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?
On
You are correct! Though to show that $*$ is not associative, you should give an explicit counterexample. Your working out should let you construct one easily, but until you have many would say it is not a proof.
I hope this helps ^_^
On
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.)
On
Subtract $\frac{a+b+c}{2}$ from both sides of your last equation, and you get an equation that's quite easy to violate.
On
Just for the sport of it, a purely abstract proof (the right proof is of course to prove that $*$ is not associative).
Clearly, the operation $*$ is idempotent ($a*a = a$), commutative and cancellative ($a * b = a * c$ implies $b = c$). Thus, if $G$ was a semigroup, it would be a cancellative semigroup and a semilattice and hence would be trivial (since in such a semigroup, $ab = aab = bab$ implies $a=b$).
by $a=n$ and $b=c=n+1$. This should be a counterexample to associativity.
so , that $∗$ is not associative
you can't continue because the Set is neither a group nor a monoid nor a semigroup.