66
$\begingroup$

I remember my professor in college challenging me with this question, which I failed to answer satisfactorily: I know there exists a bijection between the rational numbers and the natural numbers, but can anyone produce an explicit formula for such a bijection?

$\endgroup$
  • 11
    $\begingroup$ Do you need a formula or does the picture and explanation in en.wikipedia.org/wiki/… suffice? See also en.wikipedia.org/wiki/… $\endgroup$ – lhf Oct 24 '10 at 3:10
  • $\begingroup$ I wasn't familiar with pairing functions, so let me look at that more closely. My professor insisted, though, that I come up with a formula, and of course that would also require that equivalent pairs (in the rational number sense) shouldn't get counted more than once. $\endgroup$ – Alex Basson Oct 24 '10 at 4:55
  • 1
    $\begingroup$ @lhf. Maybe you should post your comment as an answer; otherwise, it's not unlikey that this question remains unanswered. $\endgroup$ – Agustí Roig Oct 24 '10 at 5:21
  • $\begingroup$ Could you provide a list of features that you consider legitimate to include in your formula? Often when these questions are posed, responses are met with "that doesn't count as a formula." $\endgroup$ – Douglas S. Stones Oct 24 '10 at 5:43
  • 3
    $\begingroup$ Don't know if it would count as "explicit" but every rational number occurs exactly one in the Calkin-Wilf sequence en.wikipedia.org/wiki/Calkin%E2%80%93Wilf_tree $\endgroup$ – Jyotirmoy Bhattacharya Oct 24 '10 at 6:42
91
$\begingroup$

We will first find a bijection $h_{+}:\mathbb Z^+\to \mathbb Q^+$. From there, we easily get a bijection $h:\mathbb Z\to \mathbb Q$ by defining: $$h(n)=\begin{cases}h_{+}(n)&n>0\\ 0&n=0\\ -h_{+}(-n)&n<0 \end{cases}$$

From there, we can use any of the bijections $\mathbb N\to\mathbb Z$ to get our bijection between $\mathbb N$ and $\mathbb Q$. (We'll need a specific such bijection below, $s$.)

Now, every positive integer can be written uniquely as $p_1^{a_1}p_2^{a_2}\cdots$, where the $p_1=2,p_2=3,p_3=5,\dots$ is the sequence of all primes, and the $a_i$ are non-negative integers, and are non-zero for only finitely many $i$s.

Similarly, every positive rational number can be written uniquely as $p_1^{b_1}p_2^{b_2}\cdots$ where the $b_i$ are integers and only finitely many of the $b_i$ are non-zero.

So define $s:\mathbb N\to\mathbb Z$ (where we take $\mathbb N$ to include $0$):

$$s(n)= (-1)^n\left\lfloor\frac{n+1}{2}\right\rfloor $$

The sequence $s(0),s(1),s(2),s(3),\dots$ would be $0,-1,1,-2,2\dots$, and this is a bijection from $\mathbb N$ to $\mathbb Z$. The only properties we really need for $s$ is that $s$ is a bijection and $s(0)=0$.

Then for any $n=p_1^{a_1}p_2^{a_2}\cdots\in\mathbb Z^+$, define $$h_{+}(n)=p_1^{s(a_1)}p_2^{s(a_2)}\cdots $$

This then defines our bijection $h_{+}:\mathbb Z^+\to \mathbb Q^{+}$.

A potientially interesting feature of $h_+$ is that it is multiplicative - that is, if $\gcd(m,n)=1$ then $h_{+}(mn)=h_+(m)h_{+}(n).$

$\endgroup$
  • 5
    $\begingroup$ Why hasn't this been upvoted more? Good answer! (+1) $\endgroup$ – fancynancy Feb 16 '15 at 17:46
  • 1
    $\begingroup$ For your bijection between $\mathbb{N}$ and $\mathbb{Q}$, why have you defined it as $\rho_1$ instead of just $\eta$ or something else (I don't see anything to indicate significance of the index)? $\endgroup$ – fancynancy Feb 16 '15 at 17:52
  • 1
    $\begingroup$ @fancynancy I think just because it was a variant of $\rho$. I think the function I now call $\rho_1$ was just called $\rho$, but then I realized I needed the intermediate function, which was more "important" in some way, so I called that $\rho$ and renamed this $\rho_1$. It's just a function name. $\endgroup$ – Thomas Andrews Feb 16 '15 at 18:13
  • 1
    $\begingroup$ That's what I thought--just wanted to make sure. Thanks for clarifying. $\endgroup$ – fancynancy Feb 16 '15 at 18:17
  • 1
    $\begingroup$ Because in each case, only finitely many of the exponents can be non-zero. @RFZ $\endgroup$ – Thomas Andrews Nov 7 '17 at 22:52
14
$\begingroup$

We will first create a bijection from $\mathbb{N}$ to $\mathbb{Q}^{+}$ and then use this to create a bijection from $\mathbb{N}$ to $\mathbb{Q}$.

Step One: Let us first define Stern's diatomic series. This process formalizes the Stern-Brocot tree mentioned above.

$a_{1} = 1 \\ a_{2k}=a_{k} \\ a_{2k+1}=a_{k}+a_{k+1}$

To get a feel for this series, let us list out the first few terms.

$a_{1}=1 \\ a_{2}=a_{1}=1 \\ a_{3}=a_{1}+a_{2}=1+1=2 \\ a_{4}=a_{2}=1 \\ a_{5}=a_{2}+a_{3}=1+2=3 \\ a_{6}=a_{3}=2 \\ a_{7}=a_{3}+a_{4}=2+1=3 \\ a_{8}=a_{4}=1$

Now to obtain the $n^{th}$ rational number, we define $f: \mathbb{N} \rightarrow \mathbb{Q}^{+}$, by $f(n)= \dfrac{a_{n}}{a_{n+1}}$.

Let us list out the first few terms.

$f(1)= a_{1}/a_{1+1} = 1/1 \\ f(2)= a_{2}/a_{2+1} = 1/2 \\ f(3)= a_{3}/a_{3+1} = 2/1 \\ f(4)= a_{4}/a_{4+1} = 1/3 \\ f(5)= a_{5}/a_{5+1} = 3/2 \\ f(6)= a_{6}/a_{6+1} = 2/3 \\ f(7)= a_{7}/a_{7+1} = 3/1 $

This function enables us to say that the $6^{th}$ rational number is $2/3$. Moreover, this function is a bijection. For proof of this, see Theorem 5.1 here http://faculty.plattsburgh.edu/sam.northshield/08-0412.pdf.

Since $f$ is a bijection this implies that $f^{-1}$ exists. That means given a rational number we can find the corresponding natural number. For example suppose you have a fraction, say it is $1/4$. Can we determine the $n$ such that $f(n)=1/4$? The answer is a resounding yes. Given a positive rational number, $q \in \mathbb{Q}$, the $n$ such that $f(n)=q$ is found by $n=f^{-1}(q)$. This function, $f^{-1}$, is given as follows:

$f^{-1}(1)=1 \\ f^{-1}(q)= 2f^{-1} \bigg(\dfrac{q}{1-q} \bigg) ~ \text{if} ~ q<1 \\ f^{-1}(q) = 2f^{-1}(q-1)+1 ~\text{if}~ q>1$

As an example, we see from above that $f(5)={3/2}$. Let us plug $(3/2)$ into $f^{-1}$ and see if we get 5.

$f^{-1}(3/2)=2f^{-1} \bigg(\dfrac{3/2}{1-(3/2)} \bigg)+1=2f^{-1} \bigg(\dfrac{1}{2} \bigg)+1.$ A quick calculation yields that $f^{-1} \bigg(\dfrac{1}{2} \bigg)=2$ and so we get $f^{-1}(3/2)=2f^{-1} \bigg(\dfrac{1}{2} \bigg)+1=2(2)+1=5$.

Step Two: We showed there exists a bijection between $\mathbb{N}$ and $\mathbb{Q}^{+}$. We now attempt to show there exists an explicit bijection between $\mathbb{N}$ and $\mathbb{Q}$. Using the work done in Step One, it appears easier to first create a bijection between $\mathbb{Z}$ and $\mathbb{Q}$. The reason for doing so is because we have already created a bijection from the positive integers (natural numbers) to the positive rationals. So it only seems natural that by adding in the negative integers, we can map them to the negative rationals and thus obtain a bijection. We do this as follows:

$$ g(z) = \begin{cases} \dfrac{a_{z}}{a_{z+1}}, & \text{if } z>0 \\ \\ - \dfrac{a_{-z}}{a_{-(z-1)}}, & \text{if } z<0 \\ \\ 0, & \text{if } z=0 \end{cases} $$ where the $a_{i}$ term refers to the $i^{th}$ term in Stern's diatomic series.

We already referenced a proof by Northshield showing that $g(z)=\dfrac{a_{z}}{a_{z+1}}$ if $z>0$ is a bijection from $\mathbb{N} \rightarrow \mathbb{Q}^{+}$. Equivalently, we may write this as $g$ is a bijection from $\mathbb{Z}^{+}$ to $\mathbb{Q}^{+}$ for $z>0$. Now, it follows by the symmetry of the problem that $g(z)=- \dfrac{a_{-z}}{a_{-(z-1)}}$ is a bijection from $\mathbb{Z}^{-}$ to $\mathbb{Q}^{-}$ if $z<0$. That is, $g$ is a bijection between the negative integers and the negative rationals. So we have covered all the positive and negative rationals. The only element in the rationals that is not accounted for is the zero element. So we shall have the integer $0$ mapping to the rational number $0$. However, $g$ is a bijection from the integers to the rationals. We wish to find a bijection from the natural numbers to the rationals. So we shall now define the well-known bijection from the natural numbers to the integers.

$$ h(n) = \begin{cases} \dfrac{n}{2}, & \text{if }n\text{ is even} \\ -\dfrac{n-1}{2}, & \text{if }n\text{ is odd} \end{cases} $$

You may check for yourself that $h$ is a bijection. It follows that $g~\circ~ h: \mathbb{N} \rightarrow \mathbb{Q}$ is a bijection since the composition of two bijections is a bijection. Thus, we have an explicit bijection from $\mathbb{N}$ to $\mathbb{Q}$.

However, given a rational number, can we find what this rational number maps to in the set of natural numbers? Although I do not prove it, the answer is yes and is given by the following piece-wise defined function which is an extension of the function defined in Step One. We first define $g^{-1}: \mathbb{Q} \rightarrow \mathbb{Z}$ as

$$ g^{-1}(q) = \begin{cases} 2f^{-1}(q-1)+1, & \text{if } q>1 \\ 1, & \text{if } q=1 \\ 2f^{-1} \bigg(\dfrac{q}{1-q} \bigg), & \text{if } 0<q<1 \\ 0, & \text{if } q=0 \\ -2 \Bigg(f^{-1} \bigg(\dfrac{-q}{1+q}\bigg) \Bigg), & \text{if } -1<q<0 \\ -1, & \text{if } q=-1 \\ -2(f^{-1}(-q-1)+1), & \text{if } q<-1 \end{cases} $$

We now define the function $h^{-1}: \mathbb{Z} \rightarrow \mathbb{N}$ as follows:

$$ h^{-1}(z)= \begin{cases} 2z, & \text{if } z>0 \\ 1, & \text{if } z=0 \\ -2z-1, & \text{if } z<0 \\ \end{cases} $$

Then $h^{-1} \circ g^{-1}: \mathbb{Q} \rightarrow \mathbb{N}$ is the bijection we are looking for.

$\endgroup$
  • $\begingroup$ Superb exposition! ... but I think you've made a little typo in the bit where you tell how to do the explicit inverse of f for arbitrary input: in the actual example you've put $\operatorname{f^{-1}}({3\over2})=2\operatorname{f^{-1}}\left(\frac{{3\over2}}{1-{3\over2}}\right)+1$; & I think it ought to be $2\operatorname{f^{-1}}({3\over2}-1)+1$. $\endgroup$ – AmbretteOrrisey Dec 12 '18 at 6:38
  • $\begingroup$ What this boils down to as an algorithm, is: commence the euclidian algorithm on the numerator & denominator, & represent the quotients as run lengths of bits from right to left, beginning with 0 if the fraction <1 & 1 if >1. Also proceed all the way to the remainder of 0, rather than only to a remainder of 1. The very last step is to OR the leftmost bit with 1. This algorithm has indeed already been expounden by Calkin & Wilf ... though I do not have a specific reference handy. $\endgroup$ – AmbretteOrrisey Dec 17 '18 at 8:04
6
$\begingroup$

Preliminaries

I will use the Continued Fraction conception. First, let us consider only rationals that are less than 1. So $$q < 1, \quad q\in\mathbb{Q}$$ So every rational $q$ can be written as continued fraction: $$q = \cfrac{1}{a_1 + \cfrac{1}{a_2 +\cfrac{1}{a_3 + ...}}} := [a_1, a_2, a_3, ...]$$ Note that none of the $a_i$ is zero and for every $q\in\mathbb{Q}$ its q.f. is of finite length. Also note that we use only that kind of q.f.'s in which all numerators are 1's.

Formula

Let us construct a bijection $\Phi$ between rationals and naturals as follows: $$\Phi: q \mapsto \prod_{i=1}^{n_q}p_{i}^{a_i - 1},$$ where $n_q$ is length of q.f. for $q$ and $p_i$ is $i$th prime number. The inverse is straightforward.

Example

$$\Phi\Big(\frac{30}{43}\Big) = 2^03^15^27^3 = 25725$$ This is because $$\frac{30}{43} = \cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4}}}} := [1,2,3,4]$$ And vice-versa: $$\Phi^{-1}(225) = \frac{10}{13} = \cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3}}}$$ This is because $$225 = 2^03^25^2$$


Of course this works iff there is bijection between those kind of continued fractions and rationals. But it is not too hard to prove.


P.S. I feel that I miss something. Please, verify.

$\endgroup$
  • $\begingroup$ It's very good! It provides a bijection between $\mathbb{Q} \cap (0,1)$ and the integers $>0$. Indeed, any such positive integer corresponds to an element of finite support in $\prod_{p } \mathbb{N}$. Any such element of finite support, increase by $1$ all the components up to the last $\ne 0$ element, Now compose the continued fraction from it (implicitly you use that the continued fraction does not end with $1$ as a last quotient). $\endgroup$ – Orest Bucicovschi Oct 11 '17 at 16:59
  • $\begingroup$ @orangeskid, It is impossible (nonsense) for q.f. to end with 1. $\endgroup$ – LRDPRDX Oct 11 '17 at 17:55
  • $\begingroup$ @Wolfgang Very interesting, and thank you! This second example, however, confuses me. As you observe, $\frac{10}{13} := [1, 3, 3]$. So shouldn't $\Phi(\frac{10}{13}) = 2^{1-1}3^{3-1}5^{3-1} = 2^03^25^2 = 225$? $\endgroup$ – Alex Basson Oct 12 '17 at 0:50
  • $\begingroup$ @AlexBasson, Good catch! Of course, should. $\endgroup$ – LRDPRDX Oct 12 '17 at 3:13
4
$\begingroup$

Recently I was reading some papers by Don Zagier and found this one most interesting. Here, you can get not only a satisfactory proof of the bijection, but also you will have the notion of the rational number immediately after, or before, a given number, which we don't have in Cantor's proof.

Theorem:

The map $$S(x)=\frac{1}{2\lfloor x\rfloor-x+1}$$ has the property that, among the sequence $S(0),S(S(0),S(S(S(0)),\cdots $ every positive rational numbers appears once and only once.

Therefore if we write $S^n(x)$ for $n^{th}$ iterate of $S$, then we obtain an explicit bijection $F:\mathbb{N}\to \mathbb{Q}^{+}$ by $F(n)=S^n(0) $. The proof is explained in the link I have mentioned above.

$\endgroup$
1
$\begingroup$

This is a bijection between the Stern-Brocot tree and the tree of Natural numbers. Every left node is given by $ L_n = [2 P_n ] $ and every right one by $ R_n= [2 P_n +1 ]$ where $P_n $ is the value of the parent node and $P_0=[1]$. We have the sequence of transformations $ P_n \rightarrow [ L_n , R_n ]$, $L_n \rightarrow P_{n+1}$, $R_n \rightarrow P^\prime_{n+1}$ .

In list notation for the tree (count the brackets) this is

$$ n = 1 \mapsto [1,[2],[3]] $$

$$n = 2 \mapsto [1,[2,[4], [5]], [3,[6], [7]]] $$ $$ n = 3 \mapsto [1,[2,[4,[8],[9]],[5,[10],[11]]],[3,[6,[12],[13]],[7,[14],[15]]]] $$ and so on.

$\endgroup$
1
$\begingroup$

The method used here cobbles together parts and pieces of the Euler's totient function to create our sequence that bijectively covers all the rational numbers.

The function is implemented using the python programming language, but the interested reader can figure out what is happening by inspecting the output.

Here is the program:

#--------*---------*---------*---------*---------*---------*---------*---------*
# Desc: Define a bijection of natural numbers to the rational numbers
#--------*---------*---------*---------*---------*---------*---------*---------*

import sys
import fractions

def moreTicks(curTick):
    for k in range(1, curTick):
        if fractions.gcd(curTick, k) == 1:
            moreTicksList.append(fractions.Fraction(k, curTick))
    return curTick + 1


#--------*---------*---------*---------*---------*---------*---------*---------#
while True:#                       M A I N L I N E                             #
#--------*---------*---------*---------*---------*---------*---------*---------#
#                                  # initialize state of function machine
    step = 0
    negSide = 0
    posSide = 0
    curTick = 2
    phiList = []
    print(0, ', ', end='')

    while True:
        #print('expand by phi(n) count on working range')
        moreTicksList = []
        curTick = moreTicks(curTick)
        for i in range(negSide, posSide + 1):
            for f in moreTicksList:
                print(f + fractions.Fraction(i, 1), ', ', end='')
        for f in moreTicksList:
            phiList.append(f)
        #print("add phiList to '-' side")
        negSide = negSide - 1
        print(negSide, ', ', end='')
        for f in phiList:
            print(f + fractions.Fraction(negSide, 1), ', ', end='')
        #print("add phiList to '+' side")
        posSide = posSide + 1
        print(posSide, ', ', end='')
        for f in phiList:
            print(f + fractions.Fraction(posSide, 1), ', ', end='')
        step = step + 1
        if step == 7:
            print('...', end='')
            sys.exit()

Here is the output sequence (you can use the slider to see further out):

0 , 1/2 , -1 , -1/2 , 1 , 3/2 , -2/3 , -1/3 , 1/3 , 2/3 , 4/3 , 5/3 , -2 , -3/2 , -5/3 , -4/3 , 2 , 5/2 , 7/3 , 8/3 , -7/4 , -5/4 , -3/4 , -1/4 , 1/4 , 3/4 , 5/4 , 7/4 , 9/4 , 11/4 , -3 , -5/2 , -8/3 , -7/3 , -11/4 , -9/4 , 3 , 7/2 , 10/3 , 11/3 , 13/4 , 15/4 , -14/5 , -13/5 , -12/5 , -11/5 , -9/5 , -8/5 , -7/5 , -6/5 , -4/5 , -3/5 , -2/5 , -1/5 , 1/5 , 2/5 , 3/5 , 4/5 , 6/5 , 7/5 , 8/5 , 9/5 , 11/5 , 12/5 , 13/5 , 14/5 , 16/5 , 17/5 , 18/5 , 19/5 , -4 , -7/2 , -11/3 , -10/3 , -15/4 , -13/4 , -19/5 , -18/5 , -17/5 , -16/5 , 4 , 9/2 , 13/3 , 14/3 , 17/4 , 19/4 , 21/5 , 22/5 , 23/5 , 24/5 , -23/6 , -19/6 , -17/6 , -13/6 , -11/6 , -7/6 , -5/6 , -1/6 , 1/6 , 5/6 , 7/6 , 11/6 , 13/6 , 17/6 , 19/6 , 23/6 , 25/6 , 29/6 , -5 , -9/2 , -14/3 , -13/3 , -19/4 , -17/4 , -24/5 , -23/5 , -22/5 , -21/5 , -29/6 , -25/6 , 5 , 11/2 , 16/3 , 17/3 , 21/4 , 23/4 , 26/5 , 27/5 , 28/5 , 29/5 , 31/6 , 35/6 , -34/7 , -33/7 , -32/7 , -31/7 , -30/7 , -29/7 , -27/7 , -26/7 , -25/7 , -24/7 , -23/7 , -22/7 , -20/7 , -19/7 , -18/7 , -17/7 , -16/7 , -15/7 , -13/7 , -12/7 , -11/7 , -10/7 , -9/7 , -8/7 , -6/7 , -5/7 , -4/7 , -3/7 , -2/7 , -1/7 , 1/7 , 2/7 , 3/7 , 4/7 , 5/7 , 6/7 , 8/7 , 9/7 , 10/7 , 11/7 , 12/7 , 13/7 , 15/7 , 16/7 , 17/7 , 18/7 , 19/7 , 20/7 , 22/7 , 23/7 , 24/7 , 25/7 , 26/7 , 27/7 , 29/7 , 30/7 , 31/7 , 32/7 , 33/7 , 34/7 , 36/7 , 37/7 , 38/7 , 39/7 , 40/7 , 41/7 , -6 , -11/2 , -17/3 , -16/3 , -23/4 , -21/4 , -29/5 , -28/5 , -27/5 , -26/5 , -35/6 , -31/6 , -41/7 , -40/7 , -39/7 , -38/7 , -37/7 , -36/7 , 6 , 13/2 , 19/3 , 20/3 , 25/4 , 27/4 , 31/5 , 32/5 , 33/5 , 34/5 , 37/6 , 41/6 , 43/7 , 44/7 , 45/7 , 46/7 , 47/7 , 48/7 , -47/8 , -45/8 , -43/8 , -41/8 , -39/8 , -37/8 , -35/8 , -33/8 , -31/8 , -29/8 , -27/8 , -25/8 , -23/8 , -21/8 , -19/8 , -17/8 , -15/8 , -13/8 , -11/8 , -9/8 , -7/8 , -5/8 , -3/8 , -1/8 , 1/8 , 3/8 , 5/8 , 7/8 , 9/8 , 11/8 , 13/8 , 15/8 , 17/8 , 19/8 , 21/8 , 23/8 , 25/8 , 27/8 , 29/8 , 31/8 , 33/8 , 35/8 , 37/8 , 39/8 , 41/8 , 43/8 , 45/8 , 47/8 , 49/8 , 51/8 , 53/8 , 55/8 , -7 , -13/2 , -20/3 , -19/3 , -27/4 , -25/4 , -34/5 , -33/5 , -32/5 , -31/5 , -41/6 , -37/6 , -48/7 , -47/7 , -46/7 , -45/7 , -44/7 , -43/7 , -55/8 , -53/8 , -51/8 , -49/8 , 7 , 15/2 , 22/3 , 23/3 , 29/4 , 31/4 , 36/5 , 37/5 , 38/5 , 39/5 , 43/6 , 47/6 , 50/7 , 51/7 , 52/7 , 53/7 , 54/7 , 55/7 , 57/8 , 59/8 , 61/8 , 63/8 , ...

The sequence 'calibrates' the rational number 'tick marks' on our ideal 'measuring rod'.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.