Is $G = (\Bbb Q^*,a*b=\frac{a+b}{2})$ a group, monoid, semigroup or none of these?
I tried to solve it by assuming $ a,b,c \in G $ such that $a*(b*c)=(a*b)*c$. Then$$\frac{a+\frac{b+c}{2}}{2} = \frac{\frac{a+b}{2}+c}{2}$$
and by taking $$ 2\times\left( \frac{a+\frac{b+c}{2}}{2} = \frac{\frac{a+b}{2}+c}{2} \right) $$
we have $$ a+\frac{b+c}{2} =\frac{a+b}{2}+c.$$
In my steps I have $(G,*)$ is not associative. Are my steps right?
Best Answer
You haven't yet proved that the operation is not associative - you need to whip up an actual counterexample. Can you think of a particular choice of $a,b$, and $c$ for which we have $$a+{b+c\over 2}\not={a+b\over 2}+c?$$ (Certainly it looks unlikely that the two expressions would always be equal, but you haven't yet proved that they're not in general.)