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Q.2 For each operation ∗ defined below, determine whether ∗ is binary, commutative
or associative.

(vi) On R - \{-1 \}, definea * b = \frac{a}{b +1}

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(iv) On R - \{-1 \}, definea * b = \frac{a}{b +1}

 

         1\ast 2 = \frac{1}{2+1}=\frac{1}{3}         and      2\ast 1 = \frac{2}{2+1}= \frac{2}{3}

             \Rightarrow               1\ast 2\neq 2\ast 1       for 1,2 \in R - \{-1 \}

            \therefore  the operation is not commutative.

         (1\ast 2)\ast 3 = (\frac{1}{2+1}) \ast 3 = \frac{1}{3} \ast 3 = \frac{\frac{1}{3}}{3+1}= \frac{1}{12}

            1\ast (2\ast 3) = 1 \ast (\frac{2}{3+1}) = 1 \ast \frac{2}{4} = 1 \ast \frac{1}{2} = \frac{1}{\frac{1}{2}+1} = \frac{2}{3}

              \therefore             (1\ast 2)\ast 3\neq 1\ast (2\ast 3) ;     where 1,2,3 \in R - \{-1 \}

\therefore  operation * is not associative.   

Posted by

seema garhwal

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