0.999... = 1 [PROOF]

This is my proof that I devised to show that 0.999... [nonterminating] is in fact equal to 1.

It will work for 2.999... = 3, 1.999... = 2

It will NOT work on anything that is between 9 and 0 (0.111..., 0.777...)

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Why Infinity Renders to Zero!

[MistyGothis]

In nonstandard analysis, the expression "0.999. .. ." designates an endless hyper-sequence of decimal digits which represents the hyper-real number 1. So "0.999. . . =1 " is true for the hyper-real numbers just as it is for the real numbers.

However we can use the string "0.999. . .9" to designate a string of decimal digits that terminates after an "infinite" number of repetitions of the digit 9, which in turn represents a hyper-real number infinitesimally less than 1.

[flaming]

Real Analysis: A Constructive Approach by Mark Bridger

In constructing the real numbers 0.999... is represented by the sequence 0.9, 0.99, 0.999, 0.9999, ... (lets call this sequence a)

The limit of this sequence as it goes to infinity is 1. Thus for every real number Epsilon>0 there exists a natural number n0 such that for all n>n0, |aN-1|<Epsilon.

However aN is never equal to 1. If we truly define 0.999... to be an infinite repetition of 9's then it is in the set of Hyperreals (see Vihart).

[MistyGothis]

@MisterrLi
Actually, even in the hyper-real numbers, 0.9999. . .. designates the number 1. To designate a number infinitesimally less than 1, you need to write 0.999. . . 9, imagining it to terminate after a "large" number of repetitions.

[TheWinnieston]

you're right about the pencil
However, the Plank constant rules that there are a finite ammount of particles in any ammount of matter. This defines that a pencil does not weigh the same as a mountain, nor do they contain the same ammount of mass

[flaming]

Reread it. As you chose a larger and larger N (in the set of natural numbers) you approach infinity. This is how a sequence is defined.

[colinjava]

Your proof is absolutely fine.

I'm not clear what you meant when you said it won't work for 0.222... and 0.111...
If you mean they won't come out as whole numbers that's obvious by looking at them, as 0.2<0.222...<0.3, similarly for 0.111...
But the technique still applies to them and gives you their fractional representation.

Would be nice to see some more interesting proofs of it.
I tend to go with the gap argument with d = 1-0.999...
Then show if d>0 that you can get contradictions.

But I did see a nice proof in base 10 and 12 arithmetic that also proved it.

[TheWinnieston]

You are right about that

and this proof shows us that infintesimally small numbers don't actually come out to what we think they do.

Like, 0.000....1 will come out to 0

[TheWinnieston]

You certainly have a point there...
I'll work on that proof now.

[TheWinnieston]

You certainly got a point there. I'll work on that one.

[flaming]

Actually the whole argument for infinitesimals is in the fact that 0.999... never terminates and goes on for infinity. And thus you get a really small number that is theoretically in existence of 0.0000... which also goes on infinitely.

[flaming]

You are missing the point. A sequence is defined as you approach infinity. You never replace N with infinity. It is as N approaches infinity. So in the way you said it above you would need to say for 2^n as a sequence as N approaches infinity the sequence approaches infinity.

[flaming]

The Cauchy sequence that represented 0.999… (namely 0.9, 0.99, 0.999, …, 0.999…9, …) by definition gets arbitrarily close to 1 as the N (the number of nines in this case) chosen gets close to infinity. And every Cauchy sequence of Real numbers is convergent to a Real number and by definition approaches 1 but is never equal to one.

[Araqius]

Then how long is your pencil?
Your pencil has a definite length (It's doesn't get longer or shorter by itself, does it?). You just cant tell its true length since you can keep zooming in and in. This means something like 0.999... (keep zooming in) can have a definite value, thus can be a number.
P.S. You're right about "0.999... doesn't mean that number go on forever." 0.999... is a number and it can't go anywhere.

[TheWinnieston]

How can 2^n = infinity even when n != infinity? If n=2, then 2^2 = 4. 2*2. Now, if you mean a geometric set or something that goes from 1-infinity, then yes, you'd eventually reach infinity.
but infinity is a concept, not an actual value.

[ericsanders]

There is an error in your 10x-x step. A just recently published paper describes why .999... is not equal to 1 in a mathematically rigorous way. The errors are detailed in an easy to read format. Go to the website PublishResearch. It is publication 136 or click on the title "Long Division and How It Reveals That.... This is the first paper that deals with this problem mathematically and by showing all the missing parts and pieces.

[Araqius]

2^n = 2, 4, 8, ...(toward infinity)
2^n = inf (This has to be true no matter what N is, not only when n = inf)
set n = 1
2^1 = inf
2 = inf
YES!!! THIS METHOD IS AWESOME!!!

[realmrkou]

What your assuming is that 1 exists to start with by stating 1x=0.999...

[TheWinnieston]

So, you're saying that I'm not wrong, its just that I proved that the hyperreal number 0.999... does not equal the real number one?

[flaming]

Is that not the point to infinitesimals?

[MisterrLi]

How can 0.999... not be a real number if it is equal to a real number, namely 1?