Infinite decadic numbers

To recap: we’ve now defined the decadic metric on integers by

$|10^k \cdot m|_{10} = \frac{1}{10^k}$

where $m$ is not divisible by 10, and also $|0|_{10} = 0$ . According to this metric, two numbers are close when their difference is decadically small. So, for example, $10587$ and $10588$ are at a distance of $1$ , but $34976$ and $20734346$ are closer, with a distance of $1/10$ . $39645$ and $99945$ are closer still, with a difference of $1/100$ . Put simply: the more final digits two integers share, the closer they are. This is a reversal of the usual situation in which integers are close when they share a lot of initial digits (like $10587$ and $10588$ ). Mark James notes that it is almost as if we were “reversing” the numbers. Jonathan notes that this is not quite right, negative and positive integers can be close without sharing digits—for example, $74$ and $-26$ are $1/100$ apart. But in this case the negative number ends with digits that are “dual” to the digits the positive number ends with.

Let’s keep thinking about negative numbers a bit. $9 - (-1) = 10$ , so $9$ and $-1$ are $1/10$ apart. $99$ and $-1$ are $1/100$ apart. $999$ and $-1$ are closer still, $1/1000$ . In fact, I hope you can see that in general, $999\dots999 = 10^k - 1$ is very close to $-1$ : specifically, $10^k - 1$ and $-1$ are always $1/10^k$ apart. So the more 9s we add, the closer we get to $-1$ .

What if we let this process go on “forever”? After all, we do this with the normal decimal numbers: $0.3$ , $0.33$ , $0.333$ , etc. get closer and closer to $1/3$ , so we define the infinite sequence $0.33333\dots = 1/3$ . We can do the same thing here: we define the infinite decadic number

$\dots 99999$

to be equal to $-1$ . Weird, huh? But it gets weirder (and cooler). The really neat thing is that arithmetic still “works” in the same way as it does for normal integers. What do I mean? Well, if we add $1$ and $-1$ we had better get $0$ , right? Let’s try it:

$\begin{array}{rr} & \dots 99999 \\ + & 1 \\ \hline & \dots 00000\end{array}$

Adding 9 and 1 gives us 10, so we write 0 and carry the 1… giving us 10 again, and so on forever.

And if we add $-1$ to itself we ought to get something that behaves like $-2$ , right?

$\begin{array}{rr} & \dots 99999 \\ + & \dots 99999 \\ \hline & \dots 99998\end{array}$

Well, adding the last two 9’s gives us 18, so the last digit of the answer is 8… we carry the 1, so the next addition gives us 19; we write down 9 and carry the 1… and this pattern is now obviously going to repeat forever, so

$\dots 99999 + \dots 99999 = \dots 99998$

which I claim represents $-2$ : you should have no trouble verifying that $\dots 99998 + 2 = \dots 00000$ .

Here are some parting exercises for you to work on before my next post:

What number is represented by $\dots 11111$ ? How do you know?
How about $\dots 34343434$ ?
Can you find an infinite decadic number which represents $-546$ ? How about $-1/7$ ? Or $-1/2$ ?

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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7 Responses to Infinite decadic numbers

Pat Muchmore says:

October 20, 2011 at 12:59 pm

OK, now I’m starting to get a sense of where these numbers lead, and it’s VERY cool. I’m guessing that I can figure out what numbers represent by seeing what addends will create 0. So is it correct to say that …11111 = –9 ? It seems like …11111 + 9 would equal 0, but this seems so strange my world feels kind of upside down. Oh wait, it doesn’t work anyway since I would need to be carrying a 9 rather than a 1 to get it to add up infinitely to 0. …11111 = –…88889 ? Is that even legal to have a negative before the reverse …?
- Brent says:
  
  October 20, 2011 at 2:54 pm
  
  Right, …11111 + 9 = …11120, not 0, and you are absolutely right that the additive inverse of …11111 is not 9 but …88889. But there are also some more familiar labels for these numbers (so seeing what adds up to zero is not sufficient to get a handle on these numbers, although it’s a good start). Keep playing around! =)
Mark James says:

October 20, 2011 at 4:13 pm

1 ) …11111 looks like -1/9 to me since 9 * …11111 = …99999, but I don’t have a good feel for what operations are valid.

2) If …111111 is -1/9 then …343434 looks like -34/99 since …343434 / 34 = …010101 and …010101 * 99 = …99999
3) -546 is the easiest since it’s just …9999454.
-1/7 is …147257142857 because this number times 7 is …99999
I can’t figure out -1/2 since it should look like 4…99999 but that’s not a valid number.
Mark James says:

October 20, 2011 at 5:53 pm

I’m guessing that -1/2 and -1/5 don’t have representations in this system since 2 and 5 are both divisible by 10.

You can figure out fractions by starting at the end and working towards the front. For example -1/7 looks like this:
Quotient Remainder
…999999
7 x 7 = 49 xxxxxxxx7 …999950
7 x 5 = 35 xxxxxxx57 …999600
7 x 8 = 56 xxxxxx857 …994000
7 x 2 = 14 xxxxx2857 …980000
….
You can’t do that with -1/2 since there are no multiples of 2 that end in 9.
- Mark James says:
  
  October 20, 2011 at 5:54 pm
  
  Sorry, my ASCII formatting didn’t work very well.
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