-1

From the context of the real numbers we know that the square root of $-1$ is not a real number. Because of this the historical way we approach this is that the concept of complex numbers were created to give meaning to the nonexistence of $\sqrt{-1} = i$ and study the potential implications of such a value. We even knew beforehand how the square root of $-1$ as it existed as a concept. It was merely that it did not exist as an intermediate vale.

When that set was created, certain properties were different from the set of real numbers and other properties (such as ordering) ceased to exist. If some superset of the complex plane were created to give meaning to another currently 'nonexistent' operation, what algebraic properties would we expect to retain? For instance, if one wanted to extend nonexsistent limits into being for which the one sided limits exist, would you only assume them equal to themselves or would you make them equal to limits with similar shapes around the point? Of course that is a hypothetical example, not to be taken as an actual request. The details of that would depend on whether geometric equivalence was the equalizing factor or the sequence used to compute the limit.

I know I have written two questions here. So to summarize:

First: what extensions upon real numbers currently exist other than the complex plane?

Second: How would one go about creating a new extension, and how do people usually work out their properties aside from creative thinking? Is there a formal method that can be used to determine the properties of an extension coming from attempting to give meaning to an intermediate value that is "undefined"?

Note: I couldn't come up with the proper tag so some guidance would be appreciated. I was thinking the tag "equality" or "extensions upon real numbers" or "monexistence" would be good, but those are not available.

user64742
  • 2,207

2 Answers2

2

Non-existence of an object $x$ that has property $P$ can be expressed as:

$\neg \exists x\ ( P(x) )$.

That has nothing to do with objects per se. Rather, it is a boolean assertion, and if you prove it then its value is "true".

You see, when you prove that an object exists, it is of course an object that you are talking about. But when you prove that no object has property $P$, you are not proving that some object does not exist, but rather you are proving that the concept of an object with property $P$ has no instance. You need to clearly distinguish between concepts (which always exist once you express it) and objects. Some concepts may have instances, meaning an object that witnesses the concept, but some concepts do not.

For example, when I claim that there is no pig with wings, I am not claiming that there is a pig with wings that does not exist, but I am claiming that the concept "pig with wings" has no instance. Symbolically I might write (and this is just one possible way to convey this concept):

$\neg \exists x \in Pigs \ ( HasWings(x) )$.

Concerning your edit about non-existent limits, you cannot directly assign a value to a non-existent limit, without changing your definition of limit. If you do so, your new interpretation of limits is incompatible with the old one. For example you might then assign the string "NULL" to an expression of the form $\lim_{x \to a} E(x)$ whenever it does not exist as a conventional limit. But then you still do not solve the problem with the limit laws, because presumably you want NULL propagation, meaning that any operation involving "NULL" will result in "NULL". Now that means that $\lim_{x \to a} ( f(x) + g(x) ) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$ still fails to hold in general because $\lim_{x \to 0} ( \frac1x + (-\frac1x) ) = 0 \ne NULL = NULL+NULL = \lim_{x \to a} \frac1x + \lim_{x \to a} -\frac1x$.

For more interesting stuff about non-existent limits see:

If you want finer control over what you say about an expression when it does not have a limit, most probably you're looking for asymptotic expansions, which usually exist for reasonably nice expressions. For example:

As $n \to \infty$:

  $\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n^2+n} \in \frac{1}{n^2} - \frac{1}{n^3} + O(\frac{1}{n^4})$.

  Note that the above gives more information than just saying $\frac{1}{n} - \frac{1}{n+1}$ does not tend to a limit.

  And we can from this recover some related limits such as $n^2 ( \frac{1}{n} - \frac{1}{n+1} ) \in 1 - \frac{1}{n} + O(\frac{1}{n^2}) \to 1$.

I gave some more involved examples at:

I personally find it better to always use asymptotic expansion instead of limit laws, because it is not only systematic but also yields much more information. One can always extract the limit from the asymptotic expansion but not necessarily the other way around. The exception is when it is not a concrete expression, in which case one might have to use abstract methods including the fundamental theorem of calculus and L'Hopital's rule.

Community
  • 1
user21820
  • 57,693
  • 9
  • 98
  • 256
  • I see... Interesting. – user64742 Jun 10 '16 at 03:56
  • 3
    This is, quite honestly, a much clearer answer than mine. – Justin Benfield Jun 10 '16 at 03:58
  • @JustinBenfield: I greatly appreciate your appreciation. See my edit for more details pertaining to limits. I've thought about such stuff before for precisely the same reason, limits. – user21820 Jun 10 '16 at 04:00
  • @TheGreatDuck: Does my edit answer your question about limits? Feel free to ask if you need clarification. – user21820 Jun 10 '16 at 04:01
  • In response to your edit, I refer to the sort of nonexistence that doesn't result from unboundedness but rather a sharp discontinuity. Infinity is already well defined. :) – user64742 Jun 10 '16 at 04:02
  • @TheGreatDuck: Just put whatever you like in place of $\frac1x$ and use its negation. – user21820 Jun 10 '16 at 04:02
  • @TheGreatDuck: And, no, infinity is not going to help you out even in this special case because $0 \ne \infty + (-\infty)$... – user21820 Jun 10 '16 at 04:03
  • No I mean that the limits aren't having nonexistence due to incompatible sides. Like when there is a jump in a graph. They are unbounded, but it's a different sort of nonexistence. I see what you mean though. – user64742 Jun 10 '16 at 04:04
  • Here's one I thought of. The limit as x approaches 0 for floor(x) and the limit as x approaches 1 for floor(x-1)? Are they considered equal? After all, it's the same sequence that would generate the limit... – user64742 Jun 10 '16 at 04:06
  • @TheGreatDuck: Ok I roughly see what you're trying to get at, but at the moment it's a bit too fuzzy. You're basically saying the shape looks the same, but even in your example it's shifted vertically. The best way to capture such things is not to use a limit. You can just use the original expression to denote the behaviour! Perhaps you're looking for a finer control over what you can say about non-existent limits, so I'll add a bit. – user21820 Jun 10 '16 at 04:11
  • Yeah, I was more curious if they lead to special extensions;like how complex numbers come from "nonexistent" square roots. Though knowing if they are comsidered equal would also be interesting to know. – user64742 Jun 10 '16 at 04:13
  • 1
    I'll have to read more on that big O notation (i keep forgetting what that means) but that edit does look interesting. You've certainly given some interesting subjects to read about! – user64742 Jun 10 '16 at 04:17
  • @TheGreatDuck: I've added some links. Have fun reading! – user21820 Jun 10 '16 at 04:23
  • 1
    @TheGreatDuck: Hyperreals do not work the way you think they do. Limits still cannot be written as arithmetic identities. This is the case no matter what construction of hyperreals is used (such as the ultrapower of reals modulo a non-principal ultrafilter over the naturals). – user21820 Oct 31 '18 at 05:32
-3

Given any collection of objects, $S$ that are considered distinct, equality is always the finest possible equivalence relation, $=$, on $S$. Which is the $(=,S,S)$ where $=$ is the set $\{ (s,s)|s\in S\}$.

Justin Benfield
  • 3,992
  • It does, just not very directly. In the case at hand, we have a situation where we have a limit that does not exist, meaning it isn't in $S$, hence it can't be equal to any limit that does exist. – Justin Benfield Jun 10 '16 at 03:49
  • I have, because the other limit exists, and is thus some distinct $s\in S$ (which has the further property that it exists), hence that $s$ is distinct from the $t$ corresponding to the limit that does not exist. Thus they cannot be equal. – Justin Benfield Jun 10 '16 at 03:52
  • It follows from the nature of what equality is as a relation, which I defined in my answer. But it does leave some work to you. The point is that because one of the limits exists, no matter what abstract object you assign to the other nonexistent limit, that object will be distinct from the existent limit, hence the corresponding equality relation on any set containing both limit objects won't have both objects in the same order pair (no $(s,t)$'s in $=$). Thus they can't be equal. – Justin Benfield Jun 10 '16 at 03:56
  • Ahh, then my answer here is inadequate to address that, I thought you had one limit that existed because that was the situation in the other thread, which I had posted in. – Justin Benfield Jun 10 '16 at 04:05