2
$\begingroup$

I'm looking to discover more fractions that have interesting* decimal expansions. (I'm asking out of curiosity, there is no particular academic reason as far as I'm concerned).

Here are a few examples:

$\dfrac{1}{99}=0.010101\dots$

$\dfrac{1}{999}=0.001001001\dots$

(and so on...)

This post talks about:

$\dfrac{1}{243}=0.\overline{004115226337448559670781893}$

This Numberphile video talks about how:

$\frac{1}{999^2}=0.00000100200300400500600...$

'generates' the 3-digit integers (except $998$). Similar patterns arise with $1/99$, $1/9999$, and so on.

*I realize that 'interesting, fun or noteworthy' might make this question a bit open-ended or subjective, hence the 'soft-question' tag. Then again, I find it difficult to be more specific about what I'm looking for.

$\endgroup$
  • 1
    $\begingroup$ I’m not sure what you find interesting, but I’ve always been fascinated by how $\frac{1}{7}=0.\overline{142857}$, and the other sevenths have the same 6 numbers repeating, in the same order, but just starting at a different number. For example, $\frac{2}{7}=0.\overline{285714}$ $\endgroup$ – Joe Sep 6 at 21:03
  • 1
    $\begingroup$ You also have $1/9^2=0.0123456\ldots$ (all the single digits except $8$) and $1/99^2 =0.00010203040506\ldots$ (can you guess the missing double digit?) $\endgroup$ – Henry Sep 6 at 21:05
3
$\begingroup$

Given any repeating decimal expansion, you can find the rational number that generates it. For example, let's say I think that $.\overline{02040608}$ is interesting, then there is a process of calculating the corresponding fraction: $$x = .\overline{02040608}$$ $$10^8x = 2040608.\overline{02040608}$$ $$10^8x - 2040608 = x$$ $$x = \frac{2040608}{10^8-1}.$$ You can generate any periodic decimal expansion you want via this method.

Edit: This method also illuminates why we see stuff like 99, 999, 9999 so much in these `interesting' decimal expansions.

$\endgroup$
1
$\begingroup$

I guess 1/7 is probably the canonical example here. Taking multiples cycles the digits in the decimal expansion.

Here's something I wrote previously that goes into considerably more detail on the topic. To excerpt part of it:

1/7 = 0.142857142857142857...

Multiplying by ten, we see that

10 * 1/7 = 1.42857 142857 142857...

On the other hand, 10 * 1/7 = 10/7 = 1 + 3/7, so

1 + 3/7 = 1.42857142857142857...

Subtracting 1 from each side, we have

3/7 = 0.42857142857142857...

If we multiply by 100 instead of ten, 14 + 2/7 = 100 * 1/7 = 14.2857142857142857... so 2/7 = .2857142857142857...

Continuing with this game, we have

142 + 6/7 = 1000/7 = 142.857142857142857... so 6/7 = 0.857142857142857...

1428 + 4/7 = 10000/7 = 1428.57142857142857... so 4/7 = 0.57142857142857...

14285 + 5/7 = 100000/7 = 14285.7142857142857... so 5/7 = 0.7142857142857...

so we see that the decimal expansions of each of 1/7, 2/7, 3/7, 4/7, 5/7, and 6/7 are cycles of the same digits.

So what were the ingredients here?

  • 1/7 is a rational number with lowest-common-terms denominator relatively prime to ten, so we get a repeating decimal that repeats right off the bat.
  • By successively multiplying 1/7 by powers of ten, we get all numbers whose fractional parts cover every multiple of 1/7 between 0 and 1.

The rest of the discussion goes on to characterize such numbers.

$\endgroup$
0
$\begingroup$

The number $$1/89 = 0.0112359550561798\ldots$$ is interesting as the digits form the Fibonacci sequence $1,1,2,3,5, 8, 13, 21,$... at least for a while. If you want more terms of the sequence you can use, e.g.,

$$1/9899 = 0.000101020305081321\ldots$$

which gives you the two-digit Fibonacci numbers or

$$1 / 998999 = 0.000001001002003005\ldots$$

which gives you the three-digit Fibonacci numbers.

Why does this happen? I cannot tell you; it is a secret you must discover yourself...

$\endgroup$

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.