Let G be a set of all non-zero real numbers. And a * b = (ab/2).
We need to show (G,*) is an abelian group.
I know I need to show ab = ba, but that seems so trivial,
Assuming the group is abelian,
a * b = b* a
ab/2 = ba/2 Crossing out 2, we get ab = ba
Is this the correct answer ?
Multiplying both sides inverse of a and b results in 1 = 1 . Am I doing something wrong ? Or am I missing something?
We have that $$a*b=\dfrac{ab}{2}=a\cdot b\cdot\dfrac{1}{2}$$form the other side$$b*a=b\cdot a\cdot\dfrac{1}{2}$$since real numbers with multiplication operator is an Abelian group (If we needed to show that the proof became more complicated) so is this group.