Nate T. Posted February 24, 2005 Report Share Posted February 24, 2005 Hal, Ocam's Razor cuts both ways. You can object to the definition of the limit as being overcomplicated and that of infinitesimals being much simpler, but it also postulates the existence of a huge number of entities to create the "superreal numbers," the vast majority of which you throw out of your final answer anyway when you take the "Standard Part" of your answer at the end, effectively ignoring the infinitesimals in your answer. Think of the philosophical conundrums that'll come up when students ask what these infinitesimals are, and you cannot answer them because their existence relies on Zorn's Lemma and other such high-level construction techniques. I'd rather teach my students some first order logic than tell them how to construct an abstract field extension with the Axiom of Choice. Not to mention, for those that go on in math, Analysis uses the idea behind limits to construct the real numbers, so it's helpful to have a background in limits. The seminal idea in topology is also based upon this definition. I'll stick with the epsilon-delta definition because I believe it integrates better into the hierarchy of knowledge. With the aid of limits, infinity is defined in terms of finite quantities and first order logic, which everyone understands. Postulating the existence of elements of an enlarged real line with certain preconstructed properties which are then "interpreted" as infinitesimals strikes me as formalistic, regardless of how intuitive infinitesimals might be. To me, it's the equivalent of saying: "Well, trying to construct calculus based on solely the real numbers is too much work. That infinitesimal idea almost worked, except for that small detail that there are no least positive numbers. But I want there to be a least positive number! Why not just say there's a least positive number and see what happens?" Finally, regarding the "intuitiveness" of infinitesimals, I don't regard the concept of an infinitesimal as being coherent. I regard it as a hand-wave, as a first attempt at getting a real idea of what the concept of infinity ought to be if we're going to base it our previous understanding of numbers. Quote Link to comment Share on other sites More sharing options...

Hal Posted February 24, 2005 Report Share Posted February 24, 2005 (edited) Hal, Ocam's Razor cuts both ways. You can object to the definition of the limit as being overcomplicated and that of infinitesimals being much simpler, but it also postulates the existence of a huge number of entities to create the "superreal numbers," the vast majority of which you throw out of your final answer anyway when you take the "Standard Part" of your answer at the end, effectively ignoring the infinitesimals in your answer. I wouldnt that say it postulates their existence at all, unless you're a Platonist of some sort - it's a formal mathematical construct (a "concept of method" if you like). Think of the philosophical conundrums that'll come up when students ask what these infinitesimals are, and you cannot answer them because their existence relies on Zorn's Lemma and other such high-level construction techniques. I'd rather teach my students some first order logic than tell them how to construct an abstract field extension with the Axiom of Choice.The axiom of choice is equivalent to Zorn's Lemma, and has the same philosophical problems. In any case, Zorn's Lemma is hardly difficult to understand. Not to mention, for those that go on in math, Analysis uses the idea behind limits to construct the real numbersI was taught the Dedekind cut construction, which doesnt involve limits. To me, it's the equivalent of saying: "Well, trying to construct calculus based on solely the real numbers is too much work. That infinitesimal idea almost worked, except for that small detail that there are no least positive numbers. But I want there to be a least positive number! Why not just say there's a least positive number and see what happens?"It's not about it being too much work - the 2 constructions are equivalent so all that really matters is the praticality issue, ie which one is easier to understand. That infinitesimal idea almost worked, except for that small detail that there are no least positive numbers. But I want there to be a least positive number! Why not just say there's a least positive number and see what happens?"How is this different from the following: That complex number idea almost worked, except for that small detail that there is no square root of -1. But I want there to be a square root of -1. Why not just say there's square root of -1 and see what happens? Edited February 24, 2005 by Hal Quote Link to comment Share on other sites More sharing options...

Nate T. Posted February 24, 2005 Report Share Posted February 24, 2005 Bryan, What is the the largest number in the sequence of natural numbers?There isn't one. If you have a natural number N, you can always add one to it to get a bigger natural number. I was talking about the defintion of a real number. So was I. That is the definition of a real number. I'm saying that the sum of i = 1 to i = [insert the largest number you can think of here]of 9/(10^i) is less than one. You're right about that, but that's beside the point. You're claiming that because every partial sum 9/10 + 9/100 + ... + 9/10^N is less than one, the number 0.999~ must also be less than one, which is a non sequitur. This is where we have a disgreement. I'm talking about a literal sum, you are talking about a limit.The problem with this is that you can't 'literally' add up an infinite list of numbers. If you try, you will never be done. What you can do, however, is determine what that sum is getting very close to. That value, which is defined to be the sum of all of the numbers in the last, is the limit of the sequence of those sums. 1/infinity is infinitely small. Once again, what is this "infinity", and why are you dividing by it like it were a number? Even supposing that "1/infinity" makes sense, you haven't answered what it means for a quanitity to be infinitely small, you've only given an example. The final answer is always an actual number, and that actual number is never 1.Okay, let's distinguish between two things. The final answer of each of the partial sums 9/10 9/10 + 9/100 9/10 + 9/100 + 9/1000 ... and the limit, which is the value to which these final answers tend. If you mean that each of the partial sums is less than one, then that's right. But 0.999~ isn't any of these partial sums. It's their limit. You can't actually obtain an actual number doing an infinite number of interations, you can describe the final answer after an infinite number of interations is 0.999~. When in the process of calculating these interations, would the answer every jump up to 1? You don't need to literally sit down forever to calculate 9/10 + 9/100 + ... in order to show it equal one. All you have to do (and this is the epsilon-delta definition of the limit) is show that no matter how close to 1 you specify, you can add up enough terms of the series to get even closer. That's what it means for that limit to be equal to one. Your proof was valid in showing that there is always a number between two numbers. Using the basis of that proof, making x a variable, and repeating it recursivly an infinite number of times, the final x you get is the largest number less than 1. You can't actually get a final x unless you put an artificial cap on infinity.I'm not sure you understand what my proof was claiming, so I'll do it in more detail. PROPOSITION. There is no greatest real number less than one. Proof. Suppose that there is a greatest real number less than one. Call it 'a'. Consider the number b = (1 + a)/2. First it is true that a < b. Second, it is true that b < 1. But a is defined to be the greatest number less than one. Since b is less than one, it follows that a is greater than it. Therefore, a is strictly less than itself, which is a contradiction. Hence no such number a exists. Tell me one more time what is means, I'll tell you if I agree. 0.999~ is the limit of the sequence (0.9, 0.99, 0.999, ...). BTW, what is your mathematical background? I'm a senior math major. Quote Link to comment Share on other sites More sharing options...

Nate T. Posted February 24, 2005 Report Share Posted February 24, 2005 Hal, I wouldn't that say it postulates their existence at all, unless you're a Platonist of some sort - it's a formal mathematical construct (a "concept of method" if you like).Yuck! No Platonist, me. I should have used the term "constructed." The axiom of choice is equivalent to Zorn's Lemma, and has the same philosophical problems. In any case, Zorn's Lemma is hardly difficult to understand. Hmm, that's another good point. I don't quite know what to think of the Axiom of Choice. But from a pedagogical standpoint, Zorn's Lemma takes a lot of terminology. "Every nonempty partially ordered set having the property that every chain is bounded from above has a maximal member." You'd have to teach them all of those terms, and never use any of them ever again in the calculus course. Not exactly tough for a college course, but I can see the high school kids' eyes glazing over right now. I was taught the Dedekind cut construction, which doesn't involve limits.They're equivalent constructions of the real line. The Dedekind cut method is awkward when it comes to multiplication, so it's generally liked less than the Cauchy Sequence construction. Although it does depend on who you ask. It's not about it being too much work - the 2 constructions are equivalent so all that really matters is the praticality issue, ie which one is easier to understand. Well, don't get me wrong-- infinitesimals are a lot easier to understand from the get-go. And I'm certainly not attacking non-standard analysis as being logically flawed or something. I just think that, (1) pedagogically, the introduction of infinitesimals will raise more questions than it answers. How is this different from the following: That complex number idea almost worked, except for that small detail that there is no square root of -1. But I want there to be a square root of -1. Why not just say there's square root of -1 and see what happens? ... hmm, that's actually a pretty good point. Except the motivation between the construction of the complex numbers and infinitesimal numbers are totally different. No one wanted to recognize the validity of imaginary numbers. Even Cardano, the man who first took them seriously, called them "false roots" and dismissed them as nonsensical, until he realised that he needed them in order to get real answers. Infinitesimals, on the other hand, are backward from this. People wanted a structure that could do something for them, and they just made one up for the sake of introducing a new kind of number into the number system (albeit with some pedagogical justification). Complex numbers were the result of closing the algebraic structure of the reals under root extraction. They also simply and unify many other branches of mathematics, such as the Fundamental Theorem of Algebra, which states that any polynomial of degree n has exactly n complex roots, and integral evaluation in the complex plane using analytic continuation. Such theorems are not possible without complex numbers. They have an elegant geometric strucure in the plane, and have been used to prove construction theorems in Euclidean Geometry. Considering infinitesimal roots of polynomials and solving infinitesimal equations can be done, but isn't really fruitful. Constructing this grand systems of infinitesimals to make the calculus work is fine, but then all evidence of the method used is wiped away in our final answer as irrelevant. Infinitesimals don't apply in any other branch of mathematics that I know of-- their sole purpose is to make taking limits conceptually easier. They don't have any beautiful geometric interpretation-- in fact, as far as I know, they behave just like formal variables. I guess it comes down to not constructing anything more than what you need to get the job done. Quote Link to comment Share on other sites More sharing options...

Bryan Posted February 24, 2005 Report Share Posted February 24, 2005 You're right about that, but that's beside the point. You're claiming that because every partial sum 9/10 + 9/100 + ... + 9/10^N is less than one, the number 0.999~ must also be less than one, which is a non sequitur. How does it not follow? Is not 0.999~ the result of the partial sum if N = infinity? The problem with this is that you can't 'literally' add up an infinite list of numbers. If you try, you will never be done. What you can do, however, is determine what that sum is getting very close to. That value, which is defined to be the sum of all of the numbers in the last, is the limit of the sequence of those sums. I would say that the limit could be defined as an approximation of the sum, but not the actual value of them. The two values are similar but not exactly the same. Once again, what is this "infinity", and why are you dividing by it like it were a number? Even supposing that "1/infinity" makes sense, you haven't answered what it means for a quanitity to be infinitely small, you've only given an example. Why do you accept the concept of infinity when it supports your argument and disregard it when it does not? Literally 1/infinity doesn't make sense, just take it to be the smallest conceivable number. You don't need to literally sit down forever to calculate 9/10 + 9/100 + ... in order to show it equal one. All you have to do (and this is the epsilon-delta definition of the limit) is show that no matter how close to 1 you specify, you can add up enough terms of the series to get even closer. That's what it means for that limit to be equal to one.Again, the limit is equal to one, but that doesn't make the actual value equal to 1. 0.999~ is the limit of the sequence (0.9, 0.99, 0.999, ...). I disagree. 1 is the limit of that sequence, 0.999~ is the value of its sum. Quote Link to comment Share on other sites More sharing options...

Nate T. Posted February 24, 2005 Report Share Posted February 24, 2005 Bryan, How does it not follow? Is not 0.999~ the result of the partial sum if N = infinity?I should explain. When mathematicians write a sum from i = 1 to i = infinity, that expression is shorthand for the limit as N goes to infinity of the sum i = 1 to i = N. This is nothing new, it's what I've been saying this whole time, that the sum of a series is the limit of the sequence of partial sums. I would say that the limit could be defined as an approximation of the sum, but not the actual value of them. The two values are similar but not exactly the same. Why the distinction between an approximation of an actual sum you can never calculate and the limit of the sums? What possible use could there be in distinguishing between them? Why do you accept the concept of infinity when it supports your argument and disregard it when it does not? Literally 1/infinity doesn't make sense, just take it to be the smallest conceivable number.First, it isn't the smallest conceivable number, not if you want to usual operations of addition and multiplication to apply to it. Just take half of "1/infinity." There, something smaller. Second, we aren't using the same concept of infinity. I can say things like lim(x to infinity) f(x) = M, because I know it really means: for all e > 0, there exists an N > 0 such that |x| >= N implies |f(x) - M| < e. When you use it, you can't fall back on the epsilon-delta definition of the limit. You're really injecting infinity into the number line as an actual number, not as the property of a sequence. There are systems in which that is valid, but I prefer to keep my number line more tied to the perceptual level. Just make sure to mention before you say that 0.999~ < 1 that you're a non-standard analyst. It'll be great for parties. Again, the limit is equal to one, but that doesn't make the actual value equal to 1. I disagree. 1 is the limit of that sequence, 0.999~ is the value of its sum. You know what-- okay. That's fine. I'll leave it to you to figure out what 0.999~ means. Quote Link to comment Share on other sites More sharing options...

Lex Posted November 16, 2005 Report Share Posted November 16, 2005 Assumption: (0.9999~) + (0.000~1) = 1 Nate’s statement: (0.9999~) = 1 Concusion: 1 + 0.000~1 = 1 Further conclusion: 1 + (0.000~1 + 0.000~1 + 0.000~1 + …) = 1 So you can add infinite sets of 0.000~1 to 1 and the sum will always be 1? Not 1.000~1, 1.000~2, 1.000~3, etc.? Quote Link to comment Share on other sites More sharing options...

Nate T. Posted November 16, 2005 Report Share Posted November 16, 2005 Assumption: (0.9999~) + (0.000~1) = 1 ... Further conclusion: 1 + (0.000~1 + 0.000~1 + 0.000~1 + …) = 1 So you can add infinite sets of 0.000~1 to 1 and the sum will always be 1? Not 1.000~1, 1.000~2, 1.000~3, etc.? These calculations (esp. the symbolism 0.000~1) make the tacit assumption that there is a "last" decimal place in a decimal expansion. That isn't the case using the standard expression of real numbers by decimal expansion. A decimal expansion 0.abcde... is merely a shorthand way of saying that the real number is equal to the series a/10 + b/100 + c/1000 + d/10000 + ..., which can either be left as it is or shown to be something simpler via calculations involving limits (as is the case for 0.999~). Therefore, each digit in the decimal expansion corresponds to exactly one natural number. Hence, assuming that there is a "last" decimal place is tantamount to assuming that there is a "last" natural number, which there isn't (at least, for the usual natural numbers learned of in grade school). Formally, what you're doing here is defining an infinitesimal (intuitively, an ideal number less than every positive number and greater than zero) using the symbol 0.000~1 to denote the unit infinitesimal and extending it to an algebraic field. Then, you're defining .999~ = 1 - 0.000~1. This is essentially non-standard analysis, which I've already said some things about in this thread. From a limit standpoint, the only way you could make sense of the notation 0.000~1 is to define it to be the limit of the sequence (10^-N) as N approaches infinity, which is zero. Similarly, 0.000~n will also be zero for any n you choose since (n 10^-N) also goes to zero as N gets large for any n, and so there are no problems with your equations above. Quote Link to comment Share on other sites More sharing options...

hunterrose Posted November 16, 2005 Report Share Posted November 16, 2005 why is it that 1/3=.333333333? I'm assuming the explanation is similar to the ones that some people gave in the other thread Exactly the same. E.g. A property of real numbers is that there is always a real number in between two real numbers. 1/3 is a real number, and if we assume 0.333... is a real number, then there must be a real number in between 0.333... and 1/3. But there can't be a real number in between 1/3 and 0.333..., right? Right? If 0.333... doesn't equal 1/3, then it really, really, really screws up math at the foundational level. Of course, Lauric can and probably will put it much better Quote Link to comment Share on other sites More sharing options...

Eternal Posted November 16, 2005 Report Share Posted November 16, 2005 At first I thought that the following statement is false: 0.999~ = 1 But after going through the posts, I now see that it is true, not only in the limit sense 0.999~ + d = 1 d = 1 - 0.999~ in order for the original statement to be true, we must have: d = 0 but instead we can see that d will be 0.00000000~1, but we will NEVER reach that 1 at the very end, so d is indeed 0 Quote Link to comment Share on other sites More sharing options...

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