# Constructing a bijection from (0,1) to the irrationals in (0,1)

How does one construct a bijection from (0,1) to the irrationals in (0,1)?

Or if I am getting my notation right, can you provide an explicit function $f:(0,1)\rightarrow(0,1)\backslash\mathbb{Q}$ such that $f$ is a bijection?

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– Jonas Meyer Oct 25 '12 at 6:41

(1) Choose an infinite countable set of irrational numbers in $(0,1)$, call them $(r_n)_{n\geqslant0}$.
(2) Enumerate the rational numbers in $(0,1)$ as $(q_n)_{n\geqslant0}$.
(3) Define $f$ by $f(q_n)=r_{2n+1}$ for every $n\geqslant0$, $f(r_n)=r_{2n}$ for every $n\geqslant0$, and $f(x)=x$ for every irrational number $x$ which does not appear in the sequence $(r_n)_{n\geqslant0}$.
Let me suggest you take it from here and show that $f$ is a bijection between $(0,1)$ and $(0,1)\setminus\mathbb Q$.
Well I found $r_n = \frac{\pi}{4n}$, but I am not so sure how to enumerate the rationals and avoid repeats like 1/2 = 2/4 – fhyve Oct 25 '12 at 7:43