## A very interesting Mathematical Paradox

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AWA
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### Re: A very interesting Mathematical Paradox

phlip wrote:So... you're trying to "construct" a number by looking at other, irrelevant numbers?

Essentially, in the same way that Zeno's paradox deals with supposedly "irrelevant" numbers.

jestingrabbit
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### Re: A very interesting Mathematical Paradox

They're not irrelevant numbers, they converge to the number in question.

But still AWA, your argument leads to the conclusion that you can't represent any number that has an infinite decimal expansion.

The meaning of an infinite decimal [imath]0.a_1 a_2 a_3 a_4 \ldots a_i \ldots[/imath] is $\lim_{N\to\infty} \sum_{i=1}^N \frac{a_i}{10^i}$ and the thing about a limit is that it isn't approaching something, it isn't a process, its the thing approached, its a constant.

Are you familiar with limits?
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AWA
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### Re: A very interesting Mathematical Paradox

Actually, I am, but even if we use the geometric limit formula, it presupposes that 0.999...=1. (On the topic of limits, I'm unconvinced that a the sum or a convergent series equals the limit; if someone could enlighten me in that regard, I'd be much obliged, though I'd need a fairly convincing proof.)

Qaanol
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### Re: A very interesting Mathematical Paradox

You have now had a chance to think about the Archimedean property. It still says that for any positive real number r, there is a positive integer n such that [imath]0<\frac{1}{n}<r[/imath].

Next, AWA, are you familiar with the trichotomy property of real numbers? It states that for any real number x, that number is exactly one of positive, negative, or zero. In particular, a real number is zero if and only if it is not positive and it is not negative. Does that make sense?

AWA wrote:Actually, I am, but even if we use the geometric limit formula, it presupposes that 0.999...=1. (On the topic of limits, I'm unconvinced that a the sum or a convergent series equals the limit; if someone could enlighten me in that regard, I'd be much obliged, though I'd need a fairly convincing proof.)

The sum of a convergent series is defined to be the limit of the partial sums.
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AWA
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### Re: A very interesting Mathematical Paradox

Qaanol wrote:You have now had a chance to think about the Archimedean property. It still says that for any positive real number r, there is a positive integer n such that [imath]0<\frac{1}{n}<r[/imath].

Next, AWA, are you familiar with the trichotomy property of real numbers? It states that for any real number x, that number is exactly one of positive, negative, or zero. In particular, a real number is zero if and only if it is not positive and it is not negative. Does that make sense?

AWA wrote:Actually, I am, but even if we use the geometric limit formula, it presupposes that 0.999...=1. (On the topic of limits, I'm unconvinced that a the sum or a convergent series equals the limit; if someone could enlighten me in that regard, I'd be much obliged, though I'd need a fairly convincing proof.)

The sum of a convergent series is defined to be the limit of the partial sums.

I am not quite sure what the Archimedean property is (aside from your given definition), and to tell the truth, I don't see how it applies here.

Assume I am an idiot.

This trichotomy property states that any real number must be negative, zero, or positive. Ok. It also states that any zero number must be neither positive nor negative. Also ok. I'll get back to this in a bit.

Keep assuming I'm an idiot. A very inquisitive idiot.

Why is the definition of the sum of a convergent series defined to be the limit? Simply saying "Because it is" isn't going to fly.

For the sake of argument:

Let X equal the sum of a convergent series.
Let Y equal the limit of that same series.

I'm told X = Y.

Basic algebra will yield that Y-X=0.

However, any number zero must be neither positive nor negative.

Will not the limit of a series, minus the sum of the constituents of the (convergent) series, yield a nontrivial, nonzero infinitesimal? If this is true, then by your own argument, X cannot equal Y.

Macbi
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### Re: A very interesting Mathematical Paradox

If we give ourselves a decimal digit for every ordinal number can we end up with a consistent system where 1 - 0.999... = 0.000...1 ?
0.000...1 / 2 = 0.000...05 and so on.
0.000...12=0.000...000...1 (with the one in the 2ω place!)
0.000...1 * 10 doesn't work though. Hmmm.
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mike-l
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### Re: A very interesting Mathematical Paradox

Macbi wrote:If we give ourselves a decimal digit for every ordinal number can we end up with a consistent system where 1 - 0.999... = 0.000...1 ?
0.000...1 / 2 = 0.000...05 and so on.
0.000...12=0.000...000...1 (with the one in the 2ω place!)
0.000...1 * 10 doesn't work though. Hmmm.

What's the ωth place of 0.9999..? Anyway, you'll run into problems at every limit ordinal, because the act of 'carrying' in addition requires a predecessor location*, and you won't be able to multiply. I imagine if you allowed arbitrary digits for limit ordinal locations, it could work, but if you just extended the regular rules to allow arbitrary ordinal positions with arbitrary digit values, I think you'd still have 1-0.99999... = 0.

*- for fractional locations.

AWA wrote:Will not the limit of a series, minus the sum of the constituents of the (convergent) series, yield a nontrivial, nonzero infinitesimal? If this is true...

How do you add up infinitely many things? We've given (and the rest of the mathematical community too) you one way, take the limit. And so the limit minus infinitely many things which converge to it, is an infinite sum. So we need to make it into a number, we do that by taking a limit. And so we get a limit minus itself, which is of course, zero.
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Gelsamel
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### Re: A very interesting Mathematical Paradox

All real infinitesimals are zero.
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AWA
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### Re: A very interesting Mathematical Paradox

Gelsamel wrote:All real infinitesimals are zero.

Lies. I refuse to believe that an inherently nonzero, albeit ridiculously small, number equals zero.

Perhaps I'm unclear. I don't want to hear some recycled "argument" which relies on "previously established" rules, because these rules presuppose either that infinitesimals equal zero, or that the sum equals the limit, or whatever.

I am challenging why this is the case.

How do you add up infinitely many things?

You can't. You can only approximate the sum. This approximation is the limit, which does not equal the sum.

Mike_Bson
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### Re: A very interesting Mathematical Paradox

AWA wrote:
Gelsamel wrote:All real infinitesimals are zero.

Lies. I refuse to believe that an inherently nonzero, albeit ridiculously small, number equals zero.

There is your mistake; it is NOT a small, nonzero number that equals zero, it is the same as zero.

mike-l
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### Re: A very interesting Mathematical Paradox

AWA wrote:How do you add up infinitely many things?

You can't.

Ok good. So 0.9999... is not a number at all. Neither is 3.1415...., 0.333... etc.

But unfortunately, using only finite decimal expansions doesn't even get us all the rational numbers. So we say ok, when I write an infinite decimal expansion, I associate that with the limit of the partial expansions. That's what it means, that's what it's defined to be. And this isn't a problem because before I've done this, it didn't mean anything at all! It wasn't a number before.

This works great, because now we can get EVERY real number. I can do addition and multiplication and all that jazz just the same way as I do with finite decimals. Unfortunately, some can be written 2 ways, but this is a small price to pay for being able to write anything whereas before I could write practically nothing.

You can only approximate the sum. This approximation is the limit, which does not equal the sum.

The sum doesn't exist, you even said so. Only the limit does.
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Gelsamel
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### Re: A very interesting Mathematical Paradox

Lies? Hehe, funny.

Math is descriptive tool that we made up as a system of interacting definitions. If you want to argue it down to the axioms then it's simply defined this way because we said so. Infinitesimals are not inherently non-zero. It's not just "ridiculously small" they are infinitely small. That is how they're defined. In the reals the only infinitely small number is zero. Just because they don't fit your intuition means that infinitesimals are defined incorrectly? You can call your own bullshit idea of infinitesimals "ridiculousimals" and you can show us how useful those are...

You say you don't want "previously established" rules because the rules "presuppose". What? How the fuck does that work? Do you want me to point to some physical phenomena which will verify some mathematical definition (rather than the math definition being used to describe the phenomena)? How about you prove to me, that 1 + 1 = 2 without using any mathematical definitions, prove it to me without presupposing or defining anything about the operations there.

You can say "I will not accept the convention of addition such that 1 + 1 = 2" all you fuckin' like it won't change the fact that our answer is "because that's how we define addition, it's useful to define an operation like that with which we can do useful things with, like describing physical systems for analysis".
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jestingrabbit
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### Re: A very interesting Mathematical Paradox

AWA wrote:
Gelsamel wrote:All real infinitesimals are zero.

Lies. I refuse to believe that an inherently nonzero, albeit ridiculously small, number equals zero.

You're free to invent a new number system if you like, but if you want to talk about the real numbers they are already defined. Now 'real' is just a word, we're not meant to see them as better resembling the universe or anything like that. But they are the set of numbers that mathematicians tend to talk about when they use decimals. So, unless you make it clear that you're not talking about the real numbers, we'll answer with respect to the real numbers.

Here are some suggestions of different systems in which you can create different interpretations of 0.999... .

http://en.wikipedia.org/wiki/0.999...#I ... er_systems

AWA wrote:I am challenging why this is the case.

The process that led to the definition of the real numbers is complicated. I can perhaps most simply justify their definition by saying that they are a good model of the number line that we see in our first years of schooling.

More abstractly, they are the unique complete ordered field. Every one of those terms is important: complete gives us limits that are always within the set; field gives us nice arithmetic; and ordered field gives meaning to positive and negative that "works". I expect qaanol was trying to get you to agree that you want a complete ordered field, and then say that you therefore have the real numbers.

AWA wrote:How do you add up infinitely many things?

You can't. You can only approximate the sum. This approximation is the limit, which does not equal the sum.

It seems that you would then be forced to say that

$0.999\ldots \neq \sum_{i=1}^{\infty} \frac{9}{10^i}$

as the RHS is undefined. Moreover, given a sequence [imath]\{a_n\}_{n=0}^\infty[/imath], with

$\lim_{n\to\infty} a_n = a$

you can usually rewrite that as a sum, with [imath]b_0 = a_0,\ b_i = a_i- a_{i-1}[/imath] and

$a = \sum_{n=0}^\infty b_n$

so I suspect that whatever you do, you're going to have to work hard to get limits back, which are the foundation of calculus, which is the main link between mathematics and the real world (where here real means what it usually means).

So, as I say, you're free to define whatever you want, but by throwing out 0.999... = 1, you are losing some of the structure that is fundamental to how most mathematicians work, and you are putting yourself outside the mainstream mathematical discourse.

And really, if you're going to do that, you should have a good reason. So, why should [imath]0.999... \neq 1[/imath]? In particular, we all know that 2/4 = 1/2, so "its written differently" isn't going to convince me that its a good idea.
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Qaanol
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### Re: A very interesting Mathematical Paradox

Okay, we have two properties defined for real numbers:

1) Archimedean: for any positive real number r, there is a positive integer n such that [imath]0<\frac{1}{n}<r[/imath].
2) Trichotomy: a real number is zero if and only if it is not positive and it is not negative.

Now we will get two more:

3) The absolute value of a real number cannot be negative. That is, [imath]|a|\geq 0[/imath].
4) Two real numbers are equal if and only if their difference is zero. That is, [imath]x=y\iff|x-y|=0[/imath].

All four of these are facts. Do they make intuitive sense to you?

(The double arrow “[imath]\iff[/imath]” is read “if and only if”, and means that the statements on either side are equivalent. If one is true so is the other, and if one is false so is the other.)

To get a sense for the Archimedean property, it is effectively the reason you can divide by any nonzero number. Since [imath]\frac{1}{n}<r[/imath], then [imath]n>\frac{1}{r}[/imath]. That is, we have bounded the reciprocal of a positive real number r by a positive integer n. You know you can’t divide by zero, but could there be some nonzero number you can’t divide by? The Archimedean property helps show there isn’t. Zero is the only real number you can’t divide by.

AWA wrote:This trichotomy property states that any real number must be negative, zero, or positive. Ok. It also states that any zero number must be neither positive nor negative. Also ok. I'll get back to this in a bit.

Good, I’m glad you were able to put this in your own words. It shows you have firm grasp of the concept.

AWA wrote:Why is the definition of the sum of a convergent series defined to be the limit? Simply saying "Because it is" isn't going to fly.

You know how to add two numbers together. By induction, you know how to add any finite number of numbers. But the grade-school method of addition doesn’t tell you how to add infinitely many numbers. You need a new definition of addition. It turns out that infinite sums would not be mathematically useful if their values were defined to be anything other than the limit of partial sums.
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lightvector
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### Re: A very interesting Mathematical Paradox

AWA wrote:Perhaps I'm unclear. I don't want to hear some recycled "argument" which relies on "previously established" rules, because these rules presuppose either that infinitesimals equal zero, or that the sum equals the limit, or whatever.

Unfortunately, these previously established rules are mostly all you are going to get.

Math is a game where we invent systems of rules and then try to discover the consequences of those rules. The real numbers are nothing more than one of these invented systems, with pre-agreed-upon rules. In this system, "0.9999..." has a specific meaning, and "1" has a specific meaning, and in the real numbers, they are equal. The real numbers are by no means special - there are plenty of other number systems. But they are the most common, and they satisfy some very nice properties, and so when you don't specify the number system, everyone assumes that you are talking about the real numbers.

Would it help you to see a precise list of the rules of the real numbers (which are called "axioms"), followed by a proof that under these rules "0.9999... = 1"?

Alternatively, try to construct your own system of numbers in which 0.9999... ≠ 1. Go ahead! Presumably, if 0.9999... ≠ 1, then you want a system in which the quantity ε = 1 - 0.9999... is nonzero and positive. So you need to create some new rules governing how this number ε behaves and how it interacts with other numbers.

Presumably, ε is the "limit" of the sequence (1, 1-0.9, 1-0.99, 1-0.999,...), which is (1, 1/10, 1/100, 1/1000,...). Is ε also the limit of the sequence (1/10,1/100,1/1000,1/10000,...)? How about (1/2,1/4,1/8,1/16,...)? Will there be another number in your system that is the limit of (ε,2ε,3ε,4ε,...), or will this sequence not have a limit? Or maybe we will do away with the general notion of a limit altogether?

Can ε be divided by other numbers? If so, is ε/10 the limit of some sequence? Is division by ε defined? If so, what is 1/ε? If it is defined, then since ε is less than 1/n for any positive integer n, presumably 1/ε is some sort of "infinity" greater than every positive integer n? Is exponentiation defined for ε? If so, what is 2^ε?

This is far from an exhaustive list, but these are the sorts of choices you must make when designing your new number system, and different choices will produce different systems. Depending on your choices of the above, you may end up with any number of things, such as the dual numbers, the set of rational expressions with real coefficients in the variable ε, the surreal numbers or some other number system entirely. If you are not careful, you could also end up with a number system that is inconsistent. And regardless, your number system will be very different than the reals, and will certainly not have all the properties that the reals do.

AWA
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### Re: A very interesting Mathematical Paradox

Two things, and I'll drop this:

Qaanol wrote:
AWA wrote:Why is the definition of the sum of a convergent series defined to be the limit? Simply saying "Because it is" isn't going to fly.

You know how to add two numbers together. By induction, you know how to add any finite number of numbers. But the grade-school method of addition doesn’t tell you how to add infinitely many numbers. You need a new definition of addition. It turns out that infinite sums would not be mathematically useful if their values were defined to be anything other than the limit of partial sums.

How can there be an infinite sum at all? Supposing one were to pick an arbitrary partial sum, there would always be another figure to add. You can't add up an infinite number of terms, because there's no end to the sequence. Which, of course, brings up the question of why the supposed "sum" is equal to the limit of the series.

Qaanol wrote:Okay, we have two properties defined for real numbers:

1) Archimedean: for any positive real number r, there is a positive integer n such that [imath]0<\frac{1}{n}<r[/imath].
2) Trichotomy: a real number is zero if and only if it is not positive and it is not negative.

Now we will get two more:

3) The absolute value of a real number cannot be negative. That is, [imath]|a|\geq 0[/imath].
4) Two real numbers are equal if and only if their difference is zero. That is, [imath]x=y\iff|x-y|=0[/imath].

All four of these are facts. Do they make intuitive sense to you?

(The double arrow “[imath]\iff[/imath]” is read “if and only if”, and means that the statements on either side are equivalent. If one is true so is the other, and if one is false so is the other.)

To get a sense for the Archimedean property, it is effectively the reason you can divide by any nonzero number. Since [imath]\frac{1}{n}<r[/imath], then [imath]n>\frac{1}{r}[/imath]. That is, we have bounded the reciprocal of a positive real number r by a positive integer n. You know you can’t divide by zero, but could there be some nonzero number you can’t divide by? The Archimedean property helps show there isn’t. Zero is the only real number you can’t divide by.

This part is, I think, what bugs me so much. If an infinitesimal is defined as [imath]\frac{1}{n}[/imath] where n=infinity, then you're arguing that there's a case where [imath]\frac{1}{n}=0[/imath]. I can't reconcile this.

Qaanol
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### Re: A very interesting Mathematical Paradox

AWA wrote:How can there be an infinite sum at all? Supposing one were to pick an arbitrary partial sum, there would always be another figure to add. You can't add up an infinite number of terms, because there's no end to the sequence. Which, of course, brings up the question of why the supposed "sum" is equal to the limit of the series.

To be precise, you can’t add up infinitely many terms if you only have addition defined for finite sums.

Motivation for why adding up infinitely many terms actually can make sense:
Spoiler:
There are plenty of reasons it would be nice (meaning mathematically useful) to add up infinitely many terms. One of them is to calculate the area of shapes bounded by curves, like the area under a parabola or the area of a circle. There are other ways to do this, but at least for the circle all of them require taking a limit somewhere.

Another way to motivate it is to consider Achilles and the Tortoise. Achilles runs exactly 10 times faster than the tortoise, and the tortoise has a 1 mile head start. We want to know where Achilles will catch up to the tortoise.

Obviously Achilles must cover that first 1 mile, to where the tortoise started. But in the time it takes for him to get there, the tortoise moves 0.1 miles, since it only goes a tenth of his speed.

Then Achilles must cover that next 0.1 miles, to where the tortoise just was. He has now gone 1 + 0.1 miles. But in that time the tortoise goes another 0.01 miles.

So Achilles also must go another 0.01 miles. He has now gone 1 + 0.1 + 0.01 miles. The tortoise goes another 0.001 miles in this time.

Obviously the amount of time going by at each interval is decreasing, but we don’t care about time, only distance. And Achilles is clearly catching up to the tortoise. He started 1 mile behind, then got to 0.1 miles behind, then 0.01, now 0.001, and soon he’ll only be 0.0001 behind.

Considering each place where Achilles reaches the point the tortoise had been the last time we checked, we see Achilles at the positions 1, 1+0.1, 1+0.1+0.01, 1+0.1+0.01+0.001, and so forth. That is, at step n, we see Achilles at position
$\sum_{i=1}^{n}\left(\frac{1}{10}\right)^{i-1}$
meanwhile, also at step n, we see the tortoise still has a lead of [imath]\left(\frac{1}{10}\right)^n[/imath] miles. We want to know where Achilles catches up to the tortoise, and that is equivalent to when there are no more steps to take in order to get to where the tortoise was last step. In other words, when we have added up all the steps. But there are infinitely many steps to add up, so we can say Achilles catches up to the tortoise at position
$\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^{i-1}$
So far we do not know how to handle the sum of infinitely many terms. And if we stop summing after finitely many terms, then Achilles is still [imath]\left(\frac{1}{10}\right)^n[/imath] miles behind the tortoise. That is, if we take a partial sum, we find Achilles has not caught up yet. So what are we to do? Well, that remains to be seen. The important point for now is we have shown it is possible for infinite sums to come up in fairly simple problems. We know Achilles does catch up, since he is faster, so it would be nice if we could define infinite sums in such a way as to show where he catches up.

AWA wrote:This part is, I think, what bugs me so much. If an infinitesimal is defined as [imath]\frac{1}{n}[/imath] where n=infinity, then you're arguing that there's a case where [imath]\frac{1}{n}=0[/imath]. I can't reconcile this.

In fact it is quite easy to reconcile. If an infinitesimal is defined as you describe, it is not a real number, because real numbers satisfy the Archimedean property. In effect the Archimedean property says, “All real numbers are finite—there are no infinite real numbers and there are no infinitesimal real numbers.”

Why this matters:
Spoiler:
For comparison, suppose someone says, “The square of any real numbers is positive or zero.” Then I look at them and say, “Well if a number is defined as [imath]\sqrt{n}[/imath] where n=-1 then you’re arguing there’s a case where -1 is positive or zero.”

In fact no, they are arguing no such thing. The number I defined is imaginary, [imath]i = \sqrt{-1}[/imath]. It is not a real number. It is still an important and useful number, but it is not a real number. Similarly, your infinitesimal is not a real number (and it is also not well-defined; but there are ways of properly defining infinitesimals: they are still not real numbers though.)

If you don’t like not having infinitesimals, consider the integers. Are there any infinitesimals in the integers? Of course not. Then consider are there any infinitesimals in the rationals? That might take a bit longer to answer, but it should be clear that there are no infinitesimals in the rationals. So we already have two infinite sets of numbers that have no infinitesimals.

Probably the most common way of defining the real numbers is (and you’re not going to like this) as the formal limits of Cauchy sequences of rational numbers. I’m not going to get into what a Cauchy sequence is (although it’s a pretty easy-to-grasp concept), but my point is that the reals are defined as limits of sequences. When viewed in that light, although it may not be completely obvious, it does follow directly that there are no infinitesimals in the reals.

Some properties of real numbers:
Qaanol wrote:1) Archimedean: for any positive real number r, there is a positive integer n such that [imath]0<\frac{1}{n}<r[/imath].
2) Trichotomy: a real number is zero if and only if it is not positive and it is not negative.
3) The absolute value of a real number cannot be negative. That is, [imath]|a|\geq 0[/imath].
4) Two real numbers are equal if and only if their difference is zero. That is, [imath]x=y\iff|x-y|=0[/imath].

It is important to note that the real numbers are closed under addition, subtraction, multiplication, and non-zero division. In particular, for any two real numbers x and y, the difference z = x-y is also a real number.

You have mentioned that you don’t like presupposing that 0.999… = 1, so I will not do that. In fact, I will do exactly the opposite. Let us suppose that 0.999… does not equal 1, and see what happens.

Well, 0.999… and 1 are both real numbers, so their difference is a real number. Let’s call it d = |0.999… - 1|. We are acting under the assumption that [imath]0.999\ldots\neq 1[/imath] so by the fourth listed property, d cannot be zero. Then by the third listed property, d also cannot be negative since it is an absolute value. So by trichotomy, d must be positive.

Now by the Archimedean property there is a positive integer whose reciprocal is greater than zero and less than d. Let us pick one such integer, call it m, and then take the next higher power of 10, which we will call N. Now, since [imath]0<\frac{1}{m}<d[/imath] and N > m, it follows that [imath]0<\frac{1}{N}<\frac{1}{m}<d[/imath]. Additionally, because N is a power of 10, say [imath]N = 10^c[/imath] for some positive integer c, the decimal expansion of [imath]\frac{1}{N}[/imath] is just 0.000(total of c-1 zeros after the decimal point)1.

Recall that 0.999… is defined as the infinite sum
$0.999\ldots = \sum_{i=1}^{\infty}9\left(\frac{1}{10}\right)^i$
Note that each term in the sum is positive. We assume 0.999… does not equal 1, so we have either 0.999… = 1 - d or else 0.999… = 1 + d. I am not going to treat the case where 0.999… > 1. I trust you are capable of reasoning through why that is unacceptable.

So we have 0.999… = 1 - d. It follows from [imath]0<\frac{1}{N}<d[/imath] that the number [imath]k = 1-\frac{1}{N}[/imath] is in between 0.999… and 1, and not equal to either. That is, [imath]0.999… < k < 1[/imath]. But we know what k is. In fact we can even say what the decimal expansion of k is, since [imath]k = 1-\frac{1}{N}[/imath].

With [imath]\frac{1}{N}[/imath] = 0.000(total of c-1 zeros)1 it follows that k = 0.999(total of c nines)9. But that is the same as the sum of the first c terms in the definition of 0.999…, and there are more terms to go beyond that. Specifically, adding one more term gives one more 9, which makes a number bigger than k, call it L. So k < L.

Since we only add positive terms, the entirety of 0.999… cannot be smaller than L, since L is the sum of only finitely many terms in the infinite sum (c+1 terms in fact). Thus [imath]L\leq 0.999\ldots[/imath].

Remember that we constructed k in a way so [imath]0.999\ldots < k[/imath], but now we have [imath]k < L \leq 0.999\ldots[/imath]. By transitivity, this gives [imath]0.999\ldots < k < L \leq 0.999\ldots[/imath], or simply [imath]0.999\ldots < 0.999\ldots[/imath]. But a number cannot be less than itself. The statement “[imath]0.999\ldots <0.999\ldots[/imath]” is false.

We began with a set of premises, reasoned logically, and arrived at a false conclusion. This is called “contradiction” in mathematics, and it means that one or more of the premises must be false. The properties of real numbers I gave are all correct, and you can verify them for yourself if you like. Both 0.999… and 1 are real numbers, as you can verify by showing, for example, that {0.9, 0.99, 0.999, …} is a Cauchy sequence of rational numbers. (Actually, if you do that, you will have shown directly that 0.999… = 1.}

It follows that the false premise was our assumption that 0.999… does not equal 1. Since that is false, its negation must be true. Therefore, 0.999… = 1.

The crux of this argument lay in showing that, if there is a real number in between 0.999… and 1, then we have a contradiction. It could be done more quickly by noting that, between any two distinct real numbers, there exists another real number (in fact there exist infinitely many, and infinitely many of them are rational). I liked doing it explicitly because this way I was able to produce a number with a finite decimal expansion that would have to be in between them.

Got that? If 0.999… were less than 1 then there would exist a real number with a finite decimal expansion, which is strictly greater than 0.999… and strictly less than 1.

Anyway, there’s nothing inherently wrong with infinitesimals, just there are none in the real number system. And since every tail-infinite decimal expansion represents a real number, therefore every such decimal expansion necessarily does not involve infinitesimals.
Last edited by Qaanol on Mon Aug 30, 2010 3:48 am UTC, edited 2 times in total.
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lightvector
Posts: 224
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### Re: A very interesting Mathematical Paradox

AWA wrote:How can there be an infinite sum at all? Supposing one were to pick an arbitrary partial sum, there would always be another figure to add. You can't add up an infinite number of terms, because there's no end to the sequence. Which, of course, brings up the question of why the supposed "sum" is equal to the limit of the series.

You're right that the ordinary operation of "sum" can only be applied to a finite number of terms. It's not that there's some magical way to add up an infinite series one term at a time. There isn't.

So, because the ordinary way doesn't work, we need a different way for infinite series. So we define a new, distinct operation, which is to take the limit of the partial sums, and we give it the name "infinite summation". Therefore, the supposed sum is equal to the limit of the series because we defined it to be that way! That is, when talking about infinite series, the phrase "the sum of the series" is a placeholder for the phrase "the limit of the partial sums".

Why would we define such a notion? Well, as Qaanol mentioned, we would like to be able to make sense of the area for shapes that are bounded by parabolas and other various types of curves. But if we try to fill unusually curved shapes them up with shapes of known area, like squares and rectangles, or if alternately we try to approximate them with finer and finer polygons, and take the areas of these polygons, we only get an infinite sequence of different numbers for each successive approximation of the area. The only thing these numbers have in common is that they are converging to some value, namely the limit, so it makes sense to define the "area" to be the limit of these values, of the partial sums.

For a more basic motivation, we can go back to Qaanol's example of the tortoise and Achilles. One way to try to calculate the location where Achilles to catch up produces an infinite sum 1 + 1/10 + 1/100 + 1/1000 + ... But our intuition says there should actually be some location at which Achilles catches up, so we'd like to define a notion of "infinite sum" to add these up and obtain this location. Why is the limit of the partial sums the "right" way to do this?

Because there's also another way to do it - by algebra. Let x be point at which Achilles catches up to the tortoise. Then, at that point, Achilles will have run a distance of x, and the tortoise will have run a distance of (x-1). Since Achilles runs 10 times faster, we know that x must be 10 times (x-1). So, x = 10(x-1). Solving, we get x = 10/9. And what is the limit of the partial sums of 1 + 1/10 + 1/100 + 1/1000 + ...? It also turns out to be 10/9. This helps give us some confidence that the limit of the partial sums is a "good" definition of "infinite sum", because it agrees with other ways of solving the same problem.

Also just to clarify, don't get hung up on the phrase "infinitesimals are not real numbers" by interpreting "real" to mean "fake" or "non-existent". To reiterate, "real number" has a specific meaning in math, roughly, the set of numbers that are expressible as the limits of fractions, or as finite or infinite decimal expansions, but not including any infinitesimals. So "real" is just an (unfortunate) name for these numbers, and not a comment on the validity or correctness of numbers that are not "real". As I mentioned in my last post, you are free to define whatever numbers you like, so long as you specify exactly how they behave.

AWA
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### Re: A very interesting Mathematical Paradox

Qaanol wrote:
AWA wrote:This part is, I think, what bugs me so much. If an infinitesimal is defined as [imath]\frac{1}{n}[/imath] where n=infinity, then you're arguing that there's a case where [imath]\frac{1}{n}=0[/imath]. I can't reconcile this.

In fact it is quite easy to reconcile. If an infinitesimal is defined as you describe, it is not a real number, because real numbers satisfy the Archimedean property. In effect the Archimedean property says, “All real numbers are finite—there are no infinite real numbers and there are no infinitesimal real numbers.”

Why this matters:
Spoiler:
For comparison, suppose someone says, “The square of any real numbers is positive or zero.” Then I look at them and say, “Well if a number is defined as [imath]\sqrt{n}[/imath] where n=-1 then you’re arguing there’s a case where -1 is positive or zero.”

In fact no, they are arguing no such thing. The number I defined is imaginary, [imath]i = \sqrt{-1}[/imath]. It is not a real number. It is still an important and useful number, but it is not a real number. Similarly, your infinitesimal is not a real number (and it is also not well-defined; but there are ways of properly defining infinitesimals: they are still not real numbers though.)

If you don’t like not having infinitesimals, consider the integers. Are there any infinitesimals in the integers? Of course not. Then consider are there any infinitesimals in the rationals? That might take a bit longer to answer, but it should be clear that there are no infinitesimals in the rationals. So we already have two infinite sets of numbers that have no infinitesimals.

Probably the most common way of defining the real numbers is (and you’re not going to like this) as the formal limits of Cauchy sequences of rational numbers. I’m not going to get into what a Cauchy sequence is (although it’s a pretty easy-to-grasp concept), but my point is that the reals are defined as limits of sequences. When viewed in that light, although it may not be completely obvious, it does follow directly that there are no infinitesimals in the reals.

Some properties of real numbers:
Qaanol wrote:1) Archimedean: for any positive real number r, there is a positive integer n such that [imath]0<\frac{1}{n}<r[/imath].
2) Trichotomy: a real number is zero if and only if it is not positive and it is not negative.
3) The absolute value of a real number cannot be negative. That is, [imath]|a|\geq 0[/imath].
4) Two real numbers are equal if and only if their difference is zero. That is, [imath]x=y\iff|x-y|=0[/imath].

It is important to note that the real numbers are closed under addition, subtraction, multiplication, and non-zero division. In particular, for any two real numbers x and y, the difference z = x-y is also a real number.

You have mentioned that you don’t like presupposing that 0.999… = 1, so I will not do that. In fact, I will do exactly the opposite. Let us suppose that 0.999… does not equal 1, and see what happens.

Well, 0.999… and 1 are both real numbers, so their difference is a real number. Let’s call it d = |0.999… - 1|. We are acting under the assumption that [imath]0.999\ldots\neq 1[/imath] so by the fourth listed property, d cannot be zero. Then by the third listed property, d also cannot be negative since it is an absolute value. So by trichotomy, d must be positive.

Now by the Archimedean property there is a positive integer whose reciprocal is greater than zero and less than d. Let us pick one such integer, call it m, and then take the next higher power of 10, which we will call N. Now, since [imath]0<\frac{1}{m}<d[/imath] and N > m, it follows that [imath]0<\frac{1}{N}<\frac{1}{m}<d[/imath]. Additionally, because N is a power of 10, say [imath]N = 10^c[/imath] for some positive integer c, the decimal expansion of [imath]\frac{1}{N}[/imath] is just 0.000(total of c-1 zeros after the decimal point)1.

Recall that 0.999… is defined as the infinite sum
$0.999\ldots = \sum_{i=1}^{\infty}9\left(\frac{1}{10}\right)^i$
Note that each term in the sum is positive. We assume 0.999… does not equal 1, so we have either 0.999… = 1 - d or else 0.999… = 1 + d. I am not going to treat the case where 0.999… > 1. I trust you are capable of reasoning through why that is unacceptable.

So we have 0.999… = 1 - d. It follows from [imath]0<\frac{1}{N}<d[/imath] that the number [imath]k = 1-\frac{1}{N}[/imath] is in between 0.999… and 1, and not equal to either. That is, [imath]0.999… < k < 1[/imath]. But we know what k is. In fact we can even say what the decimal expansion of k is, since [imath]k = 1-\frac{1}{N}[/imath].

With [imath]\frac{1}{N}[/imath] = 0.000(total of c-1 zeros)1 it follows that k = 0.999(total of c nines)9. But that is the same as the sum of the first c terms in the definition of 0.999…, and there are more terms to go beyond that. Specifically, adding one more term gives one more 9, which makes a number bigger than k, call it L. So k < L.

Since we only add positive terms, the entirety of 0.999… cannot be smaller than L, since L is the sum of only finitely many terms in the infinite sum (c+1 terms in fact). Thus [imath]L\leq 0.999\ldots[/imath].

Remember that we constructed k in a way so [imath]0.999\ldots < k[/imath], but now we have [imath]k < L \leq 0.999\ldots[/imath]. By transitivity, this gives [imath]0.999\ldots < k < L \leq 0.999\ldots[/imath], or simply [imath]0.999\ldots < 0.999\ldots[/imath]. But a number cannot be less than itself. The statement “[imath]0.999\ldots <0.999\ldots[/imath]” is false.

We began with a set of premises, reasoned logically, and arrived at a false conclusion. This is called “contradiction” in mathematics, and it means that one or more of the premises must be false. The properties of real numbers I gave are all correct, and you can verify them for yourself if you like. Both 0.999… and 1 are real numbers, as you can verify by showing, for example, that {0.9, 0.99, 0.999, …} is a Cauchy sequence of rational numbers. (Actually, if you do that, you will have shown directly that 0.999… = 1.}

It follows that the false premise was our assumption that 0.999… does not equal 1. Since that is false, its negation must be true. Therefore, 0.999… = 1.

The crux of this argument lay in showing that, if there is a real number in between 0.999… and 1, then we have a contradiction. It could be done more quickly by noting that, between any two distinct real numbers, there exists another real number (in fact there exist infinitely many, and infinitely many of them are rational). I liked doing it explicitly because this way I was able to produce a number with a finite decimal expansion that would have to be in between them.

Got that? If 0.999… were less than 1 then there would exist a real number with a finite decimal expansion, which is strictly greater than 0.999… and strictly less than 1.

Anyway, there’s nothing inherently wrong with infinitesimals, just there are none in the real number system. And since every tail-infinite decimal expansion represents a real number, therefore every such decimal expansion necessarily does not involve infinitesimals.

This, I think, is exactly what I needed: a perfectly logical, well-reasoned proof for why 0.999... must note be less than 1 (and obviously it must not be greater than 1). Thus, 0.999... must equal 1.

Only one question remains, which you began to answer, but I don't quite grasp yet:

In every definition I've read of infinitesimals, they've been described as "explicitly nonzero" (though in the same breath, also smaller than any finite number). I understand that this requires stepping away from real numbers, but can you give a quick (or as quick as possible) explanation of which set of numbers to use, and how to construct an infinitesimal so that it has mathematical significance?

Qaanol
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### Re: A very interesting Mathematical Paradox

AWA wrote:This, I think, is exactly what I needed: a perfectly logical, well-reasoned proof for why 0.999... must note be less than 1 (and obviously it must not be greater than 1). Thus, 0.999... must equal 1.

This feels really good to hear. I’m glad I was able to explain it in a way that makes sense to you.

AWA wrote:In every definition I've read of infinitesimals, they've been described as "explicitly nonzero" (though in the same breath, also smaller than any finite number). I understand that this requires stepping away from real numbers, but can you give a quick (or as quick as possible) explanation of which set of numbers to use, and how to construct an infinitesimal so that it has mathematical significance?

Well, if I were you, I’d study some college-level analysis first, to get a firm grasp of the set of real numbers, how it is constructed, and what properties it has.

But at a basic level, you can start with an infinitesimal unit u which you define to have certain properties. Then you can try making numbers of the form (x + yu) where x and y are real numbers. You can make an ordering on them by saying (a + bu) < (c + du) whenever a < c, or if a = c then whenever b < d. You can add them in the obvious manner. For multiplication you have a number of choices. Perhaps the easiest is to define u as nilpotent, so u*u = 0 even though u itself is nonzero.

The surreal numbers were already mentioned in this thread, so you could try looking at those. They are even more complicated than what I just described though. The reals are a good place to start.
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Blatm
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### Re: A very interesting Mathematical Paradox

Great post Qaanol.

xkcdfan
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### Re: A very interesting Mathematical Paradox

Why can't 0.999... be greater than 1?

Mike_Bson
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### Re: A very interesting Mathematical Paradox

xkcdfan wrote:Why can't 0.999... be greater than 1?

Because it is equal to 1.

mike-l
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### Re: A very interesting Mathematical Paradox

xkcdfan wrote:Why can't 0.999... be greater than 1?

Because it's the limit of terms less than 1.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

RonWessels
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### Re: A very interesting Mathematical Paradox

I think a significant part of the confusion occurs because of the fact that infinity is not a number, it is a concept. The phrase "digit at infinity" has no more meaning than "digit at wind".

As Qaanol quite rightly points out, strictly speaking $\sum_{i=1}^{\infty}$ would be gibberish, except that it is conventionally agreed to be a short form for $\lim_{N\to\infty}\sum_{i=1}^{N}$
If you treat infinity as a number, you get some really bizarre results, like for$f(x) = x$$g(x) = \frac{1}{x}$$h(x) = f(x) * g(x) = \frac{x}{x} = 1$we get [imath]f(\infty) = \infty[/imath], [imath]g(\infty) = 0[/imath], and [imath]h(\infty) = 1[/imath]. So [imath]\infty * 0 = 1[/imath]. As well, by looking at set cardinalities, it is easy to "prove" that [imath]\infty = \infty + 1[/imath] (so 0 = 1) and [imath]\infty = 2 * \infty[/imath] (so 1 = 2). Utter garbage!

Mike_Bson
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### Re: A very interesting Mathematical Paradox

RonWessels wrote:I think a significant part of the confusion occurs because of the fact that infinity is not a number, it is a concept. The phrase "digit at infinity" has no more meaning than "digit at wind".

As Qaanol quite rightly points out, strictly speaking $\sum_{i=1}^{\infty}$ would be gibberish, except that it is conventionally agreed to be a short form for $\lim_{N\to\infty}\sum_{i=1}^{N}$
If you treat infinity as a number, you get some really bizarre results, like for$f(x) = x$$g(x) = \frac{1}{x}$$h(x) = f(x) * g(x) = \frac{x}{x} = 1$we get [imath]f(\infty) = \infty[/imath], [imath]g(\infty) = 0[/imath], and [imath]h(\infty) = 1[/imath]. So [imath]\infty * 0 = 1[/imath]. As well, by looking at set cardinalities, it is easy to "prove" that [imath]\infty = \infty + 1[/imath] (so 0 = 1) and [imath]\infty = 2 * \infty[/imath] (so 1 = 2). Utter garbage!

Yup, which is why I die a little inside when I see people use infinity as a number instead of talking about limits.

xkcdfan
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### Re: A very interesting Mathematical Paradox

Mike_Bson wrote:
xkcdfan wrote:Why can't 0.999... be greater than 1?

Because it is equal to 1.

I want a proof.

Qaanol
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### Re: A very interesting Mathematical Paradox

xkcdfan wrote:I want a proof.

Too bad, you don’t get one. Make your own.
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Mike_Bson
Posts: 252
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### Re: A very interesting Mathematical Paradox

xkcdfan wrote:
Mike_Bson wrote:
xkcdfan wrote:Why can't 0.999... be greater than 1?

Because it is equal to 1.

I want a proof.

http://en.wikipedia.org/wiki/0.999...

RonWessels
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### Re: A very interesting Mathematical Paradox

xkcdfan wrote:I want a proof.

Because in the decimal numbering system, if x > 1, the integer part of the decimal representation of x must be 1 or larger. The integer part of 0.999... is 0. Therefore 0.999... <= 1.

This comes about because of the nature of limits. If, for all i, [imath]a_i < L[/imath], then $\lim_{i\to\infty}{a_i}\le L$So 0.999... is allowed to "reach" 1 in the limiting case, but it cannot exceed 1.

Tnarg
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### Re: A very interesting Mathematical Paradox

xkcdfan wrote:I want a proof.
The so call paradoxs on the first page are 2 proofs.
or a nice simple one:
1 =9/9
= (1/9)+(8/9)
= 0.111... + 0.888...
= 0.999...

t1mm01994
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### Re: A very interesting Mathematical Paradox

Tnarg: There are some 7 proofs in these pages, in total.. Also, you have not properly defined infinite addition, which is one of the major problems being said here.

Qaanol
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### Re: A very interesting Mathematical Paradox

Well, in order to get back in the habit of proving things, since the school year is about to start, I went ahead and wrote up a full proof here:

By definition of the number 0.999… we have,
$0.999\ldots=\sum_{n=1}^{\infty}{9\left(\frac{1}{10}\right)^n}$
By definition of infinite summation we have,
$\sum_{n=1}^{\infty}{9\left(\frac{1}{10}\right)^n}=\lim_{N\rightarrow\infty}\sum_{n=1}^{N}{9\left(\frac{1}{10}\right)^n}$
By definition of limit as N approaches infinity we have,

If there exists a number L so that, for every [imath]\varepsilon>0[/imath] there is a positive integer M such that, for all positive integers [imath]m\geq M[/imath] it is true that,
$\left|L-\sum_{n=1}^{m}{9\left(\frac{1}{10}\right)^n}\right|<\varepsilon$
then L is the only such number and,
$\lim_{N\rightarrow\infty}\sum_{n=1}^{N}{9\left(\frac{1}{10}\right)^n}=L$

Given [imath]\varepsilon>0[/imath] let [imath]M=\max\left(1,\lceil-\log_{10}{\frac{\varepsilon}{10}}\rceil\right)[/imath]. This guarantees [imath]M\geq 1[/imath] so there will always be at least one term in the sum. Let us call the absolute value of the difference between limit and sum, d. Now for [imath]m\geq M[/imath] we have,
$d=\left|L-\sum_{n=1}^{m}{9\left(\frac{1}{10}\right)^n}\right|$
$d=\left|L-\left[9\left(\frac{1}{10}\right)^1+9\left(\frac{1}{10}\right)^2+\cdots+9\left(\frac{1}{10}\right)^m\right]\right|$
$d=\left|L-9\left[\left(\frac{1}{10}\right)^1+\left(\frac{1}{10}\right)^2+\cdots+\left(\frac{1}{10}\right)^m\right]\right|$
For any [imath]x\neq 1[/imath], and in particular for [imath]x=\frac{1}{10}\neq 1[/imath], we have,
$x^1+x^2+\cdots+x^m=\frac{1}{1-x}(1-x)\left(x^1+x^2+\cdots+x^m\right)=\frac{1}{1-x}\left(x^1-x^{m+1}\right)$
Thus we have,
$d=\left|L-9\frac{1}{1-\frac{1}{10}}\left[\left(\frac{1}{10}\right)^1-\left(\frac{1}{10}\right)^{m+1}\right]\right|$
$d=\left|L-9\frac{1}{\frac{9}{10}}\left(\frac{1}{10}-\frac{1}{10^{m+1}}\right)\right|$
$d=\left|L-9\frac{10}{9}\left(\frac{1}{10}-\frac{1}{10^{m+1}}\right)\right|$
$d=\left|L-10\left(\frac{1}{10}-\frac{1}{10^{m+1}}\right)\right|$
$d=\left|L-1+\frac{1}{10^{m}}\right|$
Now we see that to eliminate the constant terms we must have [imath]L = 1[/imath]. Then we have,
$d=\left|1-1+\frac{1}{10^{m}}\right|$
$d=\left|\frac{1}{10^{m}}\right|=\frac{1}{10^{m}}=10^{-m}$
But [imath]m\geq M\geq\lceil-\log_{10}{\frac{\varepsilon}{10}}\rceil\geq\left(-\log_{10}{\frac{\varepsilon}{10}}\right)[/imath] so by the monotonicity of logarithms,
$\large d=10^{-m}\leq 10^{-M} \leq 10^{-\lceil-\log_{10}{\frac{\varepsilon}{10}}\rceil}\leq 10^{-\left(-\log_{10}{\frac{\varepsilon}{10}}\right)}=10^{\log_{10}{\frac{\varepsilon}{10}}}=\frac{\varepsilon}{10}$
Furthermore, [imath]\varepsilon>0[/imath] so we have,
$d\leq\frac{\varepsilon}{10}<\varepsilon$
But we defined d as,
$d=\left|L-\sum_{n=1}^{m}{9\left(\frac{1}{10}\right)^n}\right|$
So with [imath]L=1[/imath] we have,
$\left|1-\sum_{n=1}^{m}{9\left(\frac{1}{10}\right)^n}\right|<\varepsilon$
The existence of [imath]L=1[/imath] such that, for any [imath]\varepsilon>0[/imath] there is an M as defined above guaranteeing the previous equation holds for all positive integers [imath]m\geq M[/imath], proves that,
$\lim_{N\rightarrow\infty}\sum_{n=1}^{N}{9\left(\frac{1}{10}\right)^n}=1$
But by definition,
$0.999\ldots=\sum_{n=1}^{\infty}{9\left(\frac{1}{10}\right)^n}=\lim_{N\rightarrow\infty}\sum_{n=1}^{N}{9\left(\frac{1}{10}\right)^n}$
Therefore, by the transitivity of equality, [imath]0.999\ldots = 1[/imath].
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xkcdfan
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### Re: A very interesting Mathematical Paradox

tl;dr

Mike_Bson
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### Re: A very interesting Mathematical Paradox

xkcdfan wrote:tl;dr

Why even post that? Are you bragging about your laziness and short attention span?

Mewzle
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### Re: A very interesting Mathematical Paradox

I still can't get my head around how this works.

The two examples i've been given are as follows.

a)
x = 0.9R
10x = 9.9R
By subtraction 9X = 9
And so x = 1, and so 1 = 0.9R.

(R for the old fashioned 'Recurring')

But, if one were to multiply 0.9R by 10, the multiplied term would always have one less nine than the non multiplied. One could say 'Oh, but it's infinite, so they're the same' but There are different levels of infinity

Say infinity is defined as the number you get to if you begin counting and continue counting forever. The first term would always have a 'head start' in how many decimal places it is, no matter how far one goes, even for an infinite amount of time.

Same goes for if you substitute in the fraction for 0.9R:

x = 1-1/∞
10x = 10-10/∞
By subtraction
9x = 9-9/∞
and so x =, as we said, 1-1/∞.

We could, instead, use the first one and simply double it.

x = 0.9R.
2x = 1.9R8 (I don't know if there's a proper notation for this. Infinite number of 9s, but the last number is 8.)
By subtraction
x = 0.9R, as we've said.

b)
Using a geometric sequence, 0.9R is defined as 0.9R = 0.9 + 0.9*(1/10) + 0.9*(1/10)^2 ....

Using the equation for a sum of a geometric sequence to infinity, a/1-r = s, we get 0.9/1-0.1 = 1.

BUT! This all depends on how we get the equation. The equation for a geometric sequence normally is s = a(1-r^n)/(1-r). We get the sum to infinity by saying r*n is negligible if n is infinity, and so 1-0 = 1, and so a(1-r^∞) = a. If we don't do this, however, and say n^∞ is simply 1/∞, then we get a*0.9R/1-r.

Using the original digits, we get (0.9*0.9R)/(1-1/10) = 0.9*0.9R/0.9 = 0.9R.

0.9R = 0.9R

I may be wrong, but my limited maths doesn't let me understand the other examples set forward. I'd adore an explanation of said methods!

phlip
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### Re: A very interesting Mathematical Paradox

Mewzle wrote:One could say 'Oh, but it's infinite, so they're the same' but There are different levels of infinity

There are different levels of infinity. However, infinity and infinity+1 is the same level of infinity. That's actually one of the definitions of infinity, when it comes to sets... a set is infinite iff it's possible to remove elements from the set without changing the size. Which I guess is saying that infinity-1 = infinity, but still.

However, the proof you've quoted is not rigorous. It's supposed to just be an intuitive thing. Of course, when the result goes counter to intuition, that means it doesn't actually convince anyone. So I don't know why people keep bringing it up.

What's more important is pointing out that things like:
Mewzie wrote:2x = 1.9R8 (I don't know if there's a proper notation for this. Infinite number of 9s, but the last number is 8.)
are completely meaningless... "keep having 9's forever, and they never end. There is absolutely no end to the 9's. Then, after the 9's end, which they don't, add an 8." No, that's silly.

The way decimal expansions are defined is as a sequence of digits. Now, the way sequences are defined is that there's one position in the sequence for each natural number. So think about the natural numbers. There are infinitely many of them, and they go on forever... but each individual one of them is finite. There is no "natural number at infinity" or "natural number after infinity". There's no such thing as "count off all the natural numbers forever, and then take the number after that", because if you count off all the natural numbers forever, that's all of them.

You can picture a sequence as a mapping between the natural numbers and the things it's a sequence of... So, for instance, the decimal expansion for 1/4 is 0.25, but you could also think of the function:$f(n) = \begin{cases} 2 & n = 1 \\ 5 & n = 2 \\ 0 & n > 2 \end{cases}$That function is enough to figure out what digit should go in each place, and is enough to figure out what the number is. The decimal expansion for, say, 1/3 is even easier this way:$f(n) = 3$

Note that we're no longer talking about anything "going on forever", which is a misleading phrase, because it sounds like things are changing, that there's a process involved. Which they're isn't... we just have a function. If you come back later and look at it again, it'll still be the same function. But with the "going on forever", it feels like maybe it's 20 digits long now, but if I leave for a bit and come back, maybe it'll be up to 50 digits. Or something like that. But no - the entire sequence is fixed, all the way to infinity, right from the start.

So consider what your "0.9R8" would mean, when we're looking at the decimal expansions like this.$f(n) = \begin{cases} 9 & n < \infty \\ 8 & n = \infty \end{cases}$Or something similar. But this is a function on the natural numbers... and every natural number is less than infinity. So this is the same as$f(n) = 9$Remember that there is a digit position for each natural number... so for there to be a digit position "at infinity" or "after infinity", there would need to be a natural number "at infinity" or "after infinity", and there isn't. So your "0.9R8" is meaningless, and is the same as "0.9R".

As for the "there's always one more digit after the decimal place in x than in 10x" and "it has a head-start" objections... I'll refer you back to the fact that there's no changing or process involved, the numbers are just numbers. I'll also point you at Hilbert's Grand Hotel, to say that if you take one of the infinitely-many 9's out, and shift all the others back a space, then there are still exactly as many 9's as there was before.

The main problem with this, and your other objection, is that the real numbers, almost by definition, have the Archimedean property, which is to say that they have no infinitesimals. One way to define this is that every number greater than 0 is also greater than 1/n, for some integer n. If two numbers are "infinitely close" together, so that their difference is less than 1/n, for every integer n, then their difference is exactly 0, and the numbers are equal. The epsilon/delta definition of a limit derives pretty heavily from this fact. Or, alternatively, this fact derives directly from Cauchy sequences, depending on which (equivalent) definition you use for the real numbers.

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Mewzle
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Joined: Sat Oct 30, 2010 5:03 pm UTC

### Re: A very interesting Mathematical Paradox

Brilliant, thanks ever so much!

snowyowl
Posts: 464
Joined: Tue Jun 23, 2009 7:36 pm UTC

### Re: A very interesting Mathematical Paradox

phlip wrote:Keep having 9's forever, and they never end. There is absolutely no end to the 9's. Then, after the 9's end, which they don't, add an 8.

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Matthew21210
Posts: 5
Joined: Sun Aug 09, 2009 7:38 pm UTC

### Re:

wisnij wrote:
Marrow wrote:Also you should need the same number of digits used. if the 9's continue infinitely you need to use the same number of infinite 9's for x and 10x
So even if you cut x off at a predetermined number of decimals you are actually going to end up with 8.9991. If you used went to the ten thousanths place with x = .9999 then 10x still = 9.999 but if you say that 10x is 9.9999 and x is .9999 you can't get 10x.9999 to = 9.9999 because you didn't have that last digit when you did the math. All you are doing is rounding that last 9 up by one which has a chain effect all the way up to the first significant digit. All you are doing is proving the tendancy to round and use the closest real number to a string that will continue indeffinitely.

It doesn't work that way. The string of 9s is infinitely long; there is no last digit. Every digit 9 in x gets matched up with a digit 9 in 10x, and they all cancel just as described initially.

Teaspoon wrote:1 - 0.999... = 0.000...1

That's infinite 0s followed by a 1.

It doesn't work that way, either. You can't follow an infinite sequence of zeroes with a one... the zeroes never end, so there's no place for the one to go. More generally, there are no infinitesimal numbers other than 0 in the standard real numbers.