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In my question Are infinitesimals equal to zero? I think I learned how infinitesimals in non-standard analysis are numbers whose absolute value is smaller than any positive real number, and hence most of them are not real numbers themselves. I say most because, as an integer and thus a real number, zero is smaller than any positive real number and it can therefore be considered an infinitesimal, although not exactly equal to other infinitesimals because adding it to itself always yields itself, whereas adding non-zero infinitesimals to themselves appears to yield results that are not equal to the values being added.

If I am not mistaken, infinitesimals seem to be the reciprocal notion of infinity when thinking of the latter as a number whose absolute value is greater than any positive real number, hence not real itself. If I divide one by any real number, the result I get becomes smaller and smaller as I choose larger and larger dividers; I think it then follows that if I divide one by a number greater than any positive real value, what I get is a result smaller than any positive real number. That is why it seems to me that infinitesimals and infinities are complementary notions at both "ends" of the positive or negative real lines. Although I may be mistaken.

But if I have understood the subject correctly, then my question is regarding the number of times I would be able to fit an infinitesimally long line segment (one with a length shorter than any other expressed with a positive real number) inside another line segment one unit long. From my previous paragraph I think that if I could divide one by an infinitely small number, I would get an infinitely large result, suggesting that I would be able to fit an arbitrarily large number of infinitesimally long line segments inside a unit length; a number greater than any positive real number. Would that be correct? And if not, why not?

And assuming that the reasoning so far is valid, then I am intrigued by the seemingly special case of zero, which itself appears to be smaller than any positive infinitesimal (in other words, every positive infinitesimal except for zero appears to be greater than zero). I think I can see how, if I add zero-length "segments" to a "growing" line, I can add as many as I want before reaching one unit of length because I will never be any closer to one than I was to begin with; I will never reach one unit of length.

That seems to be in contrast to non-zero infinitesimals, which may also take an infinite number of additions but at least they seem to arrive to the unit length. Even though both procedures (dividing one by zero and dividing it by a number smaller than any real number but greater than zero) appear to yield an infinitely large result, one seems more infinite than the other, if that makes sense. When I add non-zero infinitesimals, it may take me an infinite number of steps to produce some change, but when I add zero I get no change at all even after an infinite number of iterations. Is that true? Is that a difference between zero and other infinitesimals?

Thanks!


EDIT: It has been pointed out how division by zero cannot be performed in the set of hyperreals just like it cannot in the set of real numbers and, while the subject might deserve a new question, I have indeed attempted to divide 1/0 in my original question so I will add an edit here.

I know next to nothing about hyperreals, but I think I can see why dividing by zero would yield an indeterminate result when I consider the division operation, in the context of natural numbers, as counting the number of times I can successively subtract one number from another while obtaining a remainder that is greater than or equal to zero. For example, from the starting number of 12, I can subtract 1 twelve times leaving no remainder (12 / 1 = 12), I can subtract 2 six times (12 / 2 = 6), 3 four times (12 / 3 = 4), 4 three times (12 / 4 = 3), 5 twice leaving a remainder of two and so on. How many times can I subtract 0 from 12 before yielding a remainder smaller than zero? As many as I want, because subtracting zero from a number does not change the number. So I can see how the question cannot be answered with any specific natural number. Even after a billion times I could still go on subtracting zero for as long as I lived.

It is not difficult for me to extend the idea to all integers, to rational numbers and even to real numbers if I don't mind getting approximate values from my subtractions (I don't know exactly how many times I can subtract π from 100 because I don't know what is the exact value for π). But when it comes to hyperreals, I am not sure if I can use the same logic because I don't know the properties of hyperreal numbers. Many people suggest that I must extend the notion of real numbers to hyperreals, but they never describe how that extension occurs. For example, I understand how I can extend the set of real numbers to the set of complex numbers by adding to each one a multiple of the square root of -1, and then keeping in mind that addition of terms, as well as the fact that the square root of -1 times itself is -1, when I perform operations between complex numbers. I am sure there are many other implications about complex numbers, but the point here is that at least I understand how to subtract A - B when A and B are complex numbers.

However, in hyperreals? If A is a positive real number and B a positive infinitesimal quantity, is A - B < A? I suspect it would but, by how much?

I guess my question could be reprhased as "How many times can I subtract an infinitesimal from the real number 1 before arriving to a remainder of 0?"

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    $\begingroup$ $\int_0^1 dx = 1$. $\endgroup$ – Mark Aug 30 '16 at 13:04
  • $\begingroup$ An infinitesimal is by definition some positive number $\varepsilon \neq 0$ which satisfies $$\forall n \ \ \ \varepsilon \cdot n < 1 $$ $\endgroup$ – Crostul Aug 30 '16 at 13:11
  • $\begingroup$ Using Crostul's definition, there are no "infinitesmals" because that is impossible. "Infinitesmals" are not numbers in the usual sense of "number". You have to extend the concept of "number" to the "hyper-real" numbers: en.wikipedia.org/wiki/Hyperreal_number $\endgroup$ – user247327 Aug 30 '16 at 13:47
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    $\begingroup$ @CarvoLoco If you are in an environment allowing infinitesimals, then if you make $1/\mbox{infinitesimal}$ you get (by definition) an "infinite number". Infinite numbers are exactly those which are larger than every natural number. $\endgroup$ – Crostul Aug 31 '16 at 17:16
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    $\begingroup$ Thanks for pointing that out, @MikhailKatz. I have edited the question to reflect what I think you are saying. Maybe you can shed some light on the topic? Thank you! $\endgroup$ – Carvo Loco Sep 1 '16 at 10:41
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"How many can fit" seems like a geometric question to me. The standard real line has no infinitesimals, of course, but there are fields, such as the hyperreals, which do have infinitesimals.

If you pick an infinitesimal $a>0$ in the hyperreals, then by basic algebra you can fit $1/a$ of them into the interval [0,1]. Of course, just as $a$ is smaller than every positive real number, $1/a$ will be larger than every real number. So in particular you can fit infinitely many pairwise disjoint intervals of width $a$ into the interval $[0,1]$, namely the intervals $[na,(n+1)a]$ for each natural number $n$.

Those intervals will not cover all of the interval $[0,1]$, however. In fact, no positive real number will be in any of those intervals, because if we had a real number $x> 0$ with $x \in [na, (n+1)a]$ then we would have $x < (n+2)a$ and so $x/(n+2) < a$. But $x/(n+2)$ will be a positive real, so it cannot be less than any infinitesimal. So $\bigcup_{n \in \mathbb{N}} [na, (n+1)a]$ contains only infinitesimals.

Similarly, if $x$ and $w$ are reals with $x < w$ then $x + a < w$, and so the interval $[x, x+a]$ is disjoint from $[w, w+a]$. This means we can fit uncountably many intervals of width $a$ into the interval $[0,1]$, namely all intervals of the form $[x, x+a]$ for $x \in [0,1)$.

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  • $\begingroup$ Hello @Carl, thank you and please excuse me if I misinterpret your words, because I am neither a mathematician nor a native English speaker, but I interpret what you are saying in line with my appreciation that I would be able to fit a number greater than any real number of infinitely small values, each greater than zero, into the given interval. But what if a = 0, instead of a > 0? Would the outcome be any different? And if so, how? Thanks! $\endgroup$ – Carvo Loco Aug 31 '16 at 11:13
  • $\begingroup$ Each interval $[x,x]$ is of width $0$, so I would say you can fit uncountably many of those intervals into the unit interval as well. If your hyperreal field has the same cardinality as the reals, and we take the approach I suggest in my answer, then the number of intervals of width $a$ that can be fit into $[0,1]$ seem to be the same as the number of intervals of width 0 that can be fit, and both of these are the same as the cardinality of the reals. $\endgroup$ – Carl Mummert Aug 31 '16 at 17:36
  • $\begingroup$ Carl, it could be mentioned that the intervals $[na, (n+1)a]$ as $n$ runs from $0$ to $\frac{1}{a}$ provide a partition of the unit interval. Such (hyperfinite) partitions are of course useful in applications. $\endgroup$ – Mikhail Katz Aug 31 '16 at 20:17
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The notion of "fitting infinitesimals into a unit interval" is ambiguous. As far as I can grasp you are probably referring to summation. Fitting infinitesimals into a unit is in fact a special case of summation.

There are cases of summation where infinitely (but countably) many numbers are added together. These are termed "series". There exists a whole theory on series which is part of mathematical analysis. Basically, depending on the kind of numbers that you're adding together, the following outcomes are possible:

1) Your sum converges to a finite number as you keep adding numbers; 2) Your sum stays finite but never converges to any specific number; 2) Your sum gradually grows to plus or minus infinity as you keep adding numbers; 3) Your sum grows in absolute magnitude but changes sign from time to time, never settling on some specific sign and being unbounded in magnitude.

Another case of adding uncountably infinitely many numbers together is integration. Basically any definite integral is the process of such a summation.

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  • $\begingroup$ Thanks, @Michael. I am not sure if I was referring to summation, because I am not sure how your answer addresses my question, but I am a mathematical ignorant so please don't be annoyed by my failure to understand. If I add together the same infinitesimal a > 0 infinitely many times, will the sum converge to a finite number? Stay finite without settling on a precise value? Explode to plus or minus infinity? Increase in absolute value but oscillate in sign? Or does it depend on the actual value chosen? Cheers! $\endgroup$ – Carvo Loco Aug 31 '16 at 11:11
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The OP noted that "Many people suggest that I must extend the notion of real numbers to hyperreals, but they never describe how that extension occurs." One possible construction is called the ultrapower construction and several discussions thereof can be found under the tag .

As far as division by zero is concerned, it is impossible in the hyperreals for the simple reason that the hyperreals form a field, and division by zero is impossible in a field (regardless of the specific field in question).

The hyperreals satisfy the same rules as the reals. One such rule is that if $A$ and $B$ are both positive then $A>A-B$. Therefore the same rule applies over the hyperreals, even if $B$ is (a positive) infinitesimal.

The OP asked, "How many times can I subtract an infinitesimal [say, $\alpha$] from the real number 1 before arriving to a remainder of 0?" The answer is: precisely $\frac{1}{\alpha}$ times, at least if the latter is a hyperinteger.

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  • $\begingroup$ There are several possible ways to construct the hyperreals, not just one. $\endgroup$ – Ian Sep 1 '16 at 11:08
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    $\begingroup$ @Ian, my impression of the level of the OP is that he is going to have a hard time relating to the compactness theorem of first-order logic, but perhaps you can try an alternative answer. Which construction do you have in mind? $\endgroup$ – Mikhail Katz Sep 1 '16 at 11:10
  • $\begingroup$ The IST approach is maybe hard to motivate but easy to work with. Anyway I was mostly just being pedantic, suggesting you say "can be constructed..." instead of "are constructed..." $\endgroup$ – Ian Sep 1 '16 at 13:27
  • $\begingroup$ @Ian, thanks for your comments. The IST approach is syntactic in that it assumes the existence of a model where all these things are true (i.e., a model of the ZFC+IST axioms). The proof of the existence of such a model, found in an appendix (which I have not read) to Nelson's 1977 article, is considerably more difficult than the ultrapower construction over index set $\mathbb{N}$. $\endgroup$ – Mikhail Katz Sep 1 '16 at 13:30
  • $\begingroup$ Fair point. I guess my perspective is that to an amateur, some part of the construction of the hyper reals is just not helpful. Either you have to take on faith that you have an ultrafilter with little intuition for what one "looks like", or you have to take some model theory for granted, or... It is generally more helpful to get a feel for what they enable you to do rather than what they are. That said, I don't think anyone would be surprised that ZFC+IST is equiconsistent with ZFC, even though the proof is hard. $\endgroup$ – Ian Sep 1 '16 at 13:45

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