# Show that the set G of all positive rationals forms a group under the composition $*$ defined by $a * b = (\large\frac{ab}{3})$ for all $a, b \in G.$

## 1 Answer

Toolbox:
• A non-empty set G, together with an operation $*$ i.e., $(G, *)$ is said to be a group if it satisfies the following axioms
• (1) Closure axiom : $a,b\in G\Rightarrow a*b\in G$
• (2) Associative axiom : $\forall a, b, c \in G, (a * b) * c = a * (b * c)$
• (3) Identity axiom : There exists an element $e \in G$ such that $a * e = e * a = a, \forall a ∈ G.$
• (4) Inverse axiom : $\forall a \in G$ there exists an element $a^{−1}\in G$ such that $a^{−1}*a=a*a^{−1}=e.$
• e is called the identity element of $G$ and $a^{−1}$ is called the inverse of a in $G.$
Step 1:
Let $G=Q^+$ the set of all positive rationals.Let $*$ is the operation defined by $a*b=\large\frac{ab}{3}$ for $a,b\in Q^+$
Closure : Let $a,b\in Q^+$
$a*b=\large\frac{ab}{3} $$\in Q^+ Since the product of two positive rational numbers is again a positive rational number. The closure property is satisfied. Step 2: Associativity : (a*b)*c= (\large\frac{ab}{3})*c \qquad\qquad= \large\frac{abc}{9} \qquad\qquad=a* \large\frac{bc}{3} \qquad\qquad= a*(b*c) The associative property is satisfied. Step 3: Existence of identity : Consider a,e\in Q^+ such that a*e=a \large\frac{ae}{3}$$=a$
$e=3$
Now $a*3=3*a=a$ for $a\in Q^+$
$\therefore 3$ is the identity element.
Step 4:
Existence of inverse :
Let $a\in Q^+$
Consider $a'$ such that $a*a'=3$
$\Rightarrow \large\frac{aa'}{3}$$=3\Rightarrow a'=\large\frac{9}{a}$$\in Q^+(a\neq 0)$
It can be seen that for $a'=\large\frac{9}{a}$$,a'a=aa'=3$
$\therefore$ every element in $Q^+$ has an inverse.
The four group axioms being satisfied,$(Q^+,*)$ is a group.
answered Sep 14, 2013

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