Example of a bijection from the set of real numbers to a subset of irrationals

Question

I need an example of a bijection from the set of real numbers to a subset of the irrationals. I tried something like $f(x)=x+\sqrt{2}$, but where should I map $-\sqrt{2}$?

Answer 1

Let $f(x) = \dfrac{\arctan x}{\pi}$, so $f^{-1}(x) = \tan \pi x$. $f$ maps $\mathbb{R}$ to $(-\dfrac12,\dfrac12)$.

$$g(x) = \begin{cases} x \in \mathbb{Q} & x + \sqrt{5}\\ x \notin \mathbb{Q} & x\\ \end{cases}$$

$$g^{-1}(x) = \begin{cases} x > 1 & x - \sqrt{5}\\ x \le 1 & x\\ \end{cases}$$

So $g \circ f$ maps from $\mathbb{R}$ to a certain subset of the irrationals between $-\dfrac12$ and $\dfrac12 + \sqrt{5}$, and $f^{-1} \circ g^{-1}$ maps the inverse.

Answer 2

Let $A_i = \pi^i {\mathbb Q}$ and $ A =\cup_{i\ge0} A_i$. Then the map $h:{\mathbb R}\backslash{\mathbb Q}\rightarrow {\mathbb R}$ $$ h(x) = \left\{ \begin{array}{ll} \pi x & x\in A \\ x & x \in B={\mathbb R}\backslash A\\ \end{array} \right. $$ is a bijection.

Answer 3

Enumerate $\Bbb Q$ as $q_n$ for $n\in\Bbb N$. Now define,

$$f(x)=\begin{cases}e^{2n+1} & x=q_n\\e^{2n} & x=e^n, n\in\Bbb N\\x&\text{otherwise}\end{cases}$$

Key point here is that $e$ is transcendental, so $e^n\notin\Bbb Q$ for any $n>0$. You can effectively replace $e$ by any other number with this property. For example $\pi,\sqrt2^\sqrt2$, Liouville number, and so on and so forth.

Answer 4

Consider $A$ = { $\pi + n : n \in \Bbb N$}. Clearly $A$ is countable. ( Consider the bijection $n \rightarrow \pi + n$ , $\forall n \in \Bbb N$). So elements of $A$ can be expressed as, $A = \{a_k : k \in \Bbb N\}$. And we also can have $\Bbb Q = \{b_k : k \in \Bbb N\}$

Now consider the bijection

$\phi : \Bbb R \rightarrow \Bbb R - \Bbb Q$ by,

$\phi(x) := a_{2k+1}$ if $ x = b_k , x \in \Bbb Q$.

= $a_{2k}$ if $x = a_k , x \in A$

= $x$ if $x \in \Bbb R - (A \cup \Bbb Q)$

The construction in my example can be seen as a very general bijection from a general uncountable set to a proper uncountable subset of it.