MATH questions

If I'm understanding correctly, your issue is whith the step in the proof where 9.999... - 0.999... = 9 ?

Or is it where 10 x 0.999... = 9.999... ?

What is ten times an infinitestimal?

What is 10% of infinity for that matter?

If you look at

10x-x,

x=0.999999.., for some finite number of 9s, as the number of 9s increases,

you see that the resulting expression is 8.9999...9991.

Making this substraction directly doesn't make much sense, as you can't subtract one infinitesimal 0.000000...09 from another.

Hmmm,

x = 0.111...

10x = 1.111...

10x  - x = 1

x(10 -1) = 1

x = 1/9

[Edit - in response to the deleted objection that 0.111... != 1/9]

You can't really do any sort of arithmetic operation on an infinite decimal without treating it as an infinite limit since arithmetic operations aren't clearly defined for infinitesimals.

http://www.chess.com/forum/view/community/the-last-one-to-post-is-the-biggest-fan-of-chesscom

You have to be cautious with arithmetic on limits as well.

Wonderful. This is quite true. The reason popular proofs try to conceal this is because it is one hell of a minefield to argue when you want to persuade middle/high schoolers of this truth. Vast swathes of modern mathematics arose from considering rigorously the consequences of different assumptions on infinitesimals, limits and treatment of infinities, none of which would appeal much to someone studying basic precalculus.

In any case, if we're talking about good old R and the relevant implications are clear to all parties, there really isn't any issue.

FancyKnight...I learned a lot from you today! Your link helped. Would the following be a fair synopsis of your argument for the non-mathemetician?

".999...=1 in a real number system, but does not when using a seperate system. This other system is fundamentally different in how it resolves Xeno's Paradox."

Ok, so if 0.111... isn't 1/9 then please provide the correct decimal representation of 1/9.  lol

If it's infinitesimally close to 1/9 there should be infinite infinitesimal numbers in between and yeah, it gets weird, but when using regular highschool algebra it's all easy and logical.

chess, anyone?

There are thousands of other (inferior) forum threads on chess.

The problem with .99999 = 1 is that there is a hidden transition from base number involved.

1/3 is a description in base 3.

1/10 is a description in base 10.

You can not express 1/3 properly in 10 base, nor can you express 1/10 properly in 3 base.

If you would use 10 digit numbers to express 1/3 in three base, then would the result be:

1:  1

2/3: 0.2

1/3: 0.1

2/9: 0.02

1/9: 0.01

2/27: 0.002

1/27: 0.001.

1/10 in 3 base is an infinite number starting with 0.0022222.

0.9999 is a good estimation of 3/3 in 10 base, but it is not equal to it, because 3/3 = 1. Just as 10/10 is 1, but that can not be expressed properly in 3 base.

I really get it!

It's by far simpler to avoid talking about change of base at all. It's unecessary, and increases the chance of making an error in your argument. Which people will then pick up on and sidetrack the whole thing, assuming that the oversight you made is key to your stand (which it usually isn't.)

@LoekBergman: There's a slight gap in your last line. What you say is true that 0.9999 is not equal to 1. However, what is being said is that 0.9999... = 1, with the understanding that the "..." refers to an infinite string of nines, since we can't be bothered to type all that out. When the string of nines is infinite, there ceases to be any non-infinitesimal (infinitesimal here used in the precise mathematical sense) difference between it and 1, so we say they are equal.

@LongIslandMark: only after your comment I searched back and came to page 3 and there I saw your post telling the same as I did. Even the same decimal description for 1/3 in 3 base. :-)

@Remellion: lol, there is a huge gap in your line of reasoning. LongIslandMark and others have tried to explain to you. I will do it one more time.

Let's assume that we have a 25 digit base, consisting of the numbers a-z.

Let's reformulate that .99999 = 1 into .iiiiiiii = z.

Let's now answer the question of OnStar: what is in between .99999 and 1?

A whole lot, for instance .pppqrab or .t and .suv.

9 is only a special number in a decimal system and you can not think a number in between the .99999 and 1 within a decimal system. Within a 25 base system, you can easily perceive a lot of different numbers. There are of course infinite number base systems, just as much as there are R numbers. The bigger the number base is, the more numbers can be found between the infinitesmal .99999 and 1. The gap is actually almost as big as R itself, because the total numbers that can be found between .99999 and 1 is at least (Rmax) - 10.

.9≤1 there I found an inequality to satisfy both sides of the arguement now who wants to play chess

I don't understand why you keep talking about .99999 (finite number of 9's). We're taking about .99999... (infinite number of 9's). At the moment you cut the remaining infinite tail off, clearly the number you get is smaller than 1.

Something's not quite getting through here, since I'm arguing for your (and LongIslandMark's and the others') side of the case here. Somehow you're interpreting it differently.

I'm arguing in the domain of R using base 10. If you go back a few pages, I think you'll find a comment I made that also makes clear that 0.FFFF... = 1 (hexadecimal), 0.11111... = 1 (binary) etc. I'm in no way saying that 0.9999... = 1 in all base systems or for other domains outside R (it does hold in say, Q and C but not R*) since that would be plain absurd. What I am saying is that 0.999... = 1 in decimal (since that's the main argument), and in other bases there are analogues as described above (but unnecessary to the main point of contention among most people). The underlying point being that the domain in question does not contain infinitesimals.