Rosio Pavoris

This is something most people will have seen in high school, but since this is neat and some people haven’t (like me, though obviously I skipped a year) and it’s a discussion that comes up surprisingly often for a concept so simple, I thought I’d talk about it:
The repeating decimal 0.9999999999999… (which can also be written as 0.(9), which I’ll be doing from now on), and why it’s equal to 1.

This usually comes up in the context of limits.
A lot of people have misgivings over it, usually because they don’t really understand infinity. Maybe they instinctively believe there has to be a last 9, so it doesn’t go on forever, or they don’t think an infinitely small number really is equal to 0, or they feel that a single number can only have a single decimal expansion (which, for 1, would be 1.00000…, I guess).

The simplest and least controversial way to show 0.(9) = 1 seems to be:

(Where 0.(3) is another way of writing 0.333…, as you’ll remember.)

If this seems too easy, or you feel like this just moves the problem to whether or not 0.(3) is equal to 1/3, here’s a fancier one:

The bit that may not be immediately obvious is why 0.(9) * 10 = 9.(9), in the second step.
When you multiply a number by 10 in a base 10 system, you move the decimal separator a step to the right. The number of 9s to the right of the separator is infinite, and ∞ - 1 is still ∞, so all that happens is that a nine gets moved to the position before the period.

There are other, less obvious proofs, but they’re not that important.

The consequences aren’t as fascinating as all that (numbers don’t have a unique decimal expansion, the same principle applies in every base, not just base 10 (0.11111111… = 1 in binary, for example)), and practical applications are few (I’ve been told the Cantor set makes use of it, and you could say Zeno’s paradoxes are kind of like it), but by now the question has become a classic internet argument for reasons that are entirely beyond me, so I thought it was worth addressing.
At least I got to use Mimetex again.