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I've tried to show that:

$$[0,1]\sim([0,1] ∩R-Q)$$
I know from this answer :
$$[0,1]\sim R-Q$$
But how to construct a bijection between R-Q and $([0,1]∩R-Q)$ ?

I think the function would be like $f:R-Q→[0,1]∩R-Q$:
$$f(x) = \cases{ 1/x & \text{if $ ~x∈(R-Q)-[0,1]$} \\ x & \text{if $~x ∈ (0,1)$} }$$
But I think this function is not complete ... could someone help me please to improve this?

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  • $\begingroup$ You mean $\mathbb{R}\setminus\mathbb{Q}$, not the quotient of $\mathbb{R}$ by $\mathbb{Q}$ $\endgroup$ – Stefan Octavian Apr 23 at 8:39
  • $\begingroup$ @StefanOctavian Thanks edited. $\endgroup$ – program_craft Apr 23 at 8:42
  • $\begingroup$ your function is not bijective because, for example, $f(\pi)=f\left(\frac{1}{\pi}\right)=\frac{1}{\pi}$. I don't think there is a simple way to improve your function to make it bijective without making a completely different one. $\endgroup$ – R.V.N. Apr 23 at 8:56
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Let $(r_n)_{n\in\mathbb{N}}$ be an enumeration of the elements of $\mathbb{Q}\cap[0,1]$. Let $(l_n)_{n\in\mathbb{N}}$ be a sequence of irrational numbers of $[0,1]\setminus\mathbb{Q}$, for example, we could define $l_n=\frac{1}{\sqrt{p_n}}$ where $p_n$ is the $n-th$ prime number.

Then, define $f:[0,1]\longrightarrow[0,1]\setminus\mathbb{Q}$ as: $$f(x):=\begin{cases} l_{2n} &\text{ if } x=l_n\\ l_{2n+1} &\text{ if } x=r_n \\ x &\text{ in other case} \end{cases}$$ $f$ is a bijection.

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  • $\begingroup$ Thanks! But how to prove thats bijection (surjective + injective) ? $\endgroup$ – program_craft Apr 23 at 10:02
  • $\begingroup$ I've proved thats a injection but stucked in surjection. $\endgroup$ – program_craft Apr 23 at 10:14
  • $\begingroup$ Let $y\in[0,1]\setminus\mathbb{Q}$. If $y\neq l_n$ for all $n\in\mathbb{N}$ then $f(y)=y$. If $y=l_n$ for some $n\in\mathbb{N}$ then, if $n$ is even $f\left(l_{\frac{n}{2}}\right)=l_n$; if $n$ is odd, $f\left(r_{\frac{n-1}{2}}\right)=l_n$. $\endgroup$ – R.V.N. Apr 23 at 21:41

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