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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 $a, b \in G.$

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  • 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} $$\in Q^+$
Since the product of two positive rational numbers is again a positive rational number.
The closure property is satisfied.
Step 2:
Associativity :
$(a*b)*c= (\large\frac{ab}{3})*c$
$\qquad\qquad= \large\frac{abc}{9}$
$\qquad\qquad=a* \large\frac{bc}{3}$
$\qquad\qquad= a*(b*c)$
The associative property is satisfied.
Step 3:
Existence of identity :
Consider $a,e\in Q^+$ such that
Now $a*3=3*a=a$ for $a\in Q^+$
$\therefore 3$ is the identity element.
Step 4:
Existence of inverse :
Let $a\in Q^+$
Consider $a'$ such that $a*a'=3$
$\Rightarrow \large\frac{aa'}{3}$$=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 by sreemathi.v

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