The binary operation * on R - {-1} defined by | vedclass

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(C) Given the binary operation $a * b = \frac{a}{b+1}$.Check for commutativity:$a * b = \frac{a}{b+1}$$b * a = \frac{b}{a+1}$Since $\frac{a}{b+1} 
eq

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$*$ is neither associative nor commutative

Vedclass · Answer

(C) Given the binary operation $a * b = \frac{a}{b+1}$.Check for commutativity:$a * b = \frac{a}{b+1}$$b * a = \frac{b}{a+1}$Since $\frac{a}{b+1} 
eq \frac{b}{a+1}$ for all $a, b \in R - \{-1\}$,the operation is not commutative.Check for associativity:$(a * b) * c = \left(\frac{a}{b+1}ight) * c = \frac{\frac{a}{b+1}}{c+1} = \frac{a}{(b+1)(c+1)}$$a * (b * c) = a * \left(\frac{b}{c+1}ight) = \frac{a}{\frac{b}{c+1} + 1} = \frac{a(c+1)}{b+c+1}$Since $\frac{a}{(b+1)(c+1)} 
eq \frac{a(c+1)}{b+c+1}$,the operation is not associative.Therefore,the operation $*$ is neither commutative nor associative.

$^*$	$1$	$2$	$3$	$4$	$5$
$1$	$1$	$1$	$1$	$1$	$1$
$2$	$1$	$2$	$2$	$2$	$2$
$3$	$1$	$2$	$3$	$3$	$3$
$4$	$1$	$2$	$3$	$4$	$4$
$5$	$1$	$2$	$3$	$4$	$5$

The binary operation $$ on $R - \{-1\}$ defined by $a b = \frac{a}{b+1}$ is:

Similar Questions

Determine which of the following binary operations on the set $N$ are associative and which are commutative. $a * b = \frac{a+b}{2}$ for all $a, b \in N$.

Determine whether or not each of the definitions of $$ given below gives a binary operation. In the event that $$ is not a binary operation,give justification for this. On $Z^{+}$,define $$ by $a b = a$.

Let $^$ be the binary operation on $N$ defined by $a \,^\, b = \text{H.C.F. of } a \text{ and } b$. Is $^$ commutative? Is $^$ associative? Does there exist an identity for this binary operation on $N$?

For each binary operation $^$ defined below,determine whether $^$ is commutative or associative. On $Z^+$,define $a ^* b = 2^{ab}$.

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The binary operation $*$ on $R - \{-1\}$ defined by $a * b = \frac{a}{b+1}$ is: