Waste less time on Facebook — follow Brilliant.
×

Hitting all the rational numbers

Consider the following sequence of rational numbers, starting with \( A_ 0 = 0 \) and defined recursively as

\[ A_n = \frac{ 1 } { 2 \lfloor A_{n-1} \rfloor - A_{n-1} + 1 }. \]

Show that every non-negative rational number appears in this sequence exactly once.


The sequence starts off as \( 0 , 1, \frac{1}{2}, 2, \frac{1}{3}, \frac{3}{2}, \frac{2}{3}, 3, \frac{1}{4}, \frac { 4}{3}, \ldots \).

Note by Calvin Lin
3 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Clarification : What is \(R_n\)?

Pratik Shastri - 3 years, 1 month ago

Log in to reply

Ooops, it should have been \(A_n\). Corrected.

Calvin Lin Staff - 3 years, 1 month ago

Log in to reply

numberphile just posted a video on this

Trevor Arashiro - 3 years, 1 month ago

Log in to reply

The sequence that is generated is the same, which is the one obtained through the Stern-Brocot tree. There is a pretty nice proof by continued fractions to demonstrate that this sequenece hits all the rationals exactly once.

However, it is not immediately apparent why the above recurrence relation should lead to the Stern-Brocot tree. The relationship between consecutive terms of the Stern-Brocot tree that do not have the same parents is not clear, whereas we have a very simple description in the above recurrence relation.

So yes, one possible way of approach, is to show that the above recurrence relation leads to the Stern-Brocot tree, and hence conclude that it hits each rational number exactly once.

Calvin Lin Staff - 3 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...