all $\text{a, b}\in\text{Q}.$ Now, Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then, $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{a}-\text{b}}{2}\ ^*\ \text{c}=\frac{\frac{\text{a}-\text{b}}{2}-\text{c}}{2}$ $=\frac{\text{a}-\text{b}-2\text{c}}{4}\ ....(\text{i})$ and, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \frac{\text{b}-\text{c}}{2}=\frac{\text{a}-\frac{\text{b}-\text{c}}{2}}{2}$ $=\frac{2\text{a}-\text{b}+\text{c}}{4}\ .....(\text{ii})$ From (i) and (ii),$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}\neq\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})$
Hence, '*' is not associative on Q.

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