For binary operation * defined on R − {1} such that \(\rm {a}^{*} {b}=\frac{{a}}{{b}+1}\) is

Concept Used:

Commutative: A binary operation * is commutative if for all a, b, a * b = b * a

Associative: A binary operation * is associative if for all a, b, c, (a * b) * c = a * (b * c)

Calculation:

1) Checking for Commutativity:

a * b = \(\frac{a }{(b + 1)}\)

b * a = \(\frac{b }{(a + 1)}\)

Since \(\frac{a }{(b + 1)}\) is not always equal to \(\frac{b }{(a + 1)}\), the operation is not commutative.

Example: Let a = 2 and b = 3

a * b = \(\frac{2 }{ (3 + 1)}\) = \(\frac{2 }{4}\) = \(\frac{1}{2}\)

b * a = \(\frac{3 }{ (2 + 1)}\) = 1

\(\frac{1}{2}\) ≠ 1

2) Checking for Associativity:

\((a * b) * c = \frac{a }{ (b + 1)} * c = \frac{\frac{a }{ (b + 1)}} { (c + 1) }= \frac{a }{((b + 1)(c + 1))}\)

\(a * (b * c) = a * \frac{b }{(c + 1)}= \frac{a }{ (\frac{b }{ (c + 1)} + 1)} = \frac{a }{\frac{(b + c + 1) }{ (c + 1)}}= \frac{a(c + 1) }{(b + c + 1)}\)

Since \(\frac{a }{((b + 1)(c + 1))}\) is not always equal to \(\frac{a(c + 1) }{(b + c + 1)}\), the operation is not associative.

Example: Let a = 2, b = 3, and c = 4

\((a * b) * c = \frac{2 }{ 20} = \frac{1}{10}\)

\(a * (b * c) = \frac{10 }{ 8} = \frac{5}{4}\)

\(\frac{1}{10} ≠ \frac{5}{4}\)

∴ The binary operation * is neither commutative nor associative.

Hence option 4 is correct