The mysterious fractions

What value do you get when you convert \(\frac {1}{81}\)to decimal? You get \(0.0123456790123456790...\).

What value do you get when you convert \(\frac {1}{9801}\) to decimal? You get \(0.000102030405060708091011...9697990001\).

What value do you get when you convert \(\frac {1}{998001}\) to decimal? You get \(0.000001002003...100101102...996997999000...\).

These decimals list every \(n\) digit numbers (81 is 1, 9801 is 2, 998001 is 3, etc.) apart from the second last number. There is a pattern to find one of these fractions.

Can you see something special about the denominators? \(81\) is \(9^2\), \(9801\) is \(99^2\), \(998001\) is \(999^2\).

This means that if you did \(\frac {1}{99980001}\) you would get \(0.0000000100020003...9996999799990000...\).

Can you find a fraction that, when converted to decimal, lists every \(n\) digit number?

Note by Sharky Kesa
4 years, 6 months 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

Watch the numberphile video which explains this very well at :http://www.youtube.com/watch?v=daro6K6mym8

Anish Puthuraya - 4 years, 6 months ago

Log in to reply

This is a result of generating functions, and plugging \(x=\frac{1}{10}\) into the function. For example, the generating function for the Fibonacci sequence is \(\frac{x}{1-x-x^2}\). If we plug in \(x=\frac{1}{10}\), we get \(\frac{1/10}{1-(1/10)-(1/10)^2}=\frac{10}{100-10-1}=\frac{10}{89}=0.11235\dots\).

To answer your question, sure you can. If you want to list every \(n\)-digit number, you'll want to have the sum \(\sum_{i=1}^\infty i\times10^{-in-n}\). Recall that the generating function for \(i\) is \(\frac{1}{(1-x)^2}\), so we'll have \(\frac{10^{-n}}{(1-10^{-n})^2}\).

Cody Johnson - 4 years, 6 months ago

Log in to reply

Though, for the Fibonacci sequence, note that with \( x = \frac{1}{10} \), you 'add' the tens digit to the preceding units digit, so you don't get the sequence of \(0.112358132134\ldots \), but instead \( \frac{10}{89} = 0.1123595\ldots \). Ah, if only patterns were verified by checking the first 5 terms.

Here's a spinoff question.

Is \( 0.112358132134 \ldots \) rational or irrational?

Calvin Lin Staff - 4 years, 6 months ago

Log in to reply

Irrational. I would provide a proof,but then I would be guilty of stealing your answer from MSE.

Rahul Saha - 4 years, 6 months ago

Log in to reply

You see the effects of carrying the digit, yes, but it seemed more magical to post the first 5 digits.

Cody Johnson - 4 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...