Let G be a set of rational numbers other than 1. Let * be an operation on G defined by a * b = a + b - ab for all a,b E G. prove that $(G_1 *)$ is a group.

**Solution**: Let G be a set of Rational number other than 1 and * be the binary operation then algebraic system is $(G_1 *)$

where * can be defined as

a * b = a + b - ab

**1] Closure Axiam.**

V a,b E G if a * b E G then

$(G_1 *)$ is closed.

$a * b = \frac{a+b}{R} - \frac{ab}{R}$

$\therefore$ a * b E G $\therefore$ $(G_1 *)$ is **closed.**

**2] Associative prop.**

V a,b,c E G if a*(b*c) = (a*b) *c

then $(G_1 *)$ is associative.

say a*(b*c) = (a*b) *c

a * (b+c-bc) = (a+b-ab)*c

a+(b+c-bc) - = (a+b-ab)+c - (a+b-ab) c

a(b+c-bc)

$\therefore$ a+b+c-bc - = a+b-ab+c-ac-bc+abc

ab-ac+abc

$\therefore$ **LHS = RHS**

$\therefore$ $(G_1 *)$ is associative.

**3. Identity element property.**

V a E G $\exists$ an element e E G such that

a * e = a = e * a

say a * e = a

a + e - ae = a

e = ae

$\therefore$ e = O E G

$\therefore$ Identity element exist in $(G_1 *)$

**4] Inverse element property.**

V a E G $\exists$ an element $\alpha$ E G such that

a * $\alpha$ = e = $\alpha$ * a

say a * $\alpha$ = e

a * $\alpha$ = e

a + $\alpha$ - a $\alpha$ = 0

a + $\alpha$(1-a) = 0

$\therefore$ $\alpha = \frac{-a}{1-a} E G$

$\therefore$ Inverse element exist in $(G_1 *)$

$\therefore$ $(G_1 *)$ is a **GROUP**