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Binary operation * on \( R-\{-1\} \) defined by
\( a^{*} b=\frac{a}{b+1} \) is
Options:
\( a^{*} b=\frac{a}{b+1} \) is
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Verified Answer
The correct answer is:
\( { }^{*} \) is neither associative nor commutative
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.
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