Show that * on R –{ - 1}, defined by \((a*b)=\frac{a}{(b+1)}\) is ne

1.	Show that * on R –{ - 1}, defined by \((a*b)=\frac{a}{(b+1)}\) is neither commutative nor associative.
Answer» let a = 1,b = 0 ∈ R - { - 1} ab = \(\frac{1}{0+1}\) = 1 And ba = \(\frac{1}{0+1}\) = 0 *Hence is not commutative.** Let c = 3. (ab)c = 1c = \(\frac{1}{3+1}=\frac{1}{4}\) a(bc) = a\(\frac{0}{3+1}\) = 10 = \(\frac{1}{0+1}\) = 1 Hence is not associative.**

Show that * on R –{ - 1}, defined by \((a*b)=\frac{a}{(b+1)}\) is neither commutative nor associative.

Answer»

let a = 1,b = 0 ∈ R - { - 1}

a*b = \(\frac{1}{0+1}\) = 1

And b*a = \(\frac{1}{0+1}\) = 0

Hence * is not commutative.

Let c = 3.

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

a*(b*c) = a*\(\frac{0}{3+1}\) = 1*0 = \(\frac{1}{0+1}\) = 1

Hence * is not associative.

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