Is there any rigid mathematical proof?

Question

Is there any rigid mathematical proof that 0.999... = 1? Can you show me a website that proves this definitively?

Looking at the past questions and answers, I see arguments for both sides; some more convincing than others; some unsound and some plain wrong.

Answer 1

Debunking the .999/1 proofs:

There are many so-called proofs that are
used to perpetuate the false statement
0.999... = 1. I will begin with the common
proofs and end with the most complicated.
The flaws will be clearly revealed in all
these proofs.

Let's begin with the most commonly used
proof and see why it's false.

The Multiplication Trick:
==================
If
x = 0.999... [1]
then
10x = 9.999... [2]
=> 9x = 9 [3]
=> x = 1 [4]

Our arithmetic only deals with finitely
represented numbers. For example, in [2]
above we would need to place a zero at the
end of 9.999... but this is impossible.
Although this algorithm is useful for converting decimal (or other radix) format, step 2 is not correct. It is only an approximation. Step [3] is also false since we have a subtraction taking place where the full extent (**) of both quantities is not known. i.e. 9.999... and 0.999...

(**) The extent of a number is its complete
representation in a radix format. The extent of 0.25 is known for it is finite. The extent of 0.333... is unknown because it is infinite. Can you write the whole thing out? No. The extent of numbers such as pi, e and other transcendental numbers is unknown.
Note: The extent of 0.999... is unknown.

The Geometric Sum:
===============
This trick is split into two parts. Part
A is attempt one and part B is the desparation
attempt.

Part A:

The formula for finding the LIMIT of the
sum of a geometric series with
first term a and common ratio r where
|r|<1 is a/(1-r). Note that this is not
the actual sum of the series. It is the
value that is an upper bound, i.e. a limit.

The limit of the series 9/10+9/100+9/1000+...
is in fact 1. Does this mean that 0.999...=1?
Absolutely not for the reason explained in
the previous paragraph.

Academics will now try part B:

"But the limit is defined as the sum of
the series." or "The limit of the geometric
sequence (Cauchy Sequence 0.9; 0.99; 0.999; ...)
is 1."

This is true but the limit is NOT the actual SUM.

Dedekind Cuts:
===========
As another act of desparation, academics will
try to challenge you by providing a definition
of Dedekind cuts as if it is divine law and then
challenge you to find a cut for 0.999... and 1.

You can really confound them by answering as follows:

[-inf,1) [1,+inf) Cut for 0.999...
[-inf,1] (1,+inf) Cut for 1

However, the above two cuts are really the same and they will argue that 0.999... and 1 are the same. You can immediately silence them by throwing a counter challenge: Ask them to find you a Dedekind cut for pi, e or any other irrational number. Some crafty fellows will devise a clever way to show that these Dedekind cuts are 'possible' - in fact the pi cut is really based on the definition of pi, i.e.
circumference/diameter. An actual value of pi is
never provided. If an academic says the Dedekind cut for pi is: (-inf,pi] (pi,+inf), you can respond by saying that the Dedekind cut for the NINES number is (-inf,Nines] (Nines,+inf).

Dedekind attempted to devise a method of representing any real number as a cut in the real number line. The very idea is problematic because there is a disconnect between the concept of numbers as defined in metric spaces and the design of the real number line.
This is a serious flaw in real analysis. If the
distance metric d(a,b) = 0 implies that a=b, it would be impossible to have a real number line without 'holes'. Of course this is utter nonsense.
The only problem in this approach is that Dedekind cuts assume the real line is continuous, however, real analysis states clearly there is no previous or next number. These concepts are in complete contradiction with each other.


Archimedean property:
=================
This is a property based on the 'least upper bound' property. Typical questions that arise are:

"Can you find a number between 0.999... and 1?" "Does a natural n exist such that 1/n < c for some real number c?"

So, let's examine all the corrollaries:

1) If c and b are real numbers with c > 0, then
there exists a natural number n such that nc > b.

Suppose c=1 and b=0.999..., then any n greater than 1 will do. Check this:
1 * 1 > 0.999...
2 * 1 > 0.999...

2) If c is a real number greater than 0, there exists a natural number n such that c > 1/n > 0.

Suppose that c=0.999..., then any n greater than 1 will do. Check this:
0.999... > 1/2 > 0
0.999... > 1/3 > 0

3) If c and b are real numbers with c < b, then
there exists a rational number x such that
c < x < b.

Suppose that c=0.999... and b=1. This corresponds to the question of being able to find a rational number between two numbers that are not equal.

Let A=99, then the number represented by:

A/100 + A/(100^2) + A/(100^3) + ...

lies between 0.999... and 1. Check using partial
sums:

0.9 < A/100 [0.99] < 1
0.99 < A/100+A/(100^2) [0.9999] < 1

It is easy to see that this is true for any partial sum. You can find infinitely many numbers
between 0.999... and 1.

"Does an n exist such that 1/n < c for some real
number c?"

Sure. If c = 0.999... then one such n is 2.
Check: 1/2 < 0.999...

This last question was already answered but I
repeat the full example for clarity.

Infinitesimals:
==========

This is the most strange argument of all.
An infinitesimal is a figment of the imagination.
It does not exist in other words. Why? The
word itself means infinitely small. Academics
confuse themselves horribly with this word
and the phrase "the smallest positive number
greater than zero". The two are not the same.
How small is infinitely small one might ask.
It certainly makes no sense to speak of negative infinity. Non-standard analysis cannot be used to prove anything about real numbers.

Conclusion:
1. Problems arise because of the structure of
the radix systems.
2. Misinterpretations of infinite sum and limit
of infinite sum.
3. All radix arithmetic involving numbers that
cannot be represented finitely is questionable.
Example: Show me one calculation involving
pi where the result is expressed without any
error in a radix system. You will not be able
to do this. Note that telling me Area of a
circle with unit radius is pi is unacceptable. I
want to see pi represented in some radix
system.

Answer 2

1.00000000000000000000000000000000000 is different than 0.99999999999999999999999, to any degree of accuracy demanded. As far out as one may examine, there is a 0 in a given decimal position for the real 1 and there is a 9 in the same decimal position for the approximation. This is a necessary and sufficient refutation of the premise.

Another refutation might be fun. 0.999999999999999999.... + 0.00000000000000000000000....1 will be a larger number than 0.999999999999999999.... to the same (arbitrarily large) number of decimal places, but neither will be 1.00000000000000000000.

The proof based on the sum of 1/3 = 0.333333333333 and 2/3 = 0.666666666666666, so 3/3 = 0.9999999999999999999 = 1 is a fallacy of "begging the question", since the assumption of 0.9999999999999.... = 1 is built into the proof. More to the point, 2/3 = 0.666666666666666......7, meaning that wherever you have stopped to look at the final digit, it is a 7, not a 6, and even when you look another 10,000 digits past it, the terminal digit is a 7, not a 6. This demonstrates a flaw in the so-called "proof"; the "disproof" itself is in the first paragraph.

Answer 3

This question comes up often here in Y! Answers, and it frightens me sometimes to think of how many otherwise intelligent people who think they know a lot about maths can't get this problem correct. I've had a few "Best Answers" to this question, and I've included one of them in my source below.

My guess is you'll see a number of correct proofs that 0.999... = 1 here. One of the most compelling uses the concept of betweenness. Two numbers are equal if there is no number that exists between them. That's why 0.999... (repeating forever, an infinite string of 9's) IS equal to 1. No one can argue to "Just add another 9 on the end," because there is no end to which to add another 9.

The so-called "proofs" that 0.999... ≠ 1 (and I'm sure you'll get people who'll post some of those here, too) all have flaws... one of which being the inability to grasp the concept of an infinite string of 9's. They treat it as a finite number, one that can be counted if one had the time and inclination, and show pre-Archimedean thinking.

Some others might try to use the concept of infintesimals to debunk the fact that 0.999... = 1. Their argument goes along the lines of 0.9 < 1, 0.99 < 1, 0.999 < 1, and so on. Their argument that for every expansion of this line there is an infintesimally small number that is the difference between 1 and however many 9's they choose to write is flawed, though. It's flawed because, again, they deal with finite numbers of 9's rather than an infinite number of them. Although Leibnitz and Newton both used infintesimals when first developing the calculus, it caused a great deal of disbelief as to whether these infintesimals even existed. That's why Weierstrass redefined concepts of derivatives and integrals using limits. The use of infinitesimals in proofs today is considered by most mathematicians to be unethical.

Good luck in your quest to learn more.

Answer 4

Yes there is a proof but I can promise you that unless you have had higher level college math (real analysis or modern algebra) then you won't understand it. And that's the real problem with the debate on yahoo answers about this problem. The real proof is in a language of math that most users don't understand.

All the examples that show .999.....=1 are missing one key thing and that's a theorem that basically says, in laymen terms, if you have two numbers and you can't find a distance between those two numbers then those two numbers are the same number. To prove that .999...=1 one has to prove this theorem.

Here is a website that uses that theorem but does not prove it. The text book he references might have a proof of the theorem.

http://home.comcast.net/~rossgr1/Math/one.PDF#search='proof%20.9999%3D1'

Answer 5

If you try to turn it into a fraction you can make it equal 1. However the problem associated with that is that it is still equal and yet not equal to 1 at the same time. Of course following the same logic .3+.3+.3=1 you turn them into a fraction. I think that the reason why this one in particular works is because .9999 can be thought of as X->1 which is a limit. So in other words it equal one because it is so close that it can't be clearly defined on its own. So I don't think that there is a proof so much as people found a weak spot in mathematics and try to use ways to work around it and make it true.

Answer 6

Debunking Tom:

Tom wishes to simultaneously assert:

.9999... is the sum of the infinite series .9+.09+.009+.0009...
The limit of partial sums of an infinite series is not the same as its sum.
.9999... < 1

In this case, Tom needs to tell us just what the hell he thinks an infinite sum is. Fortunately for us, he has done so elsewhere:

"You cannot find the value of ANY infinite sum" - quoted from http://answers.yahoo.com/question/index;_ylt=AsbS4OvI1w4xduMAsF_tvEzsy6IX?qid=20060902051800AAi2lgv

Now, if an infinite series does not HAVE a sum, then clearly that sum cannot be a real number, and thus CANNOT be said to be less than 1. You would think someone so moronically stupid that he cannot even avoid contradicting himself would try to avoid insulting thousands of mathematicians smarter than him, but then, this is the guy who seriously tried to argue that having multiple decimal expansions for the same number is the same as having multiple numbers for the same decimal expansion (the part where he talks about not having zero as a placeholder) - really, this is like me arguing that not all dogs are mammals because not all mammals are dogs. He concludes by saying that, and again I quote: "And yes, numbers looking different is DEFINITELY evidence of these two numbers being different!" - gee, I'll remember that the next time I'm asked to compare 1/2 and 2/4.

To answer your question: read Louise's answer again. It says all that needs to be said.

Answer 7

I remember seeing something like

.999... =.333...+.333...+.333...
=1/3+1/3+1/3=1
Thus .999...=1

This was a proof provided by my Calculus 2 instructor. I am not certain whether this would qualify as a rigid proof, but this is all i could think of for now.

Answer 8

In the real number system, every number has a definite place, and that place is occupied by only and only that number.
so in this case too, the place on the number line occupied by 1, is distinct and CANNOT be over lapped by any other REAL number.so, 0.9999.... is approximately equal to 1 and not exactly equal to it, just as 0.66666.....~0.67 or 0.667 or 0.6667....depending on precision u want.
0.999....a non terminating repeating decimal number will be lying between 0 and 1.( 0<0.99...<1)...this is SOLID and there cannot be any 'equality' sign in this inequality.
I think it is quite a SOLID proof......

Answer 9

*sigh* okay, you want rigid proof, so here it is.

in sequences and series, .9repeating can be represented by 0.9+0.09+0.009+0.0009 etc. etc. until infinity.

so, the series from 1 -> infinity of the sequence 9 * (0.1^n) where n starts at 1 (and increases to infinity)

that is the sequence. now, the formula a/(1-r), where a is the first term in the sequence (0.9) and r is the common ratio between the 2 terms (0.1)

now, plugging it in, 0.9/(1-0.1) = 0.9/(0.9) = 1

there. official proof.

Answer 10

Yes