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 $0 , 1, \frac{1}{2}, 2, \frac{1}{3}, \frac{3}{2}, \frac{2}{3}, 3, \frac{1}{4}, \frac { 4}{3}, \ldots$ .

#Algebra

Easy Math Editor

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`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`	$\sin \theta$
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Comments

Clarification : What is $R_n$ ?

Pratik Shastri - 6 years, 6 months ago

Ooops, it should have been $A_n$ . Corrected.

Calvin Lin Staff - 6 years, 6 months ago

numberphile just posted a video on this

Trevor Arashiro - 6 years, 6 months ago

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 - 6 years, 6 months ago

Math	Appears 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}`	$\boxed{123}$