Is the set of rational numbers of absolute value $\geq 1$ together with $0$ a group under addition?
My Attemtped Proof
Put $G = \{ q \in \mathbb{Q} \ | \ |q| \geq 1 \ \text{or } q =0 \}$
Let $B = (G, +)$ where $+ : G \times G \to G$. We need to prove or disprove that $B$ is a group.
Denote $+_{Q}$ to be the usual binary addition operator on $\mathbb{Q}$. In other words $+_{Q} : \mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}$.
Now take $a, b, c \in G$.
$(*)$ To show assosciativity of the $+$ operator we need to show $|a \ +_{Q} \ | b \ +_{Q} \ c || = | | a \ +_{Q} \ b | \ +_{Q} \ c|$. In other words $a + (b+c) \iff |a \ +_{Q} \ | b \ +_{Q} \ c ||$
If we take $a =3$, $b = -2$ and $C=1$, we can see that this doesn't hold, and thus $+$ is not assosciative on $G$, and thus $B$ cannot be a group. $\square$
Firstly is my proof correct? If so how rigorous is it? Any comments with regards to my proof writing is greatly appreciated. Finally is doing something like I did with the $(*)$ part of my proof a usual technique in abstract/modern algebra?
When it says "under addition", that means that the operation is just ordinary addition. That is, to add $a,b\in G$, you just take the ordinary sum that you've written $a+_Qb$, not $|a+_Qb|$. So while you've proven the latter operation is not associative, that's irrelevant to the problem.