# 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.

• $0$ is not irrational. – J.R. Jan 23 '14 at 20:38
• $\pi+(-\pi)=0$. – Did Jan 23 '14 at 20:38

$0$ is not irrational. The set of irrationals is not even closed under addition:

$$(1-\sqrt{2})+\sqrt{2}=1$$

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.

• Are you aware if it is possible to define an operation in irrationals so it becomes a group? Or more general, given a nonempty set, is there a way to put in there an operation so it becomes a group? There's a trivial operation which does the trick, choose some $e$ in the set, then do $x\ast y=e$ for any two elements in the set. – leo Jan 27 '14 at 3:54
• @Leo Yes, of course, but that is of little use, versus the above, which is widely used. – Bill Dubuque Jan 27 '14 at 3:59
• Yes, its only use is mention it exist. Do you know of some other on rationals? Some interesting? – leo Jan 27 '14 at 4:03
• Thank you, very helpful! Quick question: is it known what nonalgebraic*nonalgebraic is? – math1234567 Feb 1 '14 at 9:01
• @leo: the question of turning a non-empty set into a group is more complicated than that. The operation your proposed only works for sets of size $1$ or $2$, since otherwise $x$ would have as inverses $x$ itself together with at least another element of the set, which is a contradiction. You can browse this site to find how to define an operation. – zarathustra Jun 18 '14 at 19:59

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!)

• Is $r \in \mathbb{R}$? – math1234567 Feb 1 '14 at 9:05
• @Elise for each member of the group, $r$ is a specific element of $\mathbb{R}$, whereas the $q$ in these sets ranges over all elements of $\mathbb{Q}$ - that is, each set is a shifted version of the rationals. I'll try to make that a bit clearer in the answer when I have a good chance to edit it. – Steven Stadnicki Feb 1 '14 at 16:50