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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.

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  • $\begingroup$ What are the binary operations you are aware of, other than the two you've already mentioned? $\endgroup$
    – Naive
    Commented Jul 29, 2017 at 5:30

1 Answer 1

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

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  • $\begingroup$ I want that bijection φ ...I know it exists...so the desired group exists .But how to get the group $\endgroup$ Commented Jul 29, 2017 at 5:32
  • $\begingroup$ 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. $\endgroup$ Commented Jul 29, 2017 at 6:14

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