# Let Q+ be the set of all positive rational numbers.

Question:

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

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

So, $*$ is a binary operation from $Q^{+} \times Q^{+} \rightarrow Q^{+}$.

(ii) For commutative binary operation, $a^{*} b=b^{*} a$.

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

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

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

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

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

As $a *(b * c) \neq(a * b) * c$, hence $*$ is not associative binary operation.