Question

Show that the set G of all positive rationals forms a group the compositions * defined by $a\ast b=\frac{ab}{3}$ for all $a,b\in G$

Answer 1

G be the set of all positive rationals

let $a\ast b=\frac{ab}{3}$

(i) closure property

Let $x,y\in Q^{+}$

Now $x\ast y=\frac{xy}{3}>0$

$(\because x>0,y>0\Rightarrow xy>0\Rightarrow \frac{xy}{3}>0$

$\therefore x\ast y\in Q^{+}.\forall x,y\in Q^{+}$

(ii) Associative property :-

Let $x,y,z\in Q^{+}$

$x\ast (y\ast z)=x\ast \left(\frac{yz}{3}\right)=\frac{x\left(yz\right)}{9}=\frac{\left(xy\right)z}{9}=\frac{(x\ast y)z}{3}$

$=(x\ast y)\ast z$

$\therefore$ Associative property hold $\forall x,y,z\in Q^{+}$

(iii) Identity :-

Let a be the identity in $Q^{+}$ s.t

$a\ast x=x\forall x\in Q^{+}$

$\Rightarrow \frac{ax}{3}=x\Rightarrow ax=3x$

$\Rightarrow (a-3)x=0\Rightarrow a-3=0(\because x>0)$

$\Rightarrow a=3$

$\therefore a=3$ is the identity in $Q^{+}\forall x\in Q^{+}$ under the given composition.

(iv) Inverse :-

Let $x\in Q^{+}$, Let y be element in $Q^{+}$ s.t

$xy=3\Rightarrow y=\frac{3}{x}>0\in Q^{+}$

$\therefore$ For each $x\in Q^{+},\frac{3}{x}$ is the inverse of $x$

Hence under the given composition $a\ast b=\frac{ab}{3}$ in $Q^{+},$ is a Group.