Something I got interested in recently is 10-adic numbers and the wider p-adic numbers.

To give an idea, it's math that looks at what happens when you have infinite digits going to the left, not just infinite digits going to the right.

First think of infinite digits after the decimal point. You can represent this with a geometric series:

Code:

0.9999 ... = 9/10 + 9/100 + 9/1000 + 9/10000 ...

You can work out the sum of an infinite geometric series as

Code:

S = a / (1-r)
ratio r = 0.1 and first term a = 9/10
S = 9/10 / (0.9) = 0.9 / 0.9 = 1

There are a number of other proofs that 0.999... = 1, including

Code:

S = 0.999
S * 10 = 9.99999
S * 10 - S = 9.99999 - 0.999
S * 10 - S = 9
9 S = 9
S = 1

Ok so what do we get when S = ...9999, i.e. infinite 9s going to the LEFT?

In this case it's still an infinite series:

Code:

9 + 90 + 900 + 9000, so a = 9, and r = 10

We can use the first trick from before:

Code:

S = a / (1-r)
S = 9 / (1-10)
S = 9 / -9
S = -1

So ... fanfare if you add all 9s to the LEFT you get -1

Let's try the second method:

Code:

S = ...999
10S = ...9990
S-10S = 9
-9S = 9
S = -1

Third let's do manual addition. If a number equal -1, then adding 1 to it, should give ZERO

So we add 1 to the first 9, that becomes 0, carry the one. Then, the 1 carry adds to the next 9, turning it into zero, carry the 1, and so on. In fact adding 1 to the number DOES produce an infinite row of zeros.

So, three different ways show that infinite 9s to the left actually equal negative one.

This is a really interesting type of math, because while we might say it's undefined, it's actually well behaved compared to things like 0/0, because you can prove 0/0 equals anything, so it has no clear correct value, whereas these infinite divergent series when you try and compute an actual bound often give you the same real finite result no matter how you approach them.

So in some sense infinite ...999 can be said to be a way to *represent* the value -1. You'd expect that doing standard addition with it should give you a real result, for example adding it to 14:

Code:

14+
...999
------
000013

we add the right-most 9 to 4, and get 13. Write the 3 then carry the 1. The next 9 and the 1 add up to 10, so we add 10 to the 1, which is 11, write the 1, carry the 1. The remaining 9s then cancel each other out, leaving the value 13, the correct result if we assume infinite 9s = -1.