(Untitled)

[nicolyrezk]

1 = 0.999 recurring

Comments

[schivnorr]

Yes. Exactly.

Let x = 0.999 recurring.
Then, 10*x = 9.999 recurring.

10*x - x = 9*x = 9.000 recurring
9*x = 9 implies x = 1.

Don'tcha just love nonunique n-ary representation?

[nicolyrezk]

I do, actually. I used to be a maths geek.

[schivnorr]

High-5! _O/\o_

[nicolyrezk]

To give you an idea of how much of a maths geek I once was, I passed GCSE Maths (Which is at the level of your High School Diploma/GED) when I was 13. o_o;

Maaaaaaaaaaaaaaaaaths

[schivnorr]

Nice. :) I probably woulda done the same thing if we hadn't kept moving around (after being put in advanced math for 3rd grade and then moving halfway through the year and having to go back to regular math, I got -really- bored for the next 2 years, and then lost all will to give a fuck about the American K-12 educational system).

[nicolyrezk]

I studied on my own and just paid for the examinations.

[schivnorr]

That's how I got in the advanced class. Then, when the new school didn't have such a system, I had no motivation to continue my self study. Meh.

[nicolyrezk]

I wish I hadn't stopped. Screwed myself over education-wise. But I am trying, now, to fix it. Never too late, after all!

[psaakyrn]

Not precisely true.

x = 0.999 recurring does not directly mean that 10*x = 9.999 recurring. I would ask, where's the 0 at the end of the recurring sequence? How can you just ignore it and assume that the 0 is actually a 9?

The layman's proof would involve fractions: 1/9 = 0.111 recurring, 2/9 = 0.222 recurring, and etc up to 9/9 = 0.999 recurring.

The mathematician's proof would involve calculations involving infinity, and infinite geometric series.

[nicolyrezk]

There IS no 0 at the end of the recurring sequence. That's why it's recurring.

[psaakyrn]

Which is precisely the point. Without sufficient proof, you can't say that 0.999 recurring * 10 = 9.999 recurring, unless you can reason why there isn't a 0 at the end.

[nicolyrezk]

You can rationalise it within the confines of recurrence. It's a product of a recurring number, therefor it must ALSO be recurring.

[psaakyrn]

Or you can rationalise it within the confines of multiplication of a value by it's base. 99 X 10 = 990. 9.9 * 10 = 99.0. 0.99 * 10 = 9.90.

But mathematics doesn't care for rationalisation, only logical proofs. The first step is to prove that for -1 < x < 1, x * infinity = 0. From there, you can treat 0.999 recurring as an infinite geometric series where a = 0.9 and r = 0.1. (infinite geometric series calculations require that for -1 < x < 1, x * infinity = 0)

[schivnorr]

Or, you can treat 0.999 recurring as (in Maple notation):

sum( 9/10^k, k=1..infinity );

This, as you pointed out, is an infinite geometric series. All you need to do is multiply this series by 10 and you get

sum( 9/10^(k-1), k=1..infinity);

which shows that there is no terminal 0 in the base-10 representation (this series is, basically, the base-10 representation itself.)

[psaakyrn]

That is true, assuming that infinity - 1 is still infinity.

[schivnorr]

If you assume that infinity-1 < infinity, then you are using a different set of axioms than I.

I believe that for the idea of infinity to be consistent within most axiomatic mathematical systems, infinity-1 must be defined to be infinity, infinity+1 must be defined to be infinity, and infintiy - infinity must be undefined.