### Motivation In this paper the authors present a new way to li...
What does relatively prime mean?
The list in the beginning of the paper is just the deforestation of...
There's another way to prove this. Two numbers are relatively prime...
Here's how can we prove that every positive number occurs in the tr...
It is possible to create a similar list via a deforestation of the ...
Recounting the rationals
Neil Calkin
Department of Mathematics, Clemson University
Herbert S. Wilf
Department of Mathematics, University of Pennsylvania
We're going to arrange the positive rational numbers in a list that begins like this:
, . . .
Some of the interesting features of this list are
1. The denominator of each fraction is the numerator of the next one. That means that the
nth rational number in the list looks like f(n)/f(n + 1) (n = 0, 1, 2, . . .), where f is a certain
function of the nonnegative integers whose values are
{f(n)}
n≥0
= {1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, . . .}.
2. The function values f(n) actually count something nice. In fact, f (n) is the number of
ways of writing the integer n as a sum of powers of 2, each power being used at most twice
(i.e., once more than the legal limit for binary expansions). For instance, we can write
5 = 4 + 1 = 2 + 2 + 1, so there are two such ways to write 5, and therefore f(5) = 2. Let's
say that f(n) is the number of hyperbinary representations of the integer n.
3. Consecutive values of this function f are always relatively prime, so that each rational occurs
in reduced form when it occurs.
4. Every positive rational occurs once and only once in this list.
+
Figure 1: The tree of fractions
1 The tree of fractions
For the moment, let's forget about enumeration, and just imagine that fractions grow on the tree
that is completely described, inductively, by the following two rules:
(the root ) is at the top of the tree, and
• Each vertex
has two children: its left child is
i+j
and its right child is
We show the following properties of this tree.
1. The numerator and denominator at each vertex are relatively prime. This is certainly true at
the top vertex. Otherwise, suppose r/s be a vertex on the highest possible level of the tree
for which this is false. If r/s is a left child, then its parent is r/(s − r), which would clearly
also not be a reduced fraction, and would be on a higher level, a contradiction. If r/s is a
right child, then its parent is (r − s)/s, which leads to the same contradiction.
2. Every reduced positive rational number occurs at some vertex. The rational number 1 ce r-
tainly occurs. Otherwise, let r/s be, among all fractions that do not occur, one of smallest
denominator, and among those the one of smallest numerator. If r > s then (r − s)/s doesn't
occur either, else one of its children would be r/s, and its numerator is smaller, the denomi-
nator being the same , a contradiction. If r < s, then r/(s − r) doesn't occur either, else one
of its children would be r/s, and it has a smaller denominator, a contradiction.
3. No reduced positive rational number occurs at more than one vertex. First, the rational
number 1 occurs only at the top vertex of the tree, for if not, it would be a child of some
vertex r/s. But the children of r/s are r/(r + s) and (r + s)/s, neither of which can be 1.
Otherwise, among all reduced rationals that occur more than once, let r/s have the smallest
denominator, and among these, the smallest numerator. If r < s then r/s is a left child of
two distinct vertices, at both of which r/(s − r) lives, contradicting the minimality of the
denominator. Similarly if r > s. 2
It follows that a list of all positive rational numbers, each appearing once and only once, can be
made by writing down 1/1, then the fractions on the level just below the top of the tree, reading
from left to right, then the fractions on the next level down, reading from left to right, etc.
We claim that if that be done, then the denominator of each fraction is the numerator of its
successor. This is cle ar if the fraction is a left child and its successor is the right child, of the same
parent. If the fraction is a right child then its denominator is the same as the denominator of its
parent and the numerator of its succe ss or is the same as the numerator of the parent of its successor,
hence the result follows by downward induction on the levels of the tree. Finally, the rightmost
vertex of each row has denominator 1, as does the leftmost vertex of the next row, proving the
claim.
Thus, after we make a single sequence of the rationals by reading the successive rows of the tree
as described above, the list will be in the form {f(n)/f(n + 1)}
, for some f.
Now, as the fractions sit in the tree, the two children of f (n)/f(n + 1) are f (2n + 1)/f(2n + 2)
and f(2n + 2)/f(2n + 3). Hence from the rule of construction of the children of a parent, it must
be that
f(2n + 1) = f(n) and f(2n + 2) = f(n) + f(n + 1) (n = 0, 1, 2, . . .).
These recurrences evidently determine our function f on all nonnegative integers.
We claim that f (n) = b(n), the number of hyperbinary re presentations of n, for all n ≥ 0.
This is true for n = 0, and suppose true for all integers ≤ 2n. Now b(2n + 1) = b(n), because if
we are given a hyperbinary expansion of 2n + 1, the "1" must appear, hence by subtracting 1 from
both sides and dividing by 2, we'll get a hyperbinary representation of n. Conversely, if we have
such an expansion of n, then double each part and add a 1, to obtain a representation of 2n + 1.
Furthermore, b(2n + 2) = b(n) + b(n + 1), for a hyperbinary expansion of 2n + 2 might have
either two 1's or no 1's in it. If it has two 1's, then by deleting them and dividing by 2 we'll get an
expansion of n. If it has no 1's, then we just divide by 2 to get an expansion of n + 1. These maps
are reversible, proving the claim.
It follows that b(n) and f(n) satisfy the same recurrence formulas and take the same initial
values, hence they agree for all nonnegative integers. We state the final result as follows.
Theorem 1 The nth rational number, in reduced form, can be taken to be b(n)/b(n + 1), where
b(n) is the number of hyperbinary representations of the integer n, for n = 0, 1, 2, . . .. That is, b(n)
and b(n + 1) are relatively prime, and each positive reduced rational number occurs once and only
once in the list b(0)/b(1), b(1)/b(2), . . ..
2 Remarks
There is a large literature on the closely related subject of Stern-Brocot trees [Ste, Bro]. In particu-
lar, an excellent introduction is in [GKP], and the relationship between these trees and hyperbinary
partitions is explored in [Rez]. In Stern's original paper [Ste] of 1858 there is a structure that is es-
sentially our tree of fractions, though in a different garb, and he proved that every rational number
occurs once and only once, in reduced form. However Stern did not deal with the partition function
b(n). R eznick [Rez] studied restricted binary partition functions and observed their relationship to
Stern's sequence. Nonetheless it seemed to us worthwhile to draw these two aspects together and
explicitly note that the ratios of successive values of the partition function b(n) run through all of
the rationals.
[Bro] Achille Bro cot, Calcul des rouages par approximation, nouvelle m´ethode, Revue Chrono-
m´etrique 6 (1860), 186-194.
[GKP] Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, Addison-
Wesley, Reading, 1989.
[Leh] D. H. Lehmer, On Stern's diatomic series, this monthly 36 (1929), 59-67.
[Rez] Bruce Reznick, Some binary partition functions, in Analytic Number Theory, Proceedings
of a conference in honor of Paul T. Bateman, Birkh¨auser, Boston (1990), 451-477.
[Ste] M. A. Stern,
Uber eine zahlentheoretische Funktion, Journal f¨ur die reine und angewandte
Mathematik 55 (1858), 193-220.
There's another way to prove this. Two numbers are relatively prime if the greatest common divisor (gcd) between the two is 1 ($gcd(i,j)=1$). We can prove this by induction. 1) $gcd(1,1) = 1$, at the root vertex this is true 2) Assuming it's true for a certain vertex i,j ($gcd(i,j) = 1$) we just need to prove that it's true for the child nodes - $gcd(i,i+j) = gcd(i+j,j) = 1$ Now, it's not difficult to see that if $gcd(i,j) = k$, that means $k$ divides both $i$ and $j$, so we can write $i = ka$ and $j = kb$. Consequently, $i+j = ka + kb = k(a+b)$, so $k$ divides $i+j$ as well! $gcd(i,j)=gcd(i,j+i)=gcd(i+j,j)$ It is possible to create a similar list via a deforestation of the Stern-Brocot tree of rationals. ![](https://upload.wikimedia.org/wikipedia/commons/thumb/3/37/SternBrocotTree.svg/500px-SternBrocotTree.svg.png) For more information on that go [here](https://pdfs.semanticscholar.org/6527/441cea8228a86d61ab4bf054e2f6c1008ef5.pdf). The list in the beginning of the paper is just the deforestation of the so-called Calkin-Wilf tree, starting from top to bottom and from left to right. For 3/5 and 5 > 3, shouldn't this be a left node, rahter than right node? What does relatively prime mean? It means that the numbers share no common factors other than $1$. For instance, $6 = 2\cdot 3$ and $35 = 5 \cdot 7$ are relatively prime. this is also called "coprime" It's not clear to me how to determine whether a rational number in this tree will be a left child or a right child? From Figure 1, there are enough examples to show that just the relative magnitude of numerator and denominator isn't enough. Here's how can we prove that every positive number occurs in the tree. Let's start with a rational number, say $\frac{3}{5}$. To figure out if the number is a right or left node we just need to compare the denominator with the numerator - in this case since 5>3, this is a right node. Then to determine the parent node we just need to calculate $\frac{3}{5-3}=\frac{3}{2}$ and we can keep doing this until we get to the root node $\frac{1}{1}$. ![](https://i.imgur.com/I4HdglS.png) This means that we can find a path up to the root node from any reduced positive rational. Since there was no choice at each step, there is exactly one path from any rational up to the root of the tree and each reduced rational occurs is only one place in the tree (there are no duplicates). Actually, as noted by the authors, the elegant enumeration of the rationals was already known (see Stern-Brocot trees), but this paper makes it even more elegant, and makes explicit a relationship to the _hyperbinary_ partition function, first defined by Reznick. _Concrete Mathematics_, by Graham, Knuth and Patashnik (see references), contains the most accessible introduction I know to these trees — and to many other discrete mathematics topics. @Rishabh Can you give me an example? In a left child the denominator is always bigger than the numerator and in a right child the numerator is always bigger than the denominator. These two ways are essentially the same; the proof by induction uses the same common factor argument as the contradiction in the paper. Furthermore, the principle of induction and the well-foundedness of the integers (every non-empty collection has a least element) can be deduced from each other (with a contradiction argument like this one). ### Motivation In this paper the authors present a new way to list all the rational numbers in a more elegant fashion. A rational number can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers. Rational numbers are countable, which means there is a one-to-one correspondence between the set of rational numbers and $\mathbb{N}$ = {0, 1, 2, 3, ...} - in other words, you can put the elements of the set in order just like natural numbers are in order. Here's an example of how you can create a countable correspondence for the rationals ![](https://i.imgur.com/2wX6wa3.png) The problem with this correspondence is that you have numbers that appear multiple times $1/1,2/2,...$, making it seem a bit inelegant. The solution presented in this paper tries to fix some of these issues.