Let Q+ be the set of all positive rational numbers. (i) Show that the

1.	Let Q+ be the set of all positive rational numbers. (i) Show that the operation * on Q+ defined by \(ab=\frac{1}{2}(a+b)\) is a binary operation. (ii) Show that is commutative. (iii) Show that * is not associative.
Answer» (i) * is an operation as \(ab=\frac{1}{2}(a+b)\)where a, b ∈ Q+. Let a =1 and b = 2 two integers. \(ab=\frac{1}{2}(1+2)\)\(\Rightarrow \frac{3}{2}\in Q ^+\) So, * is a binary operation from Q⁺ x Q⁺ → Q⁺. (ii) For commutative binary operation, ab = ba. \(ba=\frac{1}{2}(2+1)\)\(\Rightarrow \frac{3}{2}\in Q^+\) Since ab = ba, hence is a commutative binary operation. (iii) For associative binary operation, a(bc) = (ab) c. \(a(bc)=a\frac{1}{2}(b+c)\)\(\Rightarrow \cfrac{1}{2}\left(a+\frac{b+c}{2}\right)\)\(=\frac{1}{4}(2a+b+c)\) \((ab)c=\frac{1}{2}(a+b)c\)\(\Rightarrow \frac{1}{2}\left(\frac{a+b}{2}+c\right)\)\(=\frac{1}{4}(a+b+2c)\) *As a(bc) ≠(ab) c, hence is not associative binary operation.**

Let Q+ be the set of all positive rational numbers. (i) Show that the operation * on Q+ defined by \(ab=\frac{1}{2}(a+b)\) is a binary operation. (ii) Show that is commutative. (iii) Show that * is not associative.

Answer»

(i) * is an operation as \(a*b=\frac{1}{2}(a+b)\)where a, b ∈ Q+. Let a =1 and b = 2 two integers.

\(a*b=\frac{1}{2}(1+2)\)\(\Rightarrow \frac{3}{2}\in Q ^+\)

So, * is a binary operation from Q⁺ x Q⁺ → Q⁺.

(ii) For commutative binary operation, a*b = b*a.

\(b*a=\frac{1}{2}(2+1)\)\(\Rightarrow \frac{3}{2}\in Q^+\)

Since a*b = b*a, hence * is a commutative binary operation.

(iii) For associative binary operation, a*(b*c) = (a*b) *c.

\(a*(b*c)=a*\frac{1}{2}(b+c)\)\(\Rightarrow \cfrac{1}{2}\left(a+\frac{b+c}{2}\right)\)\(=\frac{1}{4}(2a+b+c)\)

\((a*b)*c=\frac{1}{2}(a+b)*c\)\(\Rightarrow \frac{1}{2}\left(\frac{a+b}{2}+c\right)\)\(=\frac{1}{4}(a+b+2c)\)

As a*(b*c) ≠(a*b) *c, hence * is not associative binary operation.