-3

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 solid example.

Michael L.
  • 5,521
Supriyo Halder
  • 1,098
  • 6
  • 15
  • What are the binary operations you are aware of, other than the two you've already mentioned? – Naive Jul 29 '17 at 05:30

1 Answers1

2

Take a bijection $\varphi : (\mathbb{R} \setminus \mathbb{Q}) \to \mathbb{R}$, and then define $a*b = \varphi^{-1}(\varphi(a) \cdot \varphi(b))$.

xxxxxxxxx
  • 13,302
  • I want that bijection φ ...I know it exists...so the desired group exists .But how to get the group – Supriyo Halder Jul 29 '17 at 05:32
  • I think it's possible to specify such a bijection. We can get a bijection $f: \mathbb R \setminus {0} \to \mathbb R$ as follows: Let $S = {n\sqrt 2: n\in \mathbb N}$. Set $f(x) = x$ for $x\notin S$, and set $f(n\sqrt 2) = (n-1)\sqrt 2$ for $n\in \mathbb N$.

    Now just enumerate the rationals and do the same trick for each one, taking care that the sets $S$ do not intersect.

     – William Stagner Jul 29 '17 at 06:14