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.
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phlip wrote:So... you're trying to "construct" a number by looking at other, irrelevant numbers?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
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.)
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.
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...1^{2}=0.000...000...1 (with the one in the 2ω place!)
0.000...1 * 10 doesn't work though. Hmmm.
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...
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
Gelsamel wrote:All real infinitesimals are zero.
AWA wrote:Gelsamel wrote:All real infinitesimals are zero.
Lies. I refuse to believe that an inherently nonzero, albeit ridiculously small, number equals zero.
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.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
AWA wrote:Gelsamel wrote:All real infinitesimals are zero.
Lies. I refuse to believe that an inherently nonzero, albeit ridiculously small, number equals zero.
AWA wrote:I am challenging why this is the case.
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.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
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.
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.
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.
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.
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.
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.
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.
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].
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.
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:
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
[math]0.999\ldots = \sum_{i=1}^{\infty}9\left(\frac{1}{10}\right)^i[/math]
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.
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.
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?
xkcdfan wrote:Why can't 0.999... be greater than 1?
xkcdfan wrote:Why can't 0.999... be greater than 1?
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
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 [math]\sum_{i=1}^{\infty}[/math] would be gibberish, except that it is conventionally agreed to be a short form for [math]\lim_{N\to\infty}\sum_{i=1}^{N}[/math]
If you treat infinity as a number, you get some really bizarre results, like for[math]f(x) = x[/math][math]g(x) = \frac{1}{x}[/math][math]h(x) = f(x) * g(x) = \frac{x}{x} = 1[/math]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 wrote:xkcdfan wrote:Why can't 0.999... be greater than 1?
Because it is equal to 1.
xkcdfan wrote:I want a proof.
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.
xkcdfan wrote:I want a proof.
The so call paradoxs on the first page are 2 proofs.xkcdfan wrote:I want a proof.
xkcdfan wrote:tl;dr
Mewzle wrote:One could say 'Oh, but it's infinite, so they're the same' but There are different levels of infinity
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.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.)
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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.
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.
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