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Quite a few analysis books I've looked into start with some number system, claim some (reasonably convincing) axioms on it, assume it to be existed, and build analysis from there. Some explain why the number system exists by constructing it from some more basic number system, but I've probably never came across anything about why some number system does not exist.

Could anyone please explain how some number system does not exist by giving an example?


One of my understandings is that it might be due to the contradiction arisen from the axioms that fails the number system to be existed.

According to page 11 of Pugh's Real Mathematical Analysis (2017):

For what if a system satisfying the axioms failed to exist? Then one would be studying the empty set!

My understanding to this sentence above is as follows: let $U$ be a collection that contains everything that exists that we ever need, and $L$ be the list of axioms that are needed to be satisfied by our number system $X$. Define $X$ as a set such that for each $x \in U$, iff $x$ satisfies everything in $L$ simultaneously, then $x \in X$. Since there is a contradiction in $L$, no element in $U$ would be filled into $X$ through these implications and thus we get $X=\emptyset$. And all the subsequent analyses on $X$ are carried through vacuous implications, that is, by assuming there are elements in the empty set $X$.

May I ask if my understandings are on the right track? (it kinda confuses me...)

And here come some other questions:

  • Is that possible to have a number system that doesn't exist while the axioms on it are consistent?
  • When we say some number system does not exist, does this mean that the entities in the system, speaking of themselves, literally cannot exist? Or does this mean that those entities assumed to be inside the system indeed can exist, but they are just living in somewhere else and the system is empty? (Or are there other interpretations?)
J-A-S
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  • +1 : thought provoking. No idea why someone (else) downvoted. – user2661923 Dec 01 '20 at 05:14
  • Define "number system." – R. Burton Dec 01 '20 at 07:12
  • Here's a simpler example of a number system that doesn't exist: take the real numbers, and add the axiom that there exists a special constant $\epsilon$ with $0<\epsilon\le x$ for every positive $x$. This seems like a reasonable thing to do (and people often try it out), but it doesn't work. – MJD Dec 01 '20 at 09:00
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    Or.take the rational numbers and expand them to include $\frac10$. However you do it, yiu either end up with a system in which arithmetic is impossible, or you end up with a system that has only one number. – MJD Dec 01 '20 at 09:03
  • @MJD I haven't gone into details but is your $\epsilon$ the same one we defined in non-standard analysis? (probably not? that $x$ needs to be in $\mathbb{R}$ in non-standard analysis I think...) – J-A-S Dec 01 '20 at 09:10
  • @J-A-S In nonstandard analysis you can make an $\epsilon$ that is less than $\frac1n$ for every integer $n$, but you can't make an $\epsilon$ that is as I described. – MJD Dec 01 '20 at 11:28
  • @MJD actually, you can make just such an $\epsilon$ as long as you include the caveat that $x$ in the expression $0<\epsilon\le x$ is restricted to finite, non-infinitesimal reals (i.e. $\epsilon$ is less than any standard [positive] real) - you can even make it so that $\epsilon$ is less than the "limit" of any infinite sequence $\frac1{cn}$, with $c$ a standard real. Likewise, you can expand the rationals to include $\frac10$ by mapping $\frac10$ to a "point at infinity" (as in the one-point compactification of the reals). – R. Burton Dec 01 '20 at 12:11
  • You can say that $\frac10$ is infinity, but what you get is not useful as a number system; the arithmetic is broken. For example, it is no longer the case that $(a+b)-b =a$. – MJD Dec 01 '20 at 13:56

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Here's an answer to your first question. It's also rather topical right now:

We work over $\mathbb{R}$ throughout.

A Normed Division Algebra is a "number system" which generalizes $\mathbb{C}$ in a similar way that $\mathbb{C}$ generalizes $\mathbb{R}$. More precisely:

A division algebra is a a vector space over $\mathbb{R}$ (so it has addition and subtraction, among other things) which also admits multiplication and division (as long as you don't divide by the $0$ vector).

So you might think of a division algebra as a "number system" (since we can add/subtract/multiply/divide) which contains the reals $\mathbb{R}$ in a nice way. Indeed, $\mathbb{R}$ itself and $\mathbb{C}$ are two good examples of division algebras.

A normed division algebra is a division algebra which also admits a norm $x \mapsto |x|$ which tells you how "big" a number is. Again, $\mathbb{R}$ with the absolute value function, and $\mathbb{C}$ with its standard norm $a + bi \mapsto \sqrt{a^2 + b^2}$ are both examples of this object too.


The drama:

There is a theorem due to Hurwitz which says that if you restrict your attention to "finite dimensional" systems, the only such number systems are

So if you are interested in "number systems" which contain $\mathbb{R}$, admit a notion of "size", and "aren't too big", you are FORCED into picking one of these systems.

Rather interestingly, you lose something on each stage.

  • Passing from $\mathbb{R}$ to $\mathbb{C}$, you lose the ability to order your numbers (is $1 < i$?)
  • Passing from $\mathbb{C}$ to $\mathbb{H}$, you lose commutativity (so $ab \neq ba$ in general)
  • Passing from $\mathbb{H}$ to $\mathbb{O}$, you lose associativity! That is, $(ab)c \neq a(bc)$ in general!

What is most interesting, though, is that a few months ago, somebody published an article saying they had found a new algebra, which was like $\mathbb{O}$, but managed to preserve associativity!

That is, they claimed the existence of an $8$-dimensional, associative, normed division algebra.

This is certainly a "number system" of the kind you might be interested in. It is also nonexistent by Hurwitz' Theorem! The paper has since been retracted from the journal, though I believe the author is still fighting the decision, claiming their result is actually true.


So you see, there are cases where mathematicians accidentally study objects which don't exist. There are lots of apocryphal examples as well, but most of them don't fall under the heading of "number systems".


As a rather quick reply to your other questions:

  • Can a number system fail to exist even when its axioms are consistent?

My instinct says "no". Certainly in the case of a First Order set of axioms, a system exists if and only if the axioms are consistent. This is Godel's Completeness Theorem.

I'm notably less well versed in higher order logics, such as those needed to formalize the completeness of $\mathbb{R}$. In general, completeness/soundness/effectiveness cannot all happen at the same time for higher order logics, and I'm far from an expert in the subtleties of what is and isn't possible. Rather funnily, this is a corollary of Godel's Incompleteness Theorem.

I'm sure if you post another question on this site asking for specifics of completeness of second order logic, you'll find some more specialized logicians than me who are willing to help :)

  • What does it mean for a system to "not exist"

I think this is better asked to a room of philosophers than a room of mathematicians. For me, it "doesn't exist" if there isn't a model. That is, there is no set equipped with some functions that actually satisfy the axioms.

For instance, there is no noncyclic group of order $2$. Why does this not exist? Because you can look at a set of size $2$, and try all possible group structures, and see for yourself that none of them work. It's the same idea for other objects that "don't exist", we just need cleverer arguments since we can't check by hand that all (possibly infinitely many) options fail.


I hope this helps ^_^

HallaSurvivor
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    Thank you very much for giving these pieces of information! I think there are a lot of algebra and logic I still need to learn, they are powerful :) May I also ask how you interpret "system failed to exist as empty set" mentioned by Pugh? (Is my understanding right there?) – J-A-S Dec 01 '20 at 08:18
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    Yes! Your interpretation seems correct. That is, if $X$ is the class of all structures satisfying the axioms, and no such structures exist, then $X = \emptyset$ and we see that anything we want to know about our structures is true! Because of silly vacuous reasons like you've mentioned ^_^ – HallaSurvivor Dec 01 '20 at 08:20
  • An impressive answer (+1) – Peter Dec 01 '20 at 09:27