Math of 998001 and the Internet

If you like to have some fun listen to this, It is a video about the decimal expansion of

\(\dfrac{1}{998001} = \dfrac{1}{999^2} \)

If you like to have more fun, read the comments. The video had 231666 hits today, so I’d imagine that it is somewhat interesting.

However, if you read the comments, you rather get the impression that many people are annoyed by the uselessness of such facts or by trivial points in the video like the slang of the speaker or the decimal dots.

Obviously, or hopefully, the comment section of a blog does not reflect the average member of a community. But if you want to learn how some people tick when they see math, just inhale a few pages of comments.

If you did not care to go to the video, here is the surprising result. It is

\(\dfrac{1}{998001} = 0.000001002003\ldots996997999\ldots \)

The explanation in the video uses the infinite series

\(1+2x+3x^2+ \ldots = \dfrac{1}{(1-x)^2} \)

and an argument which is a bit too vague and hard to follow, though correct. It is nice to make things as easy as possible, but as Einstein said, you must not make it easier.

In fact I feel that the question is a case, where it is better to make things harder, or rather more abstract. Abstractness often clears the view to what is really going on. Moreover, once you master the more abstract setting, things often get easier, and along the way you will have learned something you can apply to other problems.

So why don’t we try to understand decimal expansions in general? It is not so hard. In general the expansion of a number between 0 and 1 in base b is the following.

\(0.a_1a_2a_3\ldots = a_1 b^{-1} + a_2 b^{-2} +\ldots \)

Here, the digits

\(a_1, a_2, \ldots \)

are integer numbers between 0 and b-1. The usual case is b=10, and the numbers between 0 and 9. One fact to observe is that in the decimal base b=10 we have

\(0.\overline{0\ldots01} = \dfrac{1}{999\ldots9} = \dfrac{1}{10^n-1} \)

where the bar line over the n-1 digits stands for a repetition of the pattern. From this we get e.g.

\(0.\overline{23} = \dfrac{23}{99} \)

Generalizing, we get

\(0.\overline{a_1\ldots a_{n-1}} = \dfrac{a_1\ldots a_{n-1}}{b^n-1} \)

where the numerator is in base b. I admit that understanding base representations is not easy if you have never seen that before. So let me repeat the meaning of the numerator in base b.

\(a_1\ldots,a_{n-1} = a_{n-1} + a_{n-2} b +\ldots + a_1 b^{n-1}. \)

Now, what have we gained? Using the base b=1000, our claim becomes

\(\dfrac{1}{999^2} = \dfrac{123\ldots[997][999]}{1000^{999}-1} \)

where we use digits from 0 to 999=b-1. More general,

\(\dfrac{1}{(b-1)^2} = \dfrac{123\ldots[b-3][b-1]}{b^{b-1}-1} \)

This is now a simple equation for fractions. We have to show that

\(\dfrac{b^{b-1}-1}{(b-1)^2} = b^{b-3} + 2 b^{b-2} + \ldots + (b-3) b + b-1. \)

To do this, we do something abstract once more. We substitute x=b, but not everywhere. The following can be seen by simple induction.

\(\dfrac{x^{b-1}-1}{x-1} = x^{b-2} + \ldots + x + 1 \)

It might also be known to you, since it is the so-called geometric sum. Dividing both sides by x-1 once more, we get

\(\dfrac{x^{b-1}-1}{(x-1)^2} = x^{b-3} + 2 x^{b-2} + \ldots + (b-3)x + b-2 + \dfrac{b-1}{x-1} \)

You can prove that by induction if you like, but it is easier to carry out the division. Inserting x=b again, we get the expansion in base b we were seeking.

So we see, that taking an abstract detour may bring a lot of benefits in insight and in generality.

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