Question

Let Q be the set of all positive rational numbers. 

(i) Show that the operation * on Q + defined \(a*b=\frac{1}{2}(a+b)\) by is a binary operation. 

(ii) Show that * is commutative. 

(iii) Show that * is not associative.

Answer 1

(i)Let a = 1, b = 2 ∈ Q ⁺ 

a*b = \(\frac{1}{2}(1+2)\)= 1.5 ∈ Q⁺ 

* is closed and is thus a binary operation on Q⁺ 

(ii) a*b = \(\frac{1}{2}(1+2)\) = 1.5  

And b*a = \(\frac{1}{2}(2+1)\) = 1.5 

Hence * is commutative. 

(iii)let c = 3. 

(a*b)*c = 1.5*c = \(\frac{1}{2}(1.5+3)=2.75\)

a*(b*c) = a*\(\frac{1}{2}(2+3)\) = 1*2.5 = \(\frac{1}{2}(1+2.5)\) = 1.75 

hence * is not associative.