# Do Irrationals form a group? [closed]

It's true that the irrationals do not form a group under addition or multiplication, but I want to find a binary operation "∗" such that ($\mathbb R \setminus \mathbb Q,∗$) is a group. It is possible by transpose of operation from $\mathbb R$, but I want a explicit example.

• What will be zero. – hamam_Abdallah Aug 1 '17 at 15:57
• Why isn't transposing additon/multiplication from $\mathbb{R}$ a "solid" example? – Duncan Ramage Aug 1 '17 at 16:02
• So basically you just want an explicit bijection between irrationals and a known uncountable group. – arctic tern Aug 1 '17 at 16:10
• @Fedor Yes it will. – arctic tern Aug 1 '17 at 16:20
• This question has been posted twice, by the asker. – amWhy Aug 1 '17 at 17:35

If I understand the question correctly, and as pointed out by arctic tern in the comments to the question, it would be sufficient to have an explicit bijection between the real numbers and the irrational numbers; using this bijection (and its inverse) and transpose of operation from $\mathbb{R}$, we would be done.

Such a bijection is given in this answer. Let me copy it here, since it is so simple:

Take all real numbers of the form $q+n\sqrt{2}$, with $q$ rational and $n$ a non-negative integer, to $q+(n+1)\sqrt{2}$, and fix all other real numbers.

Okay, I'm redoing this answer. To a quasi-meta-answer.

I think a lot of students frequently don't intuitively get how easy and arbitrary math definitions can be.

For example, a very common question will be: "How can I define a function $f(x)$ that returns $1$ if $x < 0$, returns $0$ if $0 \le x < 5$ and returns $-1$ if $x > 5$" and the student will not realize the answer is "You just did".

Likewise a student might ask "How can I define a binary operation on the set $\{\sqrt 2, e, \pi\}$ so that it is a group" and when told: "Let $\sqrt 2 + x = \sqrt 2$ for any $x$, let $e + \pi = \pi + e = \sqrt 2$, $e + e = \pi$, $\pi +\pi = e$; That's the modulo 3 additive group with $\sqrt 2$ serving as the identity element--- all groups with 3 elements are equivalent to the modulo 3 additive group" will somehow feel that doesn't make sense or isn't a real binary operation.

I feel this is a similar question.

I think it is probably clear to the OP that $|\mathbb R \setminus \mathbb Q| = |\mathbb R| = |(0, 1)| = |(1, \infty)|$ etc. but that the OP wonders if $+, \times$ work on $\mathbb R$ what will work on $\mathbb R \setminus \mathbb Q$; the $+, \times$ are specific to $\mathbb R$ and that $\mathbb R \setminus \mathbb Q$ will have to have an entirely different operation to work on its values.

But in actuality all that is needed is: If $x \leftarrow \rightarrow x'$ are corresponding "one-to-one" values between $\mathbb R$ and $\mathbb R \setminus \mathbb Q$. Then:

$\overline{+}$ defined as $x'\overline + y' = (x+y)'$ is the group operation.

That's it!

Pf: a) $\overline{+}$ is binary. If $x', y'\in \mathbb R \setminus Q$ then $x,y \in \mathbb R$ so $x+y \in \mathbb R$ and $(x+y) \in \mathbb R \setminus Q$.

b) $\overline{+}$ is associative. $(x' + y') + z' = (x + y)' + z' = ((x+y) + z) = (x + (y+z))' = x' + (y+z)' = x' + (y' + z')$

c) $0'$ is an identity. $0' + x' = (0 + x)' =x'$.

d) $(-x)'$ is the inverse of $x'$. $(-x)' + x' =(-x + x)' = 0'$

That's it.

So if we use the bijection in Pierre-Guy Plamondon's answer.

The binary operation is $\overline {+}$ so that:

i) if neither $a$, $b$ nor $a+ b$ are of the the form $q + n\sqrt{2}$ where $q$ is rational and $n$ is and integer greater than or equal to zero, $a \overline + b = a + b$..

ii) if neither $a$ nor $b$ is of that form bun $a+b$ is then $a \overline + b = a + b + \sqrt 2$

iii) if one of $a$ or $b$ is of the form but the other isn't and $a + b - \sqrt 2$ is not of the form then $a \overline + b = a + b - \sqrt{2}$.

iv) if one of $a$ or $b$ is of the form but the other isn't and $a+b - \sqrt 2$ is of the form, then $a \overline b = a + b$.

v) if both $a$ and $b$ are of the form then $a \overline + b = a+ b - \sqrt 2$.

will form a group on $\mathbb R \setminus \mathbb Q$ with the identity element of $\sqrt 2$.

• I wouldn't call this an explicit operation if $\phi$ isn't explicit. – arctic tern Aug 1 '17 at 16:22
• Yeah. In the comments I saw that was what the author wanted. But in the question as posted, I didn't see the the explicit definition of a bijection was relevent. Because it isn't really I got the feeling in the question that the OP had the common misconception of not getting just how arbitrary creating group operations can be. And explicit bijection may be hard to come up with at first bt that is a completely different question than an explicit group. Which is trivial with the bijection. I'll delete this answer in ten minutes now that I've read the comments and the other answer. – fleablood Aug 1 '17 at 16:33