Question:
Let Q be the set of all positive rational numbers.
(i) Show that the operation $*$ on $\mathrm{Q}^{+}$defined by $\mathrm{a} * \mathrm{~b}=\frac{1}{2}(\mathrm{a}+\mathrm{b})$ is a binary operation.
(ii) Show that $*$ is commutative.
(iii) Show that $*$ is not associative.
Solution:
(i) Let $a=1, b=2 \in Q+$
$a * b=\frac{1}{2}(1+2)=1.5 \in Q+$
* 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.