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.
 A: The irrationals are the complement $\,\overline{\Bbb Q}\,$ of the subgroup $\Bbb Q\subset \Bbb C$. But a complement of subgroup is not a subgroup since it does not contain the identity $\,0,\,$ nor is it closed under subtraction, not containing $\,\alpha -\alpha.$ However, one can do some group-like calculations with such complements, such as: rational + irrational = irrational. Such statements are a special case of the following complementary view of a subgroup.
Theorem $\ $ Let $\rm\,G\,$ be a nonempty subset of an abelian group $\rm\,H,\,$ with complement set $\rm\,\bar G = H\backslash G.\,$
Then $\rm\,G\,$ is a subgroup of $\rm\,H\iff G + \bar G\, =\, \bar G. $ 
Proof  $\ $ $\rm\,G\,$ is a subgroup of $\rm\,H\iff G\,$ is closed under subtraction, so, complementing
$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\  fails\\
&\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\
&\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\
&\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\  for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\
&\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED}
\end{eqnarray}$ 
Instances of this are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here.

A: 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!)
A: $0$ is not irrational. The set of irrationals is not even closed under addition: 
$$(1-\sqrt{2})+\sqrt{2}=1$$
