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?


1 Answer 1


(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 $(0,1)\setminus\mathbb Q$.

  • $\begingroup$ 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 $\endgroup$
    – fhyve
    Oct 25, 2012 at 7:43
  • 2
    $\begingroup$ All that matters is that such enumerations of the rationals exist. However, if really you need to enumerate the rationals in (0,1), here is a (quite classical) way: first, express every rational in the lowest terms, that is, as p/q where p and q are positive integers with no common factor other than one; then, sort the fractions in the ascending order of p+q ; in case of a tie, the smaller fraction comes first. The first few terms in this enumeration are 1/2, 1/3, 1/4, 2/3, 1/5, 1/6, 2/5... $\endgroup$
    – Did
    Oct 25, 2012 at 7:51

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