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

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a*b = 3ab = 3ba = b*a
can u do that?
of course multiplication of numbers in commutative

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mmmmm interesting detail.. i didnt know i can apply that
also you have to show Q is a group
ohhh for real.....
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
i did associative, right before u posted... omgg... i though it was easier

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