Property of real numbers: between distinct real numbers is at least one other number. Now try to find a decimal representation of a number bigger than 0.9999999... but less than 1.0. You clearly can't. They must be equal. No need for infinitesimals.

Okay, so it follows from completeness. Now prove that real numbers are complete (or provide a construction of the reals that uses completeness as an axiom) without using concepts foreign or confusing to someone with a middle school level exposure to math.

It depends on where you want to start your axioms. We can go the Whitehead/Russell route or just use this as an axiom.

We convince children that 1+1 =2 without delving into the Peano axioms. It's ok to not delve too deeply into the axiomatic structure of the reals.

You're just proving their point. None of the things you talk about are even remotely obvious or “natural” to people we're talking about. You want them to ”choose axioms”? Axiom-a-whaaa? 1+1=2 does not need Peano axioms because it's cognitively fundamentally different than 0.999...=1.0. It is immediately obvious because dealing with simple arithmetic on natural numbers is in everybody's experience constantly. In other words we do not need to convince them. Clever untrained people would be able to get 0.999... thing quickly if you gave them some explanation and let them think a bit, if they cared, but properties of real numbers are far from obvious or basic.

> Okay, so it follows from completeness.

There is nothing to do with completeness here; 0.999… = 1 is a statement about a series of rational numbers summing to a rational number, and the convergence of the series is established by the fact that it sums to the right-hand side, not by an abstract appeal to completeness.

I'd argue that the distance is 0.00000... = 0 and that distance = 0 implies that the points are equal.

Doesn't that property come from the fact that there is no infinitesimal in real number?

> find a decimal representation of a number

0.99999999...1

This denotes the abstract idea that we take 0.999... (infinite number of 9's) and add another digit.

This is no more or less abstract than 0.999... to begin with.

0.999... is a unicorn, and 0.999...9 is a unicorn with a pink ribbon on the tip of its horn.

> add another digit

But you can't. All of the places where you might want to add a digit are already occupied by 9's.

So you're saying a countable infinity can be occupied, so that there is no room for one more.

You must then disbelieve concepts such as that the even integers can be put in 1:1 correspondence with all integers; i.e. that there are exactly as many even integers as there are integers.

> So you're saying a countable infinity can be occupied, so that there is no room for one more.

No. I'm saying that a countable infinity doesn't have room for one more at the end. There is no end.

You can obviously stick a 1 in the middle, but then you have a number strictly less than 0.999...

You might be able to do something funny with a series whose terms are indexed by ordinals. The last `1` would then be indexed by `omega` (the first transfinite ordinal). I can't claim that the ordinary rules of arithmetic would make much sense, but it's certainly an idea that you can attempt to formalize.

It has been done:

https://en.wikipedia.org/wiki/Infinitesimal

Of course you can. Consider the sequence of rational numbers in (0,1). Now consider the sequence of rational numbers in (0,1]. The latter has had an extra term appended.

(This is of course flawed, but I think it illustrates that the question isn't completely trivial. It requires us to carefully distinguish between the notion of an ordered set and a sequence and even then we'll have to deal with the fact that the rationals can be made into a sequence, but not with the same ordering.)

No. There is no such thing as the sequence of rational numbers in an interval. The rationals have a total order, and they can be enumerated, but the total order of rationals does not define an enumeration of the rationals and hence the rationals are not a sequence. If you want to dispute this, tell me: what is the next rational after 1/2?

I took some liberties with the word "the", but I'm pretty sure that you know that it's easy to construct an ordered sequence of all rational numbers using any of a number of standard diagonalization arguments. I'll skip the details you already know.

Eventually, yes, you're going to be able to poke a hole in my argument. It is definitely flawed. I don't know how long it'd take us to get there, but it doesn't really matter. But we're already at the point where this cannot be considered "basic", and that is my true point here.

Attempting to demonstrate that 0.999... = 1 while meticulously avoiding any rigorous definition of what 0.999... means is not very easy and will require you to fend off all sorts of potential jabs from various directions. It's much easier to just talk about infinite sums and be done with it.

> It's much easier to just talk about infinite sums and be done with it.

Indeed. But it's a lot less fun :-)

You don't need to understand anything about infinitesimals to understand 0.999... = 1. Perhaps you meant limits? The "standard" approach would be to point out that Σ_{i=1}^∞ 9/(10^i) = 1 (that is, the sum from i = 1 to infinity of 9/(10^i) is 1), and understanding an infinite summation requires the concept of a limit. (Of course, there are simpler proofs that use only basic algebra and intuition about decimals; a limit is just the most direct approach.)

You're spot-on about needing to understand there's a distinction between a number and its decimal representation, though.

Limits use a construction that's pretty similar to an infinitesimal. The epsilon-delta definition of a limit is no joke for students.

Oddly enough, I never understood the epsilon-delta description of limits until I read David Foster Wallace's book on infinity. All through my degree in math I was taught about things without learning the historical context that created those things.

I'm always super happy that I stumbled in to taking topology before real analysis.

It meant that I understood the topological idea of limits before I had to do proofs using just the epsilon (for sequence) or epsilon-delta (for functions) definition, and so could translate the logic of showing things about neighborhoods in to the terminology of (real analysis) limits.

Limits, in the abstract, are a fairly simple concept: in the case of sequences, for any neighborhood of the limit, the entire tail of the sequence (past some point) is contained in the neighborhood; in the case of functions, for any neighborhood of the limit at f(x), there's a neighborhood around x, such that every point in that neighborhood maps to the neighborhood around the limit.

Getting a degree in math without understanding epsilon-delta is quite an achievement! I am serious.

Most people are happy to accept 0.33333... is the same as one third and from there it's a quick hop to 3*0.3333...

I didn't quite accept that until I challenged a professor in a 300-level university math class. This came up in class, and I spoke up and was like, "no, that can't be true, because 0.3333 repeated N times will never equal 1/3."

Up to that point, I had only a fuzzy notion in my mind of what 0.33333... even was or how it was defined. But the professor helped clear this up for me: that "infinitely-repeating decimals" was actually just shorthand for a limit definition, i.e. "0.333..." is defined to be the limit as N approaches infinity of (3/10^1)+(3/10^2)+...+(3/10^N).

Well, that's if they accept that an infinitely repeating number is actually EQUAL to the fraction. Some people just accept that a non-infinite number will never equal the fraction, but don't understand the concept of infinite repetition as an actual number (point on the number scale), not an estimate.

If people accept 0.33... = 1/3 but not 0.99... = 1, they are being logically inconsistent.

A "proof" that just plays a trick on people's logically inconsistent assumptions to derive a result isn't very satisfying. You're not really uncovering anything fundamental through that proof, just playing games.

Edit: what I mean to say is there are reasonable things "..." could mean such that .99... and .33... are both not equal to 1 and 1/3, respectively. But there are none for which one pair is equal and the other not, as you rightly point out. So all your proof does is show that the other person's viewpoint is inconsistent, but it doesn't give any evidence for one of the two consistent viewpoints over the other.

It's easy to imagine that eating 1 out of 3 slices of pie is "the same" in a very specific sense as eating 333333... out of 1000000... slices of the same pie (although that would be infinity out of infinity slices, which is meaningless).

What slips people up is that ignoring everything but the total pie consumed (taking the limit) is embedded in the definition of real numbers.

There's an analogous story with rationals: Suppose x1 = 1, y1 = 3, x2 = 2, and y2 = 6. If we plot them, (x1, y1) and (x2, y2) are clearly different points, but x1/y1 "equals" x2/y2 because they lie on the same line through the origin. We decide that we don't need to know about those individual points.

Its just algebra. As noted elsewhere in this thread, 10X = 9.999... so 10X - X = 9.999... - 0.999... = 9 = 9X. So X=1.

This proof is more subtle than it appears. Here's a bogus rewrite, for instance:

Let 10X = 9.9. Then 10X - X = 9.9 - 0.9 = 9 = 9X. Hence X = 1, but X is actually 0.99 in this case (not 0.9). You need 0.99 = 0.9 for this to work with the exact same structure as your version.

Your proof only works because appending a 9 to an infinite expansion of 9s does not actually add a 9. But at this point you're forced to establish meaning for an infinite expansion of 9s, at which point this is really not just algebra anymore.

What? Why -0.9?

Because that's exactly what the argument above does. It simply subtracts away the decimal part of 9.999....

This is always wrong except in the case of infinitely many repeated digits, and the proof does not explain this.

More rigorously, let 9.999{n} denote an expansion with n 9s after the decimal point, where n can also be infinity. The subtlety with the argument is that it needs X to be the same as everything after the decimal point (so that the result of the subtraction is just 9). This is never true for finite values of n, and the proof does not establish that it's true for an infinite value of n -- indeed, it can't do so without supplying a meaning in the first place.

Another way of phrasing it is that it assumes that if X = 0.999..., then 10X = 9.999..., where there are the "same number" of 9s after the decimal point in 10X as there are in X. This seems intuitive for an infinite repeating sequence of 9s, because "one less than infinity" is still infinity, but it's not very rigorous, and the argument as written certainly doesn't explain this.

Okay, this is a little late but when you say 10X = 9.999... then X = 0.999... not because you have removed the 9 but because that's what X is; a tenth of 10X! So when you say in your argument 10X = 9.9 and therefore 10X - X = 9.9 - 0.9, that's wrong. 10X = 9.9 implies X = 0.99. No wiggle room. No need to consider infinites, just one movement of a decimal point.

[And hence 10X - X = 9.9 - 0.99 = 8.91 = 9X implying that X = 0.99]

Following, as delineated above, from there, you'll see there's no contradiction. It's not so easy to break arithmetic that easily without dividing by zero :)

Sorry for dragging on this meaningless thread.

Yeah, yes and no. The premise is, that 0.9999... repeating CAN be represented as a decimal number, so that means it CAN be used in arithmetic. If we deny that 0.999... minus 0.999... is zero, then the premise is broken and the question is kind of moot.

I think I came to the conclusion that 0.9999999... = 1 when I was 12 or 13. My reasoning was that for any n, we had 0 <= 1 - 0.999999... <= 10^(-n), so the difference had to be 0.

It's also not strictly, theoretically true. More an 'for all intents and purposes' kind of thing.

The sum, 0.9 * Σ (0.1)^n from n=0 to n=N-1, is 0.999... as N goes to ∞ (write out the terms).

On the other hand, for any finite value of N, the sum [1] is equal to 0.9 * (1-(0.1)^N) / (1-0.1) = 1 - (0.1)^N. This value goes to 1 as N goes to ∞.

More rigorously, for any positive value, ε, there is a value N, such that the value of the finite sum is within ε of 1.

[1]: https://en.wikipedia.org/wiki/Geometric_series#Formula

No, it is true in the strictest sense of the word.

Hardly. We use this reduction because it's essential from a mathematical and axiomatic perspective and because limits are a fundamental construct. But more philosophically, a natural and nonterminating simulation of 0.999... (were it possible) would never strictly equal 1. You would wait an infinite amount of time for it to do so. This comes down to how you view the problem. I would never argue with this through the lens of abstract higher mathematics. I think about it from a computability perspective, which is one way in which you can discount the observation.

It's not a reduction. If you try to find where to put 0.999... on the number line, it has to go exactly where 1 is.

For one thing, 1 - 0.999... = 0.000... because you never get to have any remainder since 0.999... is infinite.

Or here's another proof:

x = 0.999...

10x = 9.999...

10x - x = 9.999... - 0.999...

9x = 9.000... = 9

9x = 9

x = 1

Your "proofs" simply assumes that 0.999... is a notation denoting 1, without examining the underpinnings which might legitimize that.

9.999.. - 0.999 is 9 no matter how we define .999... just as long as two or more occurrences of the 0.999... notation all denote the same entity, and we understand that the syntax 9.999... is 9 + 0.999...

For example, if we define 0.999... as "rubber duck" then 9.999... stands for 9 + "rubber duck", and 9.999... - 0.999... stands for 9 + "rubber duck" - "rubber duck" = 9.

While that's true, the whole key to the proof is that

0.999... * 10 = 9.999...

Because "one less 9 than infinity is still infinity" it's what really closes the loop on the proof.

You are discarding an essential part of the representation. You are moving the goal post to make yourself right. Just because people don't always understand infinitesimals doesn't make them right. It's the reason they are wrong. There just plain don't know what they are talking about.

You can't represent 0.2 exactly using IEEE floats, either, but that doesn't mean the representation 0.2 is not exactly equal to 1/5th.

Scheme has (exact->inexact x) and (inexact->exact x) for those issues.