Irrationals: A Group? I understand that the set of irrational numbers with multiplication does not form a group (clearly, $\sqrt{2}\sqrt{2}=2$, so the set is not closed). But is there a proof or a counter-example that the irrationals with addition form (or do not form) a group?
Thank you!
Edit: In particular, I am wondering if the set is closed with respect to addition.

To speak to the spirit of your question a bit: the rational numbers are a so-called normal subgroup of the reals (since the reals form an abelian group, and all subgroups of an abelian group are normal), so we can talk about the quotient group of the reals by the rationals, $\mathbb{R}\ /\ \mathbb{Q}$.  Each element of this group is a set of the form $\{r+q, q\in\mathbb{Q}\}$, and the sum of two elements $s_0=\{r_0+\mathbb{Q}\}$ and $s_1=\{r_1+\mathbb{Q}\}$ is $s_0+s_1 = \{r_0+r_1+\mathbb{Q}\}$; you can convince yourself that addition of two elements doesn't depend on which representative we choose for a given element.  The identity element of this group is just the rationals $\mathbb{Q}$ themselves.  This is a complicated object; for instance, any collection of representatives is a so-called Vitali set, and is non-measurable.  (And just building such a collection of representatives requires the Axiom of Choice!)
