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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?

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    $\begingroup$ Your proof is correct. Also this is not a binary operation take $a=1$ $b=-3/2$ and then $a+b \not\in G$. Hence it is not a group. If I were you, I would start with closure verification first before proceeding to associativity. That does not mean what you did is not correct, it's just an observation. $\endgroup$ – Anurag A Dec 25 '16 at 8:58
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    $\begingroup$ @Anurag Actually the proof is not correct since clearly the operation is associative whenever it is defined (and this was not the conclusion here). $\endgroup$ – Tobias Kildetoft Dec 25 '16 at 9:07
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    $\begingroup$ And "$a + (b+c) \iff |a \ +_{Q} \ | b \ +_{Q} \ c ||$" does not even make sense; the $\iff$ symbol should connect things with trust values, but here it is used between expressions. $\endgroup$ – Henning Makholm Dec 25 '16 at 9:48
  • $\begingroup$ @HenningMakholm. If you don't mind me asking, what exactly are trust values? I've never heard of them before and Google is of no help here $\endgroup$ – Perturbative Dec 25 '16 at 9:53
  • $\begingroup$ @Perturbative: Whoops, that was a typo -- I meant to write "truth values". $\endgroup$ – Henning Makholm Dec 25 '16 at 9:54
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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.

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