## 20080128

### Repeating 9s

So, I've once again run headlong into more nerdy math stuff.

First, I'd like to address my dear brother's comment on my previous post on the subject of 1 = .999....

"This is because trailing the end of 9.999... is an additional 0 (due to multiplying by 10), while trailing the end of 0.999... is a 9."

Except that the number 0.999 repeating doesn't *have* an end for the zero to go on. Pick any 9 you want, but there's always a 9 after it, not a zero. It's like the proof of limits in calculus, where you prove it by showing that no one can specify a place where the function will suddenly deviate from its course towards your limit. 0.999 repeating is the same thing, except that it's by definition of what repeating is.

Your argument against the rule of nines is similarly ridiculous because, again, THERE IS NO EDGE FOR SOMETHING TO TRAIL ON. Infinity doesn't "end". That's why it's infinity, that's why infinities have to be dealt with completely differently than normal numbers. Just because .111 repeating is too long of a number for us to divide by brute force doesn't mean we have to resort to remainders to say that it's equal to 1/9. I'm pretty sure it's provable that 1.0 divided by 9.0 will be .111 repeating. You want to say that somewhere along the line you'll suddenly get a different correct answer when you do the division, 10/9 = 1 with 1 remaining. Obviously, it won't, as division is a function and will give the same answer when given the same input, and in this case the input for each small division in the long chain is the same as the output. 1/9 will divide into .111 repeating And that's what .111 repeating MEANS: 1/9. There IS NO TRAILING EDGE. IT GOES ON FOREVER.

There is no trailing edge. There is no trailing edge. There is *no* trailing edge. Are you saying there's some number which is at the top of infinity, which you can then put another number on top of? No, you can't do that, because infinity has no top. In the same way, repeating numbers HAVE NO TRAILING EDGE. -- To have an edge, an end, is to not be infinite --

I'll readily admit that my solution to this issue is equal parts philosophical as it is mathematical, and I think that's where we're coming into conflict.

Allow me to represent my argument a different way.

9.999... = 9/1 + 9/10 + ... + 9/(10^n)
0.999... = 9/10 + 9/100 + ... + 9/(10^n+1)

When we plug infinity into n, we have 10^infinity versus 10^infinity+1. In my mind, and maybe I'm being childish, there's a difference between infinity and infinity+1. If there's a proof to the contrary, then me and my philosophical meanderings are washed up.

That said, I have encountered a 1=.999... proof that I can't easily counter. That is the a/(1-r) proof. See: Geometric Progression.

0.999... = 9/10 + 9/100 + ...
0.999... = E ar^k (a = 9/10, r = 1/10, |r|<1)
0.999... = a/(1-r)
0.999... = (9/10)/(1-(1/10))
0.999... = (9/10)/(9/10)
0.999... = 1

So, either I'm a complete hack, or E ar^k (where |r|<1) != a/(1-r) is wrong. In fact, it's possible to represent the 9.999... - .999... in similar notation and blow my infinity versus infinity+1 argument out of the water. So, as any credible thinker should, I make my last stand and give my opponents a chance for a coup de grace.

Looking further into it, there is a step where we have limit (n->infinity) (ar^(n+1))/(1-r). Anyone who remembers their limits knows that this evaluates to zero as r<1. Desperate for an "out", I pose the following question.

Is ar^(n+1) zero or simply infinitesimally small to the point that we really can't keep track of it any longer and in 99.999...% of situations it doesn't matter? If it's the former, and mind you I would like a "because [explanation]" following any "It's zero", then my mental exercises have done little but given me a good, if futile, stretch. If it's the latter, then I would contend we've found the hole in another argument for 1 = .999...

Thus we end my grasping at straws.