Q82 / 180 · KCET 2017 · Mathematics
Binary operation * on \( R-\{-1\} \) defined by
\( a^{*} b=\frac{a}{b+1} \) is
\( a^{*} b=\frac{a}{b+1} \) is
- A\( { }^{*} \) is associative and commutative
- B\( { }^{*} \) is associative but not commutative
- Correct answerC\( { }^{*} \) is neither associative nor commutative
- D\( { }^{*} \) is commutative but not associative
Solution
Given that
\[ \begin{array}{l} a * b=\frac{a}{b+1} \\ \text { So, } b * a=\frac{b}{a+1} \end{array} \]
Since, \( a * b \neq b^{*} a \)
It is not commutative.
Now, \( \left(a^{*} b\right)^{*} c=\left(\frac{a}{b+1}\right)^{*} c \) \( =\frac{\left(\frac{a}{b+1}\right)}{c+1}=\frac{a}{(b+1)(c+1)} \) \( a *(b * c)=a^{*} \frac{b}{c+1} \) \( =\frac{a}{\left(\frac{b}{c+1}+1\right)}=\frac{a(c+1)}{(b+c+1)} \) Since, \( (a * b)^{*} C \neq a *\left(b^{*} c\right) \) It is not associative. Therefore, \( { }^{*} \) is neither commutative nor associative.
Therefore, \( { }^{*} \) is neither commutative nor associative.
\[ \begin{array}{l} a * b=\frac{a}{b+1} \\ \text { So, } b * a=\frac{b}{a+1} \end{array} \]
Since, \( a * b \neq b^{*} a \)
It is not commutative.
Now, \( \left(a^{*} b\right)^{*} c=\left(\frac{a}{b+1}\right)^{*} c \) \( =\frac{\left(\frac{a}{b+1}\right)}{c+1}=\frac{a}{(b+1)(c+1)} \) \( a *(b * c)=a^{*} \frac{b}{c+1} \) \( =\frac{a}{\left(\frac{b}{c+1}+1\right)}=\frac{a(c+1)}{(b+c+1)} \) Since, \( (a * b)^{*} C \neq a *\left(b^{*} c\right) \) It is not associative. Therefore, \( { }^{*} \) is neither commutative nor associative.
Therefore, \( { }^{*} \) is neither commutative nor associative.
Hence, the correct answer is option: 'C'.