Question

consider the binary operation * defined on Q* by a*b=3ab for all a,b in Q*. show (Q*,*) is abelian

Answer 1

a*b = 3ab = 3ba = b*a

Answer 2

can u do that?

Answer 3

of course multiplication of numbers in commutative

Answer 4

mmmmm
interesting detail.. i didnt know i can apply that

Answer 5

also you have to show Q is a group

Answer 6

ohhh for real.....

Answer 7

closure
for all a,b belongs to Q a*b=3ab belongs to Q
thus closure property is satisfied
associativity
(a*b)*c=(3ab)*c=9abc
again a*(b*c) =9abc
thus associative property satisfied
identity
let i be the identity elemnt then i*a=a
or 3ai=a
or i=1/3 (for all not0)
and belongs to Q
inverse
let x be the inverse of a if possible
then a*x=1/3
or x=1/(3a) (for all a not 0)
and the inverse exist in Q*(as Q*=Q-{0})
now a*b=3ab=3ba=b*a
thes commutative property is satisfied
hence (Q*,*) is an abelian group

Answer 8

i did associative, right before u posted... omgg... i though it was easier