# Talk:Infinitesimal

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## What is an infinitesimal?

"When we consider numbers, the naive definition is clearly flawed: an infinitesimal is a number whose modulus is less than any non zero positive number. Considering positive numbers, the only way for a number to be less than all numbers would be to be the least positive number. If h is such a number, then what is h/2? Or if h is undividable, is it still a number?"

Then in a section called 'A definition':

"An infinitesimal number is a nonstandard number whose modulus is less than any nonzero positive standard number".

It is true that such a definition is absolute nonsense. Not only this, but there is no evidence that an infinitesimal exists. To talk about the plural is ludicrous. Isaac Newton was groping in darkness when he coined this term. He was himself uncertain how to explain the calculation of a gradient or average 'at a point'. Furthermore the article states Archimedes used infinitesimals but till this day there is no coherent definition of an 'infinitesimal' and non-standard analysis is at most wishy. How could Archimedes have used infinitesimals if they do not exist and he had no idea what these are?

However, what surprises me is that Wikipedia allows this to be published. Another article on nonstandard numbers is also dreamy. 70.120.182.243 17:55, 30 April 2007 (UTC)

You are making an ancient mistake. The only numbers we "see" are the natural numbers...starting not at zero but at one. The Romans didn't believe in zero. Europeans of the Middle Ages had no clear idea of negative numbers, perhaps because loaning money for interest was forbidden and most people lived in a cashless agrarian society. The Romans and the mediaeval Europeans would have made your argument: that the number zero or negative numbers were unreal.
The question is, are you for real? You see, there probably was a time when primitive man had no concept of number whatsoever, perhaps in a primitive communist society. Then, there may have been a long time in which people only counted up to a small integer to keep track of sheep and wives.
We can imagine a primitive guy proposing bigger integers such as "one hundred sheep" to the other guys. We can imagine a guy like you saying to the primitive Einstein, "what, you stupid, or what"?
Math progresses when we use our imagination to extend our concepts; oftentimes, a use for the extension is only found after the extension: for example, complex numbers were introduced to physics after they were invented in math. Guys like you retard progress when you claim that the mathematician should not be allowed to ask what would happen if we extended a number system and harass real mathematicians. —Preceding unsigned comment added by 202.82.33.202 (talk) 05:55, 26 April 2008 (UTC)
Let's just say that an infinitesimal is a value in a set of decreasing values whose quantity is insufficient to be named by a defined numbering system. And with the caveat that it is more than zero. And since mathematics is about the relative quantitative measure of things, and since we have been unable to infinitely subdivide physical, and other entities, we are left with quantity levels below the capability of our mathematical system to numerically evaluate.WFPM (talk) 21:53, 12 March 2012 (UTC)
Think of infinitesimals as a way of formalizing the heuristic ideas about "very small" things that physicists have. Tkuvho (talk) 18:12, 13 March 2012 (UTC)
But mathematics gets in a bind when it gets involved in physical problems like perpetual motion, where the question becomes not only about the value of dissipating factors, but the possibility that the mathematics is not correctly modeling the function of those factors in the particular situation. So if mathematicians lose sight of the physical or chemical processes involved in a certain situation the analysis can go astray related to solving the problem.WFPM (talk) 23:25, 13 March 2012 (UTC) And most of us former slide rule people are not expecting an exact solution to hardly any physical problem, but just a way to keep moving in the right direction.WFPM (talk) 03:29, 14 March 2012 (UTC)
How does your interesting comment relate to infinitesimals? Tkuvho (talk) 12:02, 14 March 2012 (UTC)
It relates to the mathematics of physical problems where infinitesimals are involved, like the perpetual possible motions of a pendulum, where we have to manage the contained kinetic energy of the process by thoroughly understanding the physical processes involved, and not letting a mathematical model control our thoughts about what is happening. And science is full of situations where multiple factors are involved in a process and the analysis process is clouded by this or that mathematical statement which is the result of some extended mathematical calculation. I've worked in the reliability calculation area, where not only the importance of factors are evaluated, but also the mathematical calculation of the probability of the correctness of those values are extensively evaluated. And sometimes the mathematics of the situation gets to be more important than the solution to the problem.WFPM (talk) 15:48, 14 March 2012 (UTC)
I couldn't agree with you more. This is precisely why infitesimals are useful: they allow you to focus on the physics rather than bother with the cumbersome technicalities of the epsilon, delta method which is the received alternative. Tkuvho (talk) 15:53, 14 March 2012 (UTC)
I also think that this discussion about infinitesimals should include about how science has overcome the problem of the infinitely small by use of the negative exponent system, which allows us to subdivide any even infinitesimal quantity into a very large quantity of much smaller entities. A lot of people don't understand this, about what you can do with the exponents of a value, and how that has relieved the situation considerably.WFPM (talk) 16:25, 14 March 2012 (UTC)

## My last edit

I think I understand Wikipedia's policy now. It does not matter what "facts" are true or false. As long as there is a publication of such facts, these qualify to be part of an article.

Is my understanding correct? If so, then I think my suggestions can be discarded. I think you should warn your readers that no information on your site can be trusted. It is also clear to me now why most academics warn their students to steer clear of Wikipedia. What I tell my students is to read everything and believe nothing unless it makes sense to them. I shall never suggest anything here again. 12.176.152.194 (talk) 18:32, 18 December 2011 (UTC)

Your first sentence is exactly the point. Spot on. Thenub314 (talk) 22:13, 18 December 2011 (UTC)
Also if you look at the bottom of the page there should be a "Disclaimers" link. I think the words printed largely on the page linked to address part of your concern. Thenub314 (talk) 22:14, 18 December 2011 (UTC)
Perhaps the "Disclaimers" link should appear in large bold font at the top of each page? I for one, never noticed it. 12.176.152.194 (talk) 17:17, 21 December 2011 (UTC)
There is no particular reason to emphasize the disclaimer on this page, as the information here is mostly correct. Tkuvho (talk) 17:22, 21 December 2011 (UTC)
See, it is this kind of attitude that is problematic. No knowledge is ever beyond investigation is all I will say and add that it is your opinion that is it "mostly correct". BTW: I read Robert Ely's research - I am not impressed. He really does not say anything that supports what you claim about students and the concept of infinitesimal. No use debating this also because I am not convinced that all real numbers can be represented in a given radix system. In fact, I am certain that real numbers are ill-defined. 12.176.152.194 (talk) 17:36, 21 December 2011 (UTC)
Interesting. The real numbers are not well-defined. But apparently a notion of derivative where x^3 is not differentiable at 0 is well-defined. Tkuvho (talk) 17:56, 21 December 2011 (UTC)
About real numbers and magnitudes: http://thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/magnitude_and_number.pdf If you use Cauchy's definition which is flawed, then a derivative exists at 0. However, using my sound New Calculus, a derivative does not exist at 0 because it is impossible to construct a finite tangent line to x^3 at 0. 12.176.152.194 (talk) 03:44, 22 December 2011 (UTC)
Since we happen to be at the "infinitesimal" talkpage, I will mention that the reason that the tangent line to y=x^3 at the origin does exist is because you can pick two values of x infinitely close to the origin, and draw the line through the corresponding points on the graph. That line is infinitely close to the x-axis, and that's enough to declare the x-axis to be the tangent line to the graph. Tkuvho (talk) 12:07, 22 December 2011 (UTC)
I don't think so. Your line would cross x^3 so it cannot be a tangent line. Before I exit this discussion, I just want to say that I have seen the original Greek and nowhere is there any mention of infinitesimal or indivisible. Now, I know your English publication states this, but it is incorrect. The modern Greek word is composed of two words "infinite" and "minimum". There was no Ancient Greek word for infinitesimal. The word for indivisible literally means that which cannot be divided into smaller parts. Archimedes most definitely did not use infinitesimals because these don't exist. Did he use indivisibles? Only in the sense of lines, areas and volumes(ala Cavalieri). The method of exhaustion relies on the approach of finding the area (e.g parabola example) of a shape which represents the area (or volume) one wishes to find. The more one can make the shape similar to the desired area, the better the approximation. You could say Archimedes anticipated "limits" [in the sense that a given approximation approaches some incommensurable magnitude (pi) or some rational number (e.g. 1/3). Not at all like the modern definition of limit], but had no idea about indivisibles or infinitesimals. The only objects he knew about were the rational numbers and the incommensurable magnitudes. In the Works of Archimedes, one finds evidence of this in many places.
And out of curiosity, please tell me which two infinitely small values (*) you can pick close to zero? See, "infinitely small" is not well-defined, therefore it is nonsense. Many questions arise... if these infinitely close values are subtracted, then the difference is 0, thus your denominator of the finite difference quotient is also 0 which is undefined. Furthermore, the ordinates corresponding to these abscissas (*) you mention also have a difference close to 0, so that your difference quotient looks like 0/0. Hmm, so what can this mean? Sorry, you should never accept any concept that is ill-defined.

12.176.152.194 (talk) 15:52, 22 December 2011 (UTC)

Archimedes did not use the word "indivisible" (this word was introduced in the middle ages), but he used indivisibles in the sense of Cavalieri all the same. Congratulations upon reading Archimedes in the original, but you might want to brush up on some Archimedes scholarship, as well, for instance the work of Reviel Netz. Tkuvho (talk) 18:50, 22 December 2011 (UTC)
Archimedes could not have used indivisibles because the method of exhaustion does not use any concept of indivisibility. Also note that an infinitesimal (even according to your understanding) is not necessarily the same as an indivisible. For example, indivisible applies to line, area and volume whereas infinitesimal applies to number. Both have entirely different meanings: infinitesimal (vaguely some magnitude close to zero) and indivisible (a line or width of disc). Reviel Netz has not revealed anything new to my knowledge except that he worked on the restoration of the palimpsest? I have not read the entire palimpsest in Greek (only parts of it). I have studied the Works of Archimedes (Thomas Heath) from cover to cover. I can tell you that there are many things in Archimedes' Works that are to this day not well-understood. 12.176.152.194 (talk) 20:41, 22 December 2011 (UTC)
There are two different methods that Archimedes used: (1) method of indivisibles, and (2) method of exhaustion. I certainly agree that there are many things that Archimedes did that you apparently don't understand, particularly his application of the method of exhaustion. Tkuvho (talk) 08:25, 23 December 2011 (UTC)
You are clueless regarding the meaning of indivisible and even more confused regarding infinitesimals but I expect this from having read your comments which are obviously wrong. This article is a joke because the method of exhaustion uses neither of these concepts, yet your article claims it does. There are none so blind as those who will not see. 12.176.152.194 (talk) 16:12, 24 December 2011 (UTC)
The point about seeing is well put. Now draw the cubic y=x^3 and the x-axis, take a deep breath, and let us know what you see. Tkuvho (talk) 11:59, 27 December 2011 (UTC)
I see the x-axis intersecting and crossing the cubic at x=0. Common sense tells me it can't be a tangent. If you can show me a secant with defined gradient that is parallel to the x-axis, then a tangent must exist for the cubic at x=0. There is no such secant - this violates the mean value theorem. Conclusion is that cubic is not differentiable at x=0. The algebra proof is somewhat longer and more complicated. 12.176.152.194 (talk) 02:33, 29 December 2011 (UTC)
If you insist that the tangent line must be a secant line through a pair of points, then you will never find any differentiable functions other than the linear ones. Even for the parabola, you have to discard a remainder term to pass from a secant line to a tangent line. A theory that denies the differentiability of the parabola may be logically consistent but it will not be very useful. Dropping the infinitesimal remainder is not something you can avoid in doing calculus. If you can't beat them, join them :) Tkuvho (talk) 08:29, 29 December 2011 (UTC)
I don't insist anything - the mean value theorem requires that there be a parallel secant for any tangent line. That you will never find any "differentiable functions other than linear ones" is outright false. I have shown that every function that is differentiable in standard calculus is also differentiable in the new calculus. Your statement about discarding a remainder is also false. In fact, every one of your sentences in the previous paragraph is false. I have beaten "them" and before long they will be joining me! 12.176.152.194 (talk) 13:54, 29 December 2011 (UTC)
You have misunderstood the mean value theorem. The mean value theorem does not assert that for every tangent line there is a parallel secant line. Rather, it asserts that for every secant line, there is a parallel tangent line. In fact, the mean value theorem clearly illustrates the problem with your approach: if y=x^3 is not differentiable at the origin as you seem to claim, then one cannot apply the mean value theorem to this polynomial. This is a major inconvenience compared to the usual approach involving dropping the infinitesimal remainder at the end of the calculation of the slope. Tkuvho (talk) 14:03, 29 December 2011 (UTC)
Seems to me the misunderstanding is on your part, not mine. The mean value theorem asserts both statements, that is, for every secant line there is a parallel tangent line and vice-versa. And of course you cannot apply the mean value theorem to x^3 at the origin because it (cubic) is not differentiable at the origin. There is no inconvenience of any sort whatsoever. If you don't like this, then it's your problem but that does not prevent it from being fact. 12.176.152.194 (talk) 14:35, 29 December 2011 (UTC)
Do you have a source for your version of the mean value theorem? All the textbooks I have seen use the version I stated above. Tkuvho (talk) 14:36, 29 December 2011 (UTC)
I don't need a source. I learned the theorem when I was 14 years old. However, your own Wikipedia entry on the mean value theorem says: For any function that is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c. 12.176.152.194 (talk) 05:04, 30 December 2011 (UTC)

────────────────────────────────────────────────────────────────────────────────────────────────────Perhaps you need to study English or mathematical reasoning more thoroughly. That states that for any secant there is a parallel tangent. You have asserted that for any tangent that there is a parallel secant. Not at all the same. — Arthur Rubin (talk) 17:02, 30 December 2011 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── The mean value states that if there is a tangent, there must be parallel secants. So for a given interval, it is exactly the same. You ought to address the previous paragraph to yourself. 12.176.152.194 (talk) 00:40, 2 January 2012 (UTC)

Cite? (Other than your files?) — Arthur Rubin (talk) 14:59, 2 January 2012 (UTC)

Netz is not an authority by any stretch on Archimedes. His main contribution is/was in the restoration effort. I would go as far as saying that very few mathematicians after Heath understand Archimedes' Works. I am one of the few who has studied and understands his works well. 12.176.152.194 (talk) 15:16, 29 December 2011 (UTC)

I would suggest you try again to go over the proof of Proposition 14 of The Method, where Archimedes uses Cavalieri's technique to compute the volume of the solid. Tkuvho (talk) 15:32, 29 December 2011 (UTC)
If you need help understanding Archimedes's works, I can provide some guidance for you. Nowhere in Proposition 14 does Archimedes use Cavalieri's principle. Do you enjoy misrepresenting facts always? I can see you have zero understanding. Reread the proposition and you will notice that it says "divide Qq into any number of equal parts". This is part of Archimedes's integration techniques that rely on natural averages. There are no infinitesimals or indivisibles mentioned anywhere. Let me give you a simple example. Suppose you wish to find the area between any curve and the x-axis. Furthermore, suppose there is no primitive function so you cannot use the fundamental theorem of calculus. So what do you do? Well, you will fall back on one of the numeric integration techniques that were taught to you. However, what you are actually doing in every case, is finding the average of the length of the ordinates in the interval and taking the product of this average with the interval width. This produces the area. You might call it rectangulation. I call it an average sum (my average sum theorem always uses equal parts or partitions whether area or volume). I define area as the product of two averages (average of ordinate lengths and average of horizontal lines which is equal to interval width). Volume as the product of 3 averages, etc. So, Archimedes used an averaging technique much as I do today. Since the area in such cases is *always* an approximation, it makes no sense to think about limits, infinitesimals or the like, because the more equal parts we divide Qq into, the better our approximation. So,once again, you need to study the Works of Archimedes in this light. Archimedes, Newton and all the great mathematicians before Cauchy would agree with me if they were able to know of this discussion. Cauchy really messed up with his definitions. Besides Archimedes, almost every other mathematician has ill-defined one or more concepts. Newton's definition of the derivative is ill-defined. Newton knew this and it was the main reason he did not publish his ideas sooner. He would have loved to know of my New Calculus. One of the characteristics of a great mathematician is the ability to well-define concepts. A mathematician is like an artist. The objects arising from concepts in a mathematician's mind are only as appealing as they are well-defined. In case you missed it earlier... 12.176.152.194 (talk) 18:54, 29 December 2011 (UTC)
I recommend the recent article by Pippinger here. Tkuvho (talk) 10:23, 30 December 2011 (UTC)
The concept of reliability that Wikipedia uses does lead to some anomalies, but using WP:FRINGE material, such as yours, would be even worse. There would be no means of distinguishing your — findings — from that of Archimedes Plutonium, other than (alleged, in both cases) expert analysis. As Wikipedia editors are not expected to be experts, neither can be used until published commentary is written. — Arthur Rubin (talk) 18:39, 30 December 2011 (UTC)
I agree. But, I am not advocating or suggesting my material be published, quite the contrary in fact. The article can be written without the author's opinion injecting his personal understanding (or lack thereof) and bias. "Archimedes used the Method of Exhaustion (a key feature that involves 'equal partitioning' in determination of approximate averages (*) that are used to compute area and volume) in his work, which eventually came to be known as the Method of Indivisibles." The previous sentence is 100% factual. If you have studied Archimedes' works, you will see that he uses the same methods over and over again as described in this sentence. Numerous proposition proofs are started with the same approach - "Let Qq be divided into equal parts,..." This is not my idea or new knowledge by any means. (*) I was the first to notice calculus is about averages, but if one does not like to mention averages, you can replace the word by areas or volumes. 12.176.152.194 (talk) 19:52, 30 December 2011 (UTC)
The method of exhaustion and the method of indivisibles are generally considered by scholars to be two distinct methods. As far as your ideas about calculus, do you also have a singularity at 0 when you integrate 3x^2? If integration goes smoothly even at the origin to produce x^3, then your calculus does not satisfy the fundamental theorem of calculus, which is a steep price to pay for a ban on dropping an infinitesimal remainder at the end of the calculation. Tkuvho (talk) 20:26, 31 December 2011 (UTC)
Perhaps what bothered the AMS is the idea that x^n is not always differentiable. It is puzzling that you blame Cauchy for this. The derivative of x^n, at all points including 0, was already known before Newton and Leibniz. I have to check whether it was Hudde, Wallis, or Barlow, but at any rate this was two centuries before Cauchy. Tkuvho (talk) 08:03, 1 January 2012 (UTC)
I don't think so. No derivative exists at an inflection point. Anyway, I don't believe they even got round to reading it. x^n always has a general derivative but even so, one cannot be certain unless one checks the derivative at a point. For example, the upper half of a circle always has a general derivative but there is no derivative when x=r(or x>r) or x=-r (or x<-r). And of course we have seen some examples with Rubin where an actual numeric derivative does not exist, e.g. f(x)=1/x at the origin. The general derivative of f(x) is g(x)=-1/(x^2) but neither f(0) nor g(0) exist; both are undefined. I 'blame' Cauchy for the definition f'(x) = (lim as h approaches 0) {f(x+h)-f(x)} / h. It is from this flawed definition that all sorts of nonsense and misconceptions have arisen, to wit, the limit and the infinitesimal - neither of which are required and have no place in differential calculus or any other mathematics. The limit is somewhat sound but the infinitesimal is absolute rubbish. I think Newton knew that he didn't know. This is what kept him from publishing anything sooner than he did. Leibniz was trying to be more precise, but he also failed. I realized their ideas are wrong when I was 14 years old. Cauchy was trying to add rigor, but he made things more complicated than they actually were. Suppose the gradient of a parallel secant is given by [f(x+n)-f(x-m)] / (m+n) = k. Now k is some ratio. It follows that f(x+n)-f(x-m) is always equal to k(m+n). Now, when we are calculating a derivative, we form the numerator of the difference ration, that is, f(x+n)-f(x-m) and its denominator is (m+n). Therefore it follows k must be given by k(m+n)/(m+n). It's really that simple. Newton, Leibniz and Cauchy missed this. In fact you and every other academic missed it, until I came along. My divisibility identities although further confirmation of these facts are interesting, but not actually required. Related distance pairs are interesting and very useful but for a general derivative all one cares is about the distance pair (0;0). There is no need to study limits for 6 months or take a course in real analysis. One finds a quotient from the difference ratio and uses (0;0) to find the general derivative at a given point. This is sound and rigorous mathematics. From there on, everything else that follows is crystal clear. 12.176.152.194 (talk) 12:15, 1 January 2012 (UTC)
The problem with your approach is that it creates difficulties in the way of applications in physics. Suppose a car is traveling a distance s(t) as a function of time t according to the law s(t)=t^3. Is there any reason to suppose that its speedometer will not show the value 0 when t=0? Do you know of any physicist that would accept this? Tkuvho (talk) 12:21, 1 January 2012 (UTC)
Well, it's not always possible to match the model to a physical situation. There are frequently questions that need to be asked. If the car was at rest at t=0, then this is a special case and v(t)=3t^2 is a new function valid at every point t>0 or t=0. When any physicist analyzes the gradient or area characteristics for given data in planar representation, you seldom have a one to one correspondence between the model and physical events. The example you have provided is simple. Consider, an optimization problem using differential equations. It's almost never a case of exact correspondence, there may be existence problems, singularity problems, etc. (*) Once again, the only reason physicists may have some difficulty accepting this, is that they are used to flawed mathematics. Old habits are hard to break. A technician who has used an old tool all his life is generally reluctant to use a better tool. Going back to the example, it is illogical to consider speed at s(0) because s(t) itself is not defined for t<0 and makes no sense. In other words, there is no data for t<0 which implies a tangent gradient (v(t)) is not possible in the first place, therefore no gradient. (*) I have found the New Calculus to be more effective in these problems than standard calculus. There is much more that can be done. You have to ask yourself whether it's more efficient to learn a flawed calculus in much longer periods of time without ever fully understanding it, or learning a robust calculus in a couple of weeks which you can understand completely. 12.176.152.194 (talk) 16:05, 1 January 2012 (UTC)
Is your calculus robust? It may suffer from a logical circularity: in order to calculate the derivative, you insist on forming a finite ratio, but you must know in advance what the value of the ratio is supposed to be! And in order to calculate that value, you have to do the usual calculation involving discarding the infinitesimal remainder at the end. Tkuvho (talk) 16:09, 1 January 2012 (UTC)
It's the only robust calculus. What makes you think you have to know anything in advance? I think you are confusing yourself. If you use the (0;0) distance pair, you don't know anything in advance. 12.176.152.194 (talk) 16:22, 1 January 2012 (UTC)
What's a (0,0) distance pair? Tkuvho (talk) 16:25, 1 January 2012 (UTC)
Read the abstract. It explains what you need to know. http://india-men.ning.com/forum/topics/meaning-of-the-differential-quotient?page=1 12.176.152.194 (talk) 16:28, 1 January 2012 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── (ec) That no sense makes. And this still has nothing to do with anything which should be on Wikipedia. As I see it, the "new calculus" definition of the derivative is:

$f'(x) \text{ exists and is equal to } L \text{ if }\bigcap_{\epsilon>0}\left\{\frac {f(x+m)-f(x-n)}{m+n} | 0 < m,n < \epsilon\right\} = \left\{L\right\},$

although, I can't come up with a definition which also requires $f(x)$ to exist. I'm pretty sure that that definition implies that $\lim_{h \to 0}f(x+h)$ exists, and that, if you then define $f(x)=\lim_{h \to 0}f(x+h)$, then $f'(x)=L$ under the usual definition. — Arthur Rubin (talk) 17:33, 1 January 2012 (UTC)

You are missing the point. No limits are required. No epsilonics. It's actually very simple: you find the tangent gradient by finding the gradient of a parallel secant in the same interval. A derivative f'(x) exists in a given interval if f(x) exists; the distance pair (0;0) satisfies the tangent gradient and infinitely many other distance pairs (m;n) exist in the same interval that satisfy the parallel secant gradients. f'(x) = [f(x+n)-f(x-n)]/(m+n) To find a general derivative one uses only the distance pair (0;0). To show that no derivative exists, one must prove that there are no distance pairs other than (0;0) if indeed (0;0) is valid.12.176.152.194 (talk) 19:00, 1 January 2012 (UTC)
In the meantime, John seems to have proposed a different definition which sounds like what Fermat did: take f(x+e)-f(x), expand in powers of e, cancel constant terms, divide by e, and look for the coefficient of e^1. In certain situations, this can be done without any infinitary processes. However, he writes this down with two variables in place of one, which seems an unnecessary complication. It has to be acknowledged that for polynomials the derivative can be calculated by a finite process. Tkuvho (talk) 17:43, 1 January 2012 (UTC)
Not even remotely the same. Fermat was trying different optimization approaches. There is no similarity between any prior mathematician's work and mine. I'll say this: if Newton had known about Euler's function notation, there may have been a chance he could have discovered my approach sooner. The problem at hand was not finding a derivative at a point - Newton showed this was easy using his approximate difference ratio of non-parallel secants. The challenge was finding a method to compute the general derivative of a given function without having to explain away division by zero. I am the first to succeed in this regard with a rigorous calculus that excludes limits. BTW: The derivative can be calculated by a finite process for any differentiable function, not just a polynomial. Every time you stumble on this, remind yourself of the (0;0) distance pair.12.176.152.194 (talk) 19:00, 1 January 2012 (UTC)
That (your reply to my formula) is nonsense. First, I was trying explain to Tkuvko that if your derivative exists, it is equal to the standard derivative (after adjusting for removable singularities, as your definition does not require that the value of the function exist at the relevant point). Second, your definition does require epsilontics; it has almost the same quantifier-complexity as the standard derivative. However, your underlying mathematics is even more restrictive than constructive mathematics; not that I think your method has any value, but you need to define your underlying mathematics and mathematical logic before anyone can determine whether it has value; it's significantly different than anything in the literature, and your "calculus" makes no sense (except in the formulation I gave it above) without modifying the underlying mathematical logic. — Arthur Rubin (talk) 02:08, 2 January 2012 (UTC)
Rubin, do tell what's nonsense about it or shut up I say. Every one of your statements in the previous paragraph are false. You do not understand my new calculus. Please don't pretend that you do. Your previous response is a whole lot of illogical rambling. Then again this is what I expect from someone who claims infinitesimals are too simple a concept for him to explain, yet is unable to provide any evidence of this. "it's significantly different than anything in the literature, and your "calculus" makes no sense" is your opinion. In fact you could do us all a favour and just hold your opinions, okay? 12.176.152.194 (talk) 04:51, 2 January 2012 (UTC)
As for infinitesimals, this article should be adequate to explain them to any mathematically trained person. My R(((ε))) is a subfield of the Levi-Civita field, which has most of the same properties, but is easier to calculate with. — Arthur Rubin (talk) 05:47, 2 January 2012 (UTC)
Nonsense. The Levi-Civita field definition assumes ε is an infinitesimal. It does not define an infinitesimal. The definition is circular. It is also a misnomer in my opinion because it follows from Cauchy's wrong ideas regarding infinitesimals. Like Cauchy, you appear to have missed this circularity in your reasoning (or lack thereof). 12.176.152.194 (talk) 15:36, 2 January 2012 (UTC)
The Levi-Civita field defines ε, and it can be shown it is an infinitesimal. — Arthur Rubin (talk) 15:45, 2 January 2012 (UTC)
That is false. To say that ε is an infinitesimal is not a definition. In order to say that something can be shown to be infinitesimal, you first have to define infinitesimal, that is, you have to know what you are talking about. Of course in your misguided thoughts this did not occur to you, did it? 12.176.152.194 (talk) 15:54, 2 January 2012 (UTC)
For your definition of derivative, it would be helpful if you wrote it out symbolically, as there does seem to be some confusion as to what you mean.
For your definition of function, you will have to explain why the absolute value function isn't well-defined, as even intuitionists seem to accept it. Also, whether the function $x^2 |x|$ has a derivative at 0 in your system. — Arthur Rubin (talk) 06:07, 2 January 2012 (UTC)
The NewCalculusAbstract-Part1.pdf contains a perfect definition in it that uses symbols. Your previous definition is wrong. m and n can only take on the value of (0;0) pair (in addition to infinitely many other pairs before and after reduction) after the difference quotient is reduced in my calculus, but this is illegal in standard calculus even though it works. Cauchy's Kludge explains this. I am not answering any more questions regarding my Calculus on this web page. 12.176.152.194 (talk) 14:13, 2 January 2012 (UTC)

## Tkuvho Responses 3

I thought about your comment regarding circularity and it occurred to me that you are getting confused. So I will try to help you understand this. Gradient = rise/run. Rise = f(x+n)-f(x-m) Run = m+n So, gradient k = f(x+n)-f(x-m)/(m+n) => f(x+n)-f(x-m) = k(m+n). We don't know k but we know both rise and run so we can find k. To convince yourself there is no "infinitesimal remainder", divide both sides of f(x+n)-f(x-m) = k(m+n) by (m+n). On the left you have what you started out with and on the right you have k. For any difference quotient, you work with the left hand side so that after cancellation you will have one term without any m or n in it. This term denotes the gradient when m=n=0, that is, both distances on the side of the tangent at the tangent point are zero. Study diagrams in files to get a better understanding. Now, if you wish to find k for any one of the other secant lines that are parallel to the tangent, then you must know their (m,n) pairs. Finding a relationship between m and n helps. However, k will be the same for all the secant lines that are parallel to the tangent. BTW: The terms in m and n are not remainders! But their sum is always zero because the secants are parallel to the tangent. Each secant has its own (m,n) pair which makes all these terms zero. For example, consider f(x)=x^2. f'(x)=2x+(n-m). This is exactly the derivative. 2x+(n-m)=2x always. If x=1, then all the following are valid gradients: 2(1)+(0-0); 2(1)+(0.005-0.005); 2(1)+(3-3); 2(1)+(m-n) Note that m=n in the case of the parabola. This is not always true for every function. In fact, it's hardly ever true for most other functions. 12.176.152.194 (talk) 17:05, 1 January 2012 (UTC)

You refer to "terms". I assume therefore that you are working with polynomials. The technique you outlined is interesting, but it seems to be what Pierre de Fermat did a few decades before Newton and Leibniz, in developing his method of adequality. If you figured this out on your own that's certainly brilliant. But, believe me, calculus has come a long way since then. In particular, dealing with "terms" can only be done in the context of polynomials. Alternatively, you need power series which would allow you to account for analytic functions. Already in power series you would need to delete the infinitesimal remainder at the end of the calculations. Furthermore, to apply this to functions that are not analytic, you would need the usual differential quotient and the standard part function. Tkuvho (talk) 17:14, 1 January 2012 (UTC)
Not at all the same. It does not matter what you are working with, polynomials or any other function. If the function is smooth and continuous, it will work. No, Fermat had no idea about this. In fact no mathematician before me knew any of it. And although I am trying to encourage you to study it, there is some learning to be done. If your neurons are firing connections with any previous mathematics, you are not getting it. There are no remainders - infinitesimal or otherwise. I hope you will somehow understand this. It seems to me that you have a big stumbling block where this is concerned. There are no infinitesimals. Not in theory and not in reality. 12.176.152.194 (talk) 17:17, 1 January 2012 (UTC)
A curve ball for you - the only objects we know about are the rational numbers and incommensurable magnitudes. Nothing has changed since Archimedes. Most of what you learned in real analysis is either unsound or just plain wrong. Usually the latter is true. 12.176.152.194 (talk) 17:26, 1 January 2012 (UTC)
You will find some here and very well explained, too. Tkuvho (talk) 17:26, 1 January 2012 (UTC)
Has lots of misconceptions and errors. Not much different from anything else like it. Are you Keisler?12.176.152.194 (talk) 17:29, 1 January 2012 (UTC)

## Understanding Cauchy's Kludge

An important step in learning the New Calculus is first realizing where the standard calculus is wrong. You cannot divide by h ever in the standard difference ratio. You can divide by (m+n) always in the New Calculus. Why? Every term of the numerator f(x+n)-f(x-m) contains a factor of (m+n). After cancellation (taking the quotient), exactly one term will be the gradient of the tangent line [distance pair (0;0)]. To find the gradients of all the parallel secants we use the terms in m and n if we want to be "devout". However, there is no need to do this because their gradients are all equal to the tangent gradient. Now we can find distance pairs in (m,n) for other reasons and there are many interesting reasons - especially in the theory of differential equations which I have researched using the new calculus. So, what you have to do is forget everything you learned and interpret what you read literally. It will take a while even if you are extremely smart. I have found that it's much easier to teach someone who has not learned standard calculus. 12.176.152.194 (talk) 19:47, 1 January 2012 (UTC)

## A word of warning

This discussion is not about my New Calculus. Now, although I do not mind whether you mention my New Calculus or not, I will mind if you mention it without proper attribution (my name and web page). I will win any argument in a court of law if it comes to this. Not threatening, just warning. Cauchy's kludge, secant method, distance pairs, etc are also my copyright phrases, not to be mentioned without correct attribution. What I have noticed about academics is that they are cynical until they understand and then they think it's no big deal. Well, it is a big deal because I was the first to think of it. It is also a big deal that I have corrected three great mathematicians: Newton, Leibniz and Cauchy. Although I can't stop you from quoting my work with the correct attribution, I would prefer that you do not quote my work at all. 12.176.152.194 (talk) 19:05, 1 January 2012 (UTC)

I do believe that is a legal threat. You can stop us quoting your work if it's under copyright, except for "fair use", which our discussion trying to find out whether it has any possible validity seems to fall under. If you don't want it discussed, you shouldn't have brought it up. — Arthur Rubin (talk) 02:11, 2 January 2012 (UTC)
I can stop you quoting it if you do not use correct attribution and if you do quote it without correct attribution, I will stop you. Look Rubin, you annoy me intensely. I have repeatedly informed you that the original discussion was regarding infinitesimals. You kept coming back to my New Calculus. Thukvo continued to ask me about it and the dialogue is primarily with Thukvo, not you. I do not agree with most of your views because they are wrong. If you think this is a threat, that's your problem. 12.176.152.194 (talk) 04:57, 2 January 2012 (UTC)

Dear 12.176.152.194, Wikipedia should not be used for self-promotion. Neither in the articles nor in the talk pages. If you have your own version of the Calculus I urge you to get it published in a peer reviewed journal. In any case, Wikipedia is definitely the wrong place for publishing or discussing original research. Please respect that. iNic (talk) 04:39, 2 January 2012 (UTC)

Self-promotion? Have you read any of my comments? I have been trying to contribute to this article. It is full of non-factual statements. Although Thukvo asks me questions regarding my calculus, I keep returning to the main topic. I am prepared to stop right here if Thukvo ceases to ask me questions. I even recommended he contact me via private email if he wishes to continue the discussion. Rubin is an annoying trouble maker with a lot of time on his hands. As for self-promotion, what do you call Rubin's page on Wikipedia? It has nothing notable or remarkable.12.176.152.194 (talk) 05:00, 2 January 2012 (UTC)
The only non-factual statements made are by you, and possibly those attempting to interpret your <redacted> "New Calculus". — Arthur Rubin (talk) 05:38, 2 January 2012 (UTC)
You are the main reason I would prefer no references are made to my work. Your previous <redacted> definition is your wrong interpretation of my definition. 12.176.152.194 (talk) 14:00, 2 January 2012 (UTC)

Agreed. The relevant OR is Cauchy's Kludge. I will not respond to any more questions on my New Calculus or any mathematics not related to this topic. 12.176.152.194 (talk) 14:00, 2 January 2012 (UTC)
"Cauchy's Kludge" (the name, that it is a "kludge", and the alleged error in Cauchy's work) is also your original research. — Arthur Rubin (talk) 15:03, 2 January 2012 (UTC)
And so? I am not disputing this fact. iNic asked me what OR and I responded. What's your problem Rubin? Perhaps a new pair of reading glasses is in order? Still living with your mother? 12.176.152.194 (talk) 15:20, 2 January 2012 (UTC)

If you have good arguments you should not have to resort to personal attacks like this. By the way, it's not allowed here and you can be banned from wikipedia if you continue like this. iNic (talk) 15:34, 2 January 2012 (UTC)

Good arguments I have. Patience for Rubin I do not. Rubin and I go back a long way. There is no love lost between us. I can assure you he probably dislikes me more than I dislike him. What do you say about the tone of Rubin's comments? Do you think he is not attacking me? He is very disdainful and continues to accuse me falsely. A normal person experiences what's called annoyance. 12.176.152.194 (talk) 15:44, 2 January 2012 (UTC)

How can he attack you if you stop talking about your own ideas? iNic (talk) 16:09, 2 January 2012 (UTC)

Yes, it is a kludge and it feels good to see you squirming because just about everything you think is knowledge is based on this Kludge. I almost feel sorry for you Rubin. Let's see - Einstein proved to be wrong. Next to follow will be your fake hero Abraham Robinson? 12.176.152.194 (talk) 15:23, 2 January 2012 (UTC)

Aha so you proved Einstein wrong too? Did you publish it? iNic (talk) 15:34, 2 January 2012 (UTC)

Don't tell me you suffer from reading problems also? Sorry, I meant to say he has been proved wrong. 12.176.152.194 (talk) 15:41, 2 January 2012 (UTC)

This is very much off topic but please tell me when and in what context Einstein was proved wrong? iNic (talk) 15:48, 2 January 2012 (UTC)

Have you been following the news lately? I think we must discuss only the topic here - this article. Practice what you preach! 12.176.152.194 (talk) 18:14, 2 January 2012 (UTC)

## Rubin - Legal Threats

My work has been published online. That I have a website means it is copyrighted. Furthermore, it is dated so no one can say it's not original. Don't give me that nonsense regarding your knowledge of legal matters. One more thing - I did not bring up the topic, I have been asked several questions and referred those readers to the material. They did not have to read it or continue to ask me further questions. 12.176.152.194 (talk) 05:03, 2 January 2012 (UTC)

You did bring it up, because it is the only source you have given for the assertion that infinitesimals are even problematic. I still don't know what you have against R(((ε)))). — Arthur Rubin (talk) 05:36, 2 January 2012 (UTC)
Rubbish. I did not bring it up. I was asserting that the modern theory of infinitesimals started with Cauchy's Kludge. Therefore in this respect, the file I referred to (Cauchy's Kludge on my web site) is indeed not only relevant but central to the discussion. There are other sources, many of which are published online, not necessarily also published in the form of a physical book. I have already explained that R(((ε)))) is a figment of your troubled imagination. We had this discussion years ago and neither you nor Michael Hardy could not see the light then. What makes you think you will see it now? The entire theory is absolute rot. But I can't argue against this because Abraham Robinson, bless his dead little Jewish heart, actually got to print his wrong ideas. 12.176.152.194 (talk) 14:07, 2 January 2012 (UTC)

So why don't you publish your own ideas proving Abraham Robinson to be wrong? Why wasting your time here while you have your important mission? Wikipedia can only take into account already published work and so far Robinson has published his ideas whereas you haven't. In the meantime please only talk about work that is not your own and that is published. "Published online" doesn't count I'm afraid, unless it's in a peer reviewed online journal. iNic (talk) 15:43, 2 January 2012 (UTC)

I have spent sufficient time satisfying myself that Robinson was an idiot. As for wasting time, it's not really a waste of time pointing out that your article contains many non-factual statements regarding Archimedes and infinitesimals. These have been changed but are still not correct. I don't really care one way or the other so no need to respond to me again. If you want to have the last word, be my guest. 12.176.152.194 (talk) 15:48, 2 January 2012 (UTC)

Claiming that Robinson was an idiot is just stupid. Period. iNic (talk) 10:23, 3 January 2012 (UTC)

## Rubin tries to define infinitesimal

As for infinitesimals, this article should be adequate to explain them to any mathematically trained person. My R(((ε))) is a subfield of the Levi-Civita field, which has most of the same properties, but is easier to calculate with. — Arthur Rubin (talk) 05:47, 2 January 2012 (UTC)

Nonsense. The Levi-Civita field definition assumes ε is an infinitesimal. It does not define an infinitesimal. The definition is circular. It is also a misnomer in my opinion because it follows from Cauchy's wrong ideas regarding infinitesimals. Like Cauchy, you appear to have missed this circularity in your reasoning (or lack thereof). 12.176.152.194 (talk) 15:36, 2 January 2012 (UTC)

The Levi-Civita field defines ε, and it can be shown it is an infinitesimal. — Arthur Rubin (talk) 15:45, 2 January 2012 (UTC)

That is false. Care to define ε? Care to define "mathematically trained" person? (*) To say that ε is an infinitesimal is not a definition. In order to say that something can be shown to be infinitesimal, you first have to define infinitesimal, that is, you have to know what you are talking about. Of course in your misguided thoughts this did not occur to you, did it? 12.176.152.194 (talk) 15:54, 2 January 2012 (UTC) — Preceding unsigned comment added by 12.176.152.194 (talk)

(*) Mathematically trained according to you would be someone who believes the same rot as you do? Hmm, Archimedes and most great mathematicians did not possess degrees. So please do tell what this means? 12.176.152.194 (talk) 16:35, 2 January 2012 (UTC)

In the Levi-Civita field, ε is an element of the field, corresponding to the function "a" which maps 1 to 1 and all other rationals to 0. That it is infinitesimal follows from the definition of "<" in the field (which isn't specified in the article, but is specified in the definition of the field.) I suppose you're going to tell me that the definition of the field (the collection of all functions a from the rationals to the reals such that the set of indices of the nonvanishing coefficients $\{q\in\mathbb{Q}:a_q\neq 0\}$ is be a left-finite set) is not a definition, either.
If I recall correctly, the definition of addition, multiplication, and less that, where a and b are in the field, are as follows:
$(a+b)_q = a_q + b_q$
$(a \cdot b)_q = \sum_{r+s=q} (a_r \cdot b_s)$
$a < b \text{ if the least rational } q \text { such that } a_q \neq b_q \text{ satisfies } a_q < b_q$
where the sum and the existence of "the least rational" follow from the left-finite properties. — Arthur Rubin (talk) 16:53, 2 January 2012 (UTC)
"My" field R(((ε))) can be defined the same way, except that the support $\{q\in\mathbb{Q}:a_q\neq 0\}$ is required to be bounded below and have a common denominator, making the verification of closure easier. — Arthur Rubin (talk) 16:59, 2 January 2012 (UTC)
For what it's worth, for the purpose of this argument, I define mathematically trained as anyone who can understand the definition of the Levi-Civita field, as defined in our article. — Arthur Rubin (talk) 17:03, 2 January 2012 (UTC)
There are so many problems with what you have written, that it is difficult to know where I can begin to show how wrong you are.

      f(x)  = 1     if    x=1
f(x)  = 0     if   x=a/b      and   a/b is rational


then x is infinitesimal. I don't think so.

"That it is infinitesimal follows from the definition of "<" in the field." - is the most ridiculous nonsense I have ever read.

I define one mathematically trained if one can see immediately that what you've written is absolute rot.

You are correct about your left-finite set definition - it is a non-definition. Aside from being completely irrelevant, it only makes your attempt to define an infinitesimal more complex. Furthermore, the fact that your imaginary set of infinitesimals has no LUB tells me immediately it is ill-defined even in terms of set theory. I don't care about the transfer principle because it is BS and there are mathematicians who agree with me on this.

Rubin, no well-trained mathematician will honestly believe in infinitesimals. What you have written is such nonsense that it's almost laughable. I'll go one step further: any mathematician who thinks infinitesimals are a sound concept is not a mathematician. More like a fool.

I suppose you are going to tell me this is just my opinion. Well, I'll tell you, anyone who claims infinitesimal theory borders on being a moron.

If Robinson is an idiot I'm happily a moron in your view. It sounds like a great offer. iNic (talk) 10:29, 3 January 2012 (UTC)

Rubin, I am sorry to say this (really) but you may be a bigger moron than I thought, if you sincerely believe in the garbage you've written.

One more thing: I can tell that you don't understand the theory very well. Most mathematicians will simply allow you to pull the wool over their dull eyes. Perhaps you should get your buddy Hardy to help you? But he is a statistician who claims that dy/dx is not a ratio. Tsk, tsk. 12.176.152.194 (talk) 18:03, 2 January 2012 (UTC)

I see you don't understand the concept of abstract algebra. The function f, as you call it, considered as an element of the field, is an infinitesimal. And the set of infinitesimals never has an LUB, in the Levi-Civita field, because there is no attempt to claim the field is complete, and in the hyperreals or ultraproduct analysis, because the set is not "internal" or "standard". I wasn't going to say that you are a moron, but anyone who doesn't understand the article Levi-Civita field, whether or not you agree with it, is not a mathematician. It's clear that you don't understand it. — Arthur Rubin (talk) 20:37, 2 January 2012 (UTC)
I think I understand very well. What I am saying is you are wrong and these are two different things. You can call the function f anything you like, but it does not remotely resemble anything close to the idea associated with the infinitesimal concept. In fact you can call your subset (which is ill-defined because it has no LUB) of infinitesimals "rubins" if you like, but it does not change the fact that it's all feces. There is absolutely ZERO relation between your abstract little set and infinitesimals in the traditional meaning thereof. The Levi-Cevita field is ill-defined. In plain words, it's a load of rubbish. And you are no mathematician. Also, no honest mathematician who has completed studies in Abstract Algebra (as I have) will agree with what you claim. The difference between us is this: I am a real mathematician without a PhD. You have a PhD but are a fake mathematician. 12.176.152.194 (talk) 22:07, 2 January 2012 (UTC)

Dear 12.176.152.194, why do you write in the talk page section of an article if you don't understand the subject? There are many wikipedia articles and I'm sure you can contribute in a positive way to Wikipedia if you find a topic that you understand. Good luck! iNic (talk) 00:29, 3 January 2012 (UTC)

Inic: I know this is all too overwhelming for you. I suggest you follow your advice. Perhaps you can contribute in a more positive way. What do you say? 12.176.152.194 (talk) 02:53, 3 January 2012 (UTC)

Please take my advice and leave this page. You are just making a fool of yourself. iNic (talk) 10:38, 3 January 2012 (UTC)

## Another non-factual and misleading claim in the article

It has applications to numerical differentiation in cases that are intractable by symbolic differentiation or finite-difference methods. This is outright false. The reference is subject to opinion and debate. Khodr Shamseddine, "Analysis on the Levi-Civia Field: A Brief Overview," http://www.uwec.edu/surepam/media/RS-Overview.pdf

12.176.152.194 (talk) 23:15, 2 January 2012 (UTC)


}}

## lede

The lede was recently shortened in a drastic way. Wiki policy allows for the statutory 4 paragraphs. Is there any reason to make the lede much shorter than provided by policy? Tkuvho (talk) 16:18, 8 January 2012 (UTC)

## n extensions of R

Hello We would like to contibute to the infinitesimals article. It the next article n- extensions of R are constructed, each n-extension has cardinality $\aleph_n$ℵn, so every time happens that the next extension has more numbers than before, so every time we have more holes than numbers in the real line.

Sélem Avila, Elías Proper $n-extensions of${}^\ast \bold R$∗R with cardinalities$\aleph_n\$ℵn. (Spanish) XXIX National Congress of the Mexican Mathematical Society (Spanish) (San Luis Potosí, 1996), 13–24, Aportaciones Mat. Comun., 20, Soc. Mat. Mexicana, México, 1997.

I'm (Nselem (talk) 14:38, 25 January 2012 (UTC)) and my father is the author of the article, I'm helping him with typing and translations, so please be patiente with us, we really want to discuss this subject and colaborate if possible.

I'm afraid I don't see the benefits to Wikipedia of the chain of hyperreal fields. Assuming the Generalized Continuum Hypothesis, then, using either the compactness theorem or the ultrapower construction, given any real-closed field *R, there is an extension **R, with infinitesimals over *R, and any specified (non-limit) cardinality greater than or equal to the cardinality of *R. (I'm not sure it needs to be a non-limit cardinal, and I'm pretty sure the compactness theorem method doesn't require it, but the ultrapower method does require that it be the cardinal of a power set.) But I also don't see how to work that into the article. — Arthur Rubin (talk) 15:14, 25 January 2012 (UTC)
What might be of interest is the construction of a maximal hyperreal field that contains all of the above, recently developed by Philip Ehrlich, to appear in BFL, and available at his homepage. The field (which is a class) is isomorphic to the maximal surreals that he describes there. Tkuvho (talk) 15:23, 25 January 2012 (UTC)

(Nselem (talk) 03:36, 26 January 2012 (UTC)) For Rubin : The compacity theorem does not aplply to jump from *R to **R, because *N is not numerable (required condition), it is neither usable for any other construction eçwith a greater cardinality. About the second option that it is mentioned (ultrapowers),even when it is true what is said (It is what it is done in the article) the fact is that it does not work any ultrafilter that contains the fréchet filter (isomorphic extentions to *R are obtained) and this is the usual way to construct *R starting with R; in this way, the extentions are "existentials". For the explicit construction it is required an ultrafilter that contains the filter of the co-bounded sets (with bounded complement ) about *N, as is done in the article; then it is possible to do all the proper extentions *R, **R,..... ****...***R, with incressing cardinals aleph-2, aleph-3, ... aleph-n; with infinitesimals every time smaller, limitless (and each time bigger infinit numbers, limitless). This ultrafilter co bounded over R, works for extend R to *R. And the concept of infinitesimal becomes relative in each extention. This article was reviewed in Current Mathematical Publications, American Mathematical Society,Number 4, March 19, 1999; and Zentralblatt MATH, European Mathematical Society, FIZ Karlsruhe & Springer-VErlag, 0945.03097

You may be right about the ultrafilter construction: *(*R) appears to be quasi-isomorphic to *R, and I can't see immediately whether it has infinitesimals over the embedded *R. Still, the compactness argument will produce, for any field X, a larger field Y with infinitesimals over X, with cardinality at least α as follows:
Define constant symbols $c_x$ for each x in X, constant symbols $d_\beta$ for each β < α, and constant symbol ε with axioms:
$c_x + c_y = c_{x+y}$ and $c_x \cdot c_y = c_{x \cdot y}$ for x, y in X
$c_x < c_y$ for x < y in X
$0 < \epsilon$
$\epsilon < c_x$ for x > 0 in X
$d_\beta \neq d_\gamma$ for β < γ < α
Arthur Rubin (talk) 06:58, 26 January 2012 (UTC)

## Subject to opinion

I removed the last sentence from this paragraph on Levi Field - "It has applications to numerical differentiation in cases that are intractable by symbolic differentiation or finite-difference methods." It is a matter of opinion. The stated reference (8) does not provide any evidence this is true. 166.249.134.226 (talk) 17:51, 17 June 2012 (UTC)

## An infinitesimal number by itself is useless

I added the following statements to the second paragraph:

"An infinitesimal object by itself is often useless and not very well defined; in order to give it a meaning it usually has to be compared to another infinitesimal object in the same context (as in a derivative) or added together with an extremely large (an infinite) amount of other infinitesimal objects (as in an integral)."

I know that there maybe exists other ways to give infinitesimal numbers a meaning, but I didn't really know how to continue the lasts sentence. "Or in any other way give it a meaning" does just not sound right. Feel free to extend this statement to complete it. —Kri (talk) 22:42, 17 June 2012 (UTC)

Do you have a source for your "uselessness" claim? Tkuvho (talk) 16:35, 25 December 2012 (UTC)

## Lede edits

Since when does 1 - .999... = 1/x? As stated [1] its incorrect because 1/x <>0 and with limits it can be shown that .999... equals one, thus 1 - .999... = 0 (and not 1/x). In any case, even if you can show its properly sourced with some twisted interpetation of .999... (verifiable, not truth and all that), as I said in my edit summary, the first paragraph is supposed to define and summarize the article, not inadequately go into the minutia of how students are taught, per wp:lede. Thus, it needs to be removed. I just removed it again[2]. I removed it thinking the maths were sourced to a primary source, but I was mistaken on that. The authors referenced (Katz & Katz, 2010). I've not yet looked at that reference, but from another source I see that the expansion being referenced does not involve standard notation, since 0.999...;...999... is the hyperreal version of .999..., so the claim that .999... is different from 1 is either misleading or missing appropriate context, thus it does not belong in the lede. -Modocc (talk) 21:55, 27 April 2013 (UTC)

I am not sure I understand your question "Since when does 1 - .999... = 1/x?" How does "1/x" come into the picture? The point is that students naturally relate to the string "0.999..." as being infinitely close to 1, so that 1-"0.999..." is a kind of a "naturally occurring" infinitesimal. As has been explained in a number of recent articles including the ones cited in the lede, student intuitions of an infinite string of 9s falling just short of 1 can be rigorously justified. As a "naturally occurring" infinitesimal, 1-"0.999..." is helpful in setting the stage in this article, because it is an immediate way of giving the reader an idea of what we are talking about, that he or she can perfectly well relate to. It therefore fits nicely in the lede. Tkuvho (talk) 10:05, 28 April 2013 (UTC)
The point of my question is that, according to all sources, that for the same reason there is absolutely no infinitesimal of any kind between .333... and 1/3, there is also absolutely no infinitesimal between .999... and (1/3)*3. That a single student journal points out a couple of fringe worthy guys that dispute these facts and references a math wizard's hyperreal expansion (without even actually discussing it) and then having something perhaps vaguely like it placed prominently here? No. Sure, its vital that students and teachers can and should talk about the infinitesimal difference between .999...;...999 and 1, since .999...;...999 < .999...;...999... = 1. Of course, they need to be astute and explicit with their math when going about this too and to not be so vague or inaccurate that they mistakenly promulgate misconceptions in the process. -Modocc (talk) 13:54, 28 April 2013 (UTC)
The references in question are: (1) [1] as well as (2) [2]. These are by no means "student journals". Thus, Journal for Research in Mathematics Education is a leading (some would say, the leading) education journal. Meanwhile, The Mathematics Educator is published by students, but it is a refereed journal that publishes established education scholars, as reference (2) clearly illustrates. As far as your claim of "promulgating misconceptions" is concerned, I believe wiki policy is dictated by standards of verifiability rather than personal opinions of individual editors. Tkuvho (talk) 14:33, 28 April 2013 (UTC)
1. ^ Ely, Robert (2010). "Nonstandard student conceptions about infinitesimals". Journal for Research in Mathematics Education 41 (2): 117–146.
2. ^ Norton, A.; Baldwin, M. (2011/2012), "Does 0.999... really equal 1?", The Mathematics Educator, here 21 (2): 58–67
The relevant text with regard to your specific claim is: "Nonstandard analysis provides a sound basis for treating infinitesimals like real numbers and for rejecting the equality of 0.999... and 1(Katz & Katz, 2010)." Thus verifiable, but it is from a student journal and its not clear to me that this is the accepted position. Per wp:due, if you can provide other statements like this one by others, I'd like to see them (I don't have access to all the text in question). -Modocc (talk) 15:09, 28 April 2013 (UTC)
I am not sure why you choose to focus on the article in The Mathematics Educator and to emphasize the fact that it is student edited (again, it is not a "student journal" as you call it). The article in Journal for Research in Mathematics Education is more notable, makes the point about usefulness of an "0.999..." entity falling short of 1 as far as student learning is concerned (based on a field study), and is cited by no fewer than 24 articles by a wide variety of authors, see here Given the hostile tone of your earlier messages concerning this issue, it is hard to understand your claims to the effect that "more references are needed". Does one need 48 references to get this past this particular editor? 96? Tkuvho (talk) 12:08, 29 April 2013 (UTC)
To begin with, do any of the articles actually refer to the "difference" between 0.999... and 1 as "infinitesimal"? If not, the discussion do not belong here. If so, I would need to see the context to see if it is appropriate, and need to do research to see if it's "fringe". — Arthur Rubin (talk) 12:39, 29 April 2013 (UTC)
Certainly, Ely's article specifically discusses nonstandard student intuitions of an infinitesimal number 0.000... ...1 and its usefulness in learning calculus. To the extent that there is a crystal clear formalisation of this in the context of the hyperreals, it seems to escape the "fringe" label. The point, of course, is not that the students should be taught about ultrafilters, but that their intuitions are "nonstandard" rather than "erroneous". User:Modocc obviously adheres to the latter view, and is moreover perfectly within his rights to view such intuitions as "erroneous". As soon as he publishes his research, we will be able to cite it. Tkuvho (talk) 12:45, 29 April 2013 (UTC)
I focused on the authors' math statement because that is precisely what you want inserted into the article (I refer to the editors involved because they do bear responsibility to correct mistakes). Since 0.999... = 1, you need statements that clearly and unambiguously state otherwise (you need specific statements from math articles like 0.999... <> 1 to make that claim!). The alternative (and not to be placed in the lede please) is to write about how the hyperreal conception of ".999...;X" falls short and makes rigorous the students' misconception of ".999...". --Modocc (talk) 13:01, 29 April 2013 (UTC)
It's unsuitable for the lead, because the students' intuition is erroneous. Reliable sources making statements contrary to fact may be disregarded. It might still be helpful in student learning, although I dislike the concept of teaching using erroneous concepts. The statement "The argument that 0.999… only approximates 1 has grounding in formal mathematics.", in Norton & Baldwin, is erroneous. I'm sure we can find reliable sources, if necessary. Neither Norton nor Baldwin appears to have any published papers in non-standard analysis, nor in any field related to it, nor is non-standard analysis within the primary field of study of either journal. I could argue that my statement here that the intuition is erroneous might quailify as a WP:RS, but I won't; my field being mathematical logic and set theory, not specifically including non-standard analysis. — Arthur Rubin (talk) 13:45, 29 April 2013 (UTC)
Arthur, I don't think the leading math education journal would have published a paper with such a glaring mathematical error, nor would such a paper have gotten widely cited. Obviously, the author is referring to student intuitions of "0.999..." being a nonstandard number, rather than a real number. After all, before the students have been taught anything about the real numbers, they can legitimately hold opinions about a variety of possible numbers. We have a large subsection in 0.999... explaining this. Your point that the citation is unsuitable for the lede may be well-taken, but we should agree about the mathematics before we go on to discuss notability! Tkuvho (talk) 14:18, 29 April 2013 (UTC)
Thanks for the new reference that I can access. Inclusion of this proposed reassignment of 0.999... is fine in the body of this article (not the lede) iff its shown with the explicit context that was included in the 0.999... article. Actually, since this article only briefly summarizes the hyperreals, it should only be noted that such a reassignment is proposed by these authors. --Modocc (talk) 16:16, 29 April 2013 (UTC)
We're not discussing the proposed addition. I have little objection to the proposed text appearing in some article, but I don't think students could have an useful intuitive understanding unless they could also see "0.999..." < 1 < 2 − "0.999...", which I see no reference to in the papers I've been able to read. I think it should more likely be in a hyperreal article than in infinitesimalArthur Rubin (talk) 16:58, 29 April 2013 (UTC)
The proposed addition, wherever it lives, does not show that "0.999..." (Katz & Katz) is the reassignment of "00.999..." to a number different than 1, nor does it make it clear that it was recently proposed. I can see the motivation for this though, because if the rank goes to infinity then the limit of the reciprocal term equals zero, and one ends up with the standard real 0.999... = 1. Like I said at the beginning, context matters and this is missing from the proposed addition. My apologizes if I seemed arrogant, indifferent, or "in the way" of the contribution I removed. My goal now is to improve the text so I and others have a better understanding on a first read without trudging though discussions such as this. In addition, the reason I wrote 1/x is because I didn't type out the markup for 1/infinity when 1/x works, because the reciprocal of any quantity, of any size, is always nonzero and the reason infinitesimals (and limits) exist. -Modocc (talk) 17:19, 29 April 2013 (UTC)
@Arthur: I am not sure I understand your query with regard to the inequality [1 < 2 − "0.999..."]. By very simple algebra this reduces to ["0.999..." < 1] so a student who can relate to the one, can also relate to the other. What does the other inequality add? The comment about a "naturally occurring" infinitesimal is accessible to a large audience which is certainly broader than the readership of hyperreal number. Tkuvho (talk) 12:35, 30 April 2013 (UTC)
@User:Modocc: I re-read your comment on 1/x several times, but I am still not sure what you are getting at. If you are asking what the reciprocal of such a non-standard [1 - "0.999..."] is, the answer is precisely 10H where H is the (infinite) rank where the last "9" occurs. Tkuvho (talk) 14:29, 30 April 2013 (UTC)
I'm not asking what the reciprocal of the non-standard [1 - "0.999..."] is, but I am implying that the math of that is incorrect for an infinite rank, but I'm not a mathematician and I have not studied hyperreals: thus I might be gulping down someone's porridge if its not to hot, or not to cold, thus bear with me please, :), and I'll do my best to explain my rationale: I'm claiming that for the set of hyperreals and its subset of reals, the formula from the 0.999... article (attributed to Karin Katz and Mikhail Katz) does not consistently hold for an infinite rank H. To show this (with a rigor that is only sufficient for me at the moment) consider that a standard number is 3/10 + 3/100 + ... = 0.333... = 1/3 (this should be true whether we are working with reals or hyperreals, since hyperreals include all reals). Therefore: 0.999... = 9/10 + 9/100 + ... = 3*(3/10 + 3/100 + ...) = 3*(1/3) = 1 = 1 - 0 <> 1 - |1/x| for all x because the reciprocal function f(x)=1/x<>0 can not be zero for any and all x including all possible infinitely small numbers given by the infinitesimals such as the specific one given by Katz&Katz: 1/10H. QED. In short, the standard summation of "9/10 + 9/100 + ..." does not equate with the summation of 1 and an infinitesimal. Now like I said, I am not a mathematician, but if the maths I learned for the reals somehow do not apply to the hyperreals, and I need to study that article too I'm possibly in the wrong house and ... -Modocc (talk) 16:59, 30 April 2013 (UTC)
@User:Modocc: You apparently feel that there is an error in these articles published in refereed journals. This is certainly possible. However, there is a detailed discussion of this calculation in the "infinitesimal" section of 0.999.... A number of editors have gone over that section without detecting errors. My impression is that you are taking the expression "infinite sum" too literally. You could raise this issue at Talk:0.999... if you still feel there is an error. Tkuvho (talk) 19:32, 30 April 2013 (UTC)
I've no intention of trying to contravene policy (see my comments below). --Modocc (talk) 20:43, 30 April 2013 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── (ec; I think I agree with Modocc, but expressed it differently.) The claim being made in the papers is that students understand "0.999..." < 1 as an infinitesimal difference. If they don't understand 1 − ε < 1 < 1 + &epsilon (where ε = 1 − "0.999..."), then I would argue that what they "understand" doesn't act like an infinitesimal. Although mathematical education is not my field, if the papers don't comment on that one way or the other, then they seem questionable. — Arthur Rubin (talk) 17:02, 30 April 2013 (UTC)

The student R. Ely interviewed described this ε as 0.000... ...1 so she would certainly not have any problem describing 1+ε as 1.000... ...1 --what exactly are you getting at? Tkuvho (talk) 19:09, 30 April 2013 (UTC)

Tkuvho, I am fine with the Wikipedia's verifiability not truth policy and I intend to keep my editor's hat on when it comes to deciding appropriate placement of content. Essentially, Katz & Katz is proposing to subtract terms in a way that with the hyperreal notation, defines some sets of numbers that according to their rank, gives numbers such that 1 - infinitesimal. But, as my demonstration above shows, I don't see how this allows for consistency across-the-board for all the reals, thus its a pedagogically nightmare for me to understand, and I wouldn't want to try to teach such a major conceptual revision without learning it for myself. But putting my opinion of this wiki's coverage of these papers (which I will read once I get a chance to visit the library) aside, there shouldn't be any need for me to discuss the content's veracity further. Anyway, this article is not about teaching students Katz & Katz's work on hyperreals! Since its nonstandard, its not a notable "introduction"! Further, without any tertiary sources (which cover well-seasoned research) actually discussing the K&K's recent proposal, it does not yet come close to meeting wp:DUE to warrant being placed in any article's lede. -Modocc (talk) 20:43, 30 April 2013 (UTC)

Please see Talk:0.999...#Consistency_across-the-board_for_all_the_reals where I responded in more detail. Tkuvho (talk) 08:38, 1 May 2013 (UTC)

### Partially resolved

I clarified the nonstandard reinterpretation of "0.999..." that is being taught, with this edit.[3]. And I started a new subsection regarding the Norton and Baldwin reference that led to this discussion. -Modocc (talk) 21:12, 5 May 2013 (UTC)

### Norton and Baldwin

What is the justification for including the Norton and Baldwin reference? Clicking on the one citation [4] in google scholar brings up an article that doesn't even appear to cite it. Which leaves no citations for it. Since it presents arguments that the standard number is somehow incorrect (its not), it seems prudent that this research paper be removed per wp:fringe since "exceptional claims require high-quality reliable sources". The emphasis in bold is mine. -Modocc (talk) 22:20, 5 May 2013 (UTC)

Thanks for your interest in this page. I don't think the Norton-Baldwin paper is either "incorrect" or "fringe" but the citation issue makes it hard to insist on its inclusion. Let me take this opportunity to explain why the term "infinite sum" cannot be taken too literally. Consider the infinite sum 1 + 1/2 + 1/4 + 1/8 + etc. The partial sums are clearly less than 2, whereas the series yields 2. Now consider the following "thought experiment": suppose you are working inside a system containing infinitesimals. Let e be a positive infinitesimal. The number 2-e is infinitely close to 2. Now notice that 2-e is in fact an upper bound for each of the partial sums! Yes the "infinite sum" is bigger than 2-e, namely 2. You get the same "paradox" if you concatenate intervals of lengths 1, 1/2, 1/4, 1/8, etc. At each stage the resulting interval has length less than 2-e, yet the "infinite concatenation" (if you think of the series in such terms) somehow overcomes 2-e. There is of course no paradox here if one interprets "infinite sum" as it should be, namely as a series defined via the concept of limit. There is another way of seeing this by decomposing the "taking the limit" procedure into two steps. Tkuvho (talk) 12:38, 6 May 2013 (UTC
If students apply nonstandard analysis to the problem, they still must accept the fact that the standard non-terminating number 0.999...;...999... is equal to 1 regardless of the partial sums they might consider instead (a rank of 2 if they like or a terminating infinite sequence) when learning about real numbers and limits. These facts are straight forward enough, thus 1)my reasons for not keeping the Norton and Baldwin reference are unchanged, 2)we seem to have better references regarding these partial sums than it, and 3)we agree that its lack of citations is a problem for it. -Modocc (talk) 13:49, 12 May 2013 (UTC)
Thuvho: I think you may have overlooked something which may mitigate the need to think in terms of limits. All the finite partial sums are less than 2-e but if you are viewing this from inside the system which contains the infinitesimal, you need to carry the sum out "all the way" to get 2, and that will include infinite terms which bring the partial sums above 2-e. In particular, if N is an infinite hyperinteger larger than 1/e, then the partial sum 1 + 1/2 + 1/4 + ... + 1/N will be larger than 2 - e. Note further that, if you're doing this from inside the nonstandard system, you have no way of taking the sum over the reals without including the nonstandard terms, since, from inside the system, you can't tell the difference between a standard and a nonstandard term. Similar objections apply to the idea that 0.999... could be considered less than 1, of course -- inside the system, you have partial sums, you have the whole series, but you have no way to specify that you will include only standard terms. Salaw (talk) 16:01, 11 February 2014 (UTC)
I believe the Norton-Baldwin reference was removed in the end, so I am not sure what we are discussing exactly. I fully agree that 0.999...=1 :-) Tkuvho (talk) 16:07, 11 February 2014 (UTC)

## First-order properties

I presume some legitimate points are being made in this section, but it is such a muddle.

Perhaps the subject head should be "elementary properties". The term "elementary" is introduced here, and it is true that the Archimedean property is a consequence of the completeness property of the real numbers, and stating the completeness property as the LUB property does involve making statements about sets of real numbers. So it evidently is not an elementary property of the real numbers.

The section also refers repeatedly to first-order logic (FOL), and that somehow non-elementary properties cannot be stated in FOL. The section states that logic with quantification restricted to elements and not sets "is referred to as first-order logic". But FOL is not only compatible with set theory, but widely used in combination with it throughout mathematics;

The second paragraph seems to contradict itself, and I do not have the energy to try to sort it out. It states that the Archimedean property can be expressed by quantification over sets. The LUB property is indeed expressed using sets, but the Archimedean property can be expressed as:

∀x, y ∈ R. x > 0 ∧ y > 0 ⇒ ∃ n ∈ N. n ⋅ x > y

and this does not involve quantification over sets.

The second paragraph then goes on to make further points, but I am unable to follow the argument. — Preceding unsigned comment added by Crisperdue (talkcontribs) 20:09, 12 June 2013 (UTC)

Comments in this talk page section so far are by me, sorry I omitted the explicit signature originally. Crisperdue (talk) 21:06, 12 June 2013 (UTC)

Actually, the Archimedean property is in $L_{\omega_1 \omega}$, as the quantification over N is in the metalanguage. (see Infinitary logic#Definition of Hilbert-type infinitary logics for notation.) You'll note that hyperreals satisfy the displayed definition of the Archimedean property, with N being the nonnegative hyperintegers. — Arthur Rubin (talk) 23:05, 12 June 2013 (UTC)
I spent some time digging through the history of this section last night. In case anyone's interested in what went wrong with this section, here's the deal. It started out as a completely conventional exposition of first-order infinitesimal theory, just like I learned it in school. In particular, it asserted that not all properties of the hyperreals are identical to the reals (after all, they'd be kinda useless if they were totally identical to the reals, eh?), and it asserted that, since the reals are the only complete ordered field, we can't expect the hyperreals to be complete. An IP address then came in and added some rather angry (my description) and ungrammatical (objective judgement) edits to the section to assert that one can extend the reals in a way that preserves all properties, and also added the last sentence of paragraph 2 that begins "This is also wrong..." which contradicts the sentence just before it.
Some long-suffering editor came in afterwards and cleaned up the grammar and made things read a little better, but the last sentence in the paragraph is still standing there as a direct (and clearly intentional) contradiction of the earlier assertions.
I'd try to fix it up but unfortunately my knowledge of this subject is too soft for me to be sure what's correct here. I know that simple expositions of the hyperreals carry over only first order properties. I also know that Robinson spent what seemed to me to be a large (and difficult) amount of space in his book working to carry forward the second order properties of the reals as well (at least, I think that's what he was doing), which would seem to imply he was striving for a version of the hyperreals which were complete. That agrees with some things I've read elsewhere on the web. Presumably, the catch is that from inside the hyperreals, you can't tell the difference between a real and a hyperreal, so you can't form, say, the set of all values infinitesimally close to 1, which obviously doesn't have either a GLB or LUB. Salaw (talk) 15:41, 11 February 2014 (UTC)
Thanks for your interest. The IP is probably influenced by the internal set theory viewpoint, where they really are identical to the reals (in fact, they are the reals). I remember a few years ago there was an editor making mistakes of this sort that had to be corrected by other users. He apparently moonlighted as an IP without anybody noticing it. If the last sentence is incorrect the simplest thing would be to remove it. Tkuvho (talk) 15:56, 11 February 2014 (UTC)
If you check the French wiki page for non-standard analysis you will notice that it is dominated by the IST outlook. Checking through its history may help identify the editor but who cares? Tkuvho (talk) 15:57, 11 February 2014 (UTC)
By the way, the IST viewpoint is fabulous, but obviously it should not be confused with the NSA viewpoint. Tkuvho (talk) 16:01, 11 February 2014 (UTC)

## Infinitesimals in teaching

Perhaps this is not a good question to ask here, but given the very small set of published calculus books using Robinson's methods, I'm curious as to why Henle & Kleinberg's Infinitesimal Calculus isn't mentioned. I'm no doubt biased, since, as a college student some decades ago, I was totally entranced by Kleinberg's lectures on the subject. None the less, though perhaps not a complete calculus course, the text seems like a pretty nice exposition of the use of hyperreals in elementary calculus. Is the issue that its circulation was too small to make it notable, or that it's not widely enough used -- or was it just an oversight? Salaw (talk) 05:58, 11 February 2014 (UTC)