Let G x epsilon R x 1 and define a b ab a b 2 for all a

Let G:= {x epsilon R: x > 1}, and define a * b:= ab - a - b + 2 for all a, b epsilon R. Show that: G is closed under *; the set G under the operation * forms an abelian group.

Solution

a) it is closed under * since

if a,b >1

then (a-1),(b-1)>0

and so

a*b = ab- a -b +2 = a(b-1)-1(b-1) + 1 = (a-1)(b-1) +1 >1

so a*b is in G

G is closed

b) abelian group

closed is done

a*b = b*a by definition (abelian)

a*2 =2*a = 2a - a- 2 +2 = a

So 2 is an identity element

inverse

a>1 => a/(a-1) >1

a*(a/a-1) = a^2/(a-1) - a - a/(a-1) + 2 = 2

so a/(a-1) is an inverse element

a*(b*c) = a(bc - b - c +2) - a - (bc - b - c +2) +2 = abc - ab -ac +2a -a - bc +b + c = abc - ab - ac - bc +a+b+c= (a*b)*c

associative

So G is a group

done