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Suppose you're trying to teach analysis to a stubborn algebraist who refuses to acknowledge the existence of any characteristic $0$ field other than $\mathbb{Q}$. How ugly are things going to get for him?

The algebraist argues that the real numbers are a silly construction because any real number can be approximated to arbitrarily high precision by the rational numbers - i.e., given any real number $r$ and any $\epsilon>0$, the set $\left\{x\in\mathbb{Q}:\left|x-r\right|<\epsilon\right\}$ is nonempty, thus sating the mad gods of algebra.

As @J.M. and @75064 pointed out to me in chat, we do start having some topology problems, for example that $f(x)=x^2$ and $g(x)=2$ are nonintersecting functions in $\mathbb{Q}$. They do, however, come arbitrarily close to intersecting, i.e. given any $\epsilon>0$ there exist rational solutions to $\left|2-x^2\right|<\epsilon$. The algebraist doesn't find this totally unsatisfying.

Where is this guy really going to start running into trouble? Are there definitions in analysis which simply can't be reasonably formulated without leaving the rational numbers? Which concepts would be particularly difficult to understand without the rest of the reals?

Alexander Gruber
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    It sounds to me like this algebraist is happy enough to effectively be working with Cauchy sequences of rational numbers. His stubbornness seems to be more a linguistic point than anything else. – Qiaochu Yuan May 10 '13 at 04:06
  • He's definitely going to run into trouble if he refuses to acknowledge the existence of $\overline{\mathbb{Q}}$ -- or $\mathbb{Q}(i)$ -- or $\mathbb{Q}_p$... – Cam McLeman May 10 '13 at 04:06
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    Why not turn the assertion around, and ask why he accepts the existence of 0.1, given that it can be approximated to any desired accuracy by a fraction of the form $\frac{a}{2^b}$ (where $a$ is an integer and $b$ is a positive integer)? – Glen O May 10 '13 at 04:11
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    I think the correct way to formulate this question is "how far can one get in analysis without leaving constructive logic?" (otherwise, as Qiaochu pointed out, it's merely a taboo on certain definitions and notations, which are no worse than any others). And then it's a very broad logic/TCS problem. In my intuition, which is completely uninformed (it's analysis, after all) and therefore should not be trusted, problems start when you try to exploit compactness and similar properties which constructively run afoul of the halting problem. – darij grinberg May 10 '13 at 04:17
  • @darijgrinberg That's a good point, I've put some logic tags on this. – Alexander Gruber May 10 '13 at 04:18
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    @QiaochuYuan Not necessarily. He could, for example, be opposed to the power set axiom. – Quinn Culver May 10 '13 at 05:34
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    I wrote [a short essay about this question](http://plover.com/~mjd/misc/irrational) in a different context a couple of years back. It is a recurrent trope in science fiction. – MJD May 10 '13 at 06:10
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    Tim Gowers wrote something about the need for the real numbers once: https://www.dpmms.cam.ac.uk/~wtg10/reals.html – kahen May 10 '13 at 11:13
  • Norman J Wildberger develops analytic geometry around the rationals, see [rational trigonometry](https://en.wikipedia.org/wiki/Rational_trigonometry), this might be relevant, too. – k.stm May 12 '13 at 07:22
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    See also "Is the diagonal of the unit square truly irrational?" for a geometric quasi-proof for the algebraist. There it's shown that it is not merely a notion of linguistics and terms. – Marcos Jun 07 '13 at 03:39
  • You definitely need to restate the IVT. By the way, correct me if I'm wrong, but I think any function on $\Bbb Q$ can be continuously extended to $\Bbb R$ iff it's uniformly continuous on any closed and bounded interval. Is this true? That is, for any $N$ and $\epsilon$, there exists a $\delta$, such that for any $x$ and $y$, we have $|x| – Akiva Weinberger Sep 01 '15 at 13:07
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    @MJD It seems like you're suggesting that, if we stubbornly refuse to accept irrationals, we have to eventually stumble upon the concept of Cauchy sequences or Dedekind cuts. If that's true, though, how come these ideas are relatively recent? If they're as natural as you make it seem, they should have been discovered centuries or millennia ago. (Perhaps it's because no one would be so stubborn as to say that the diagonal of my field isn't a number.) – Akiva Weinberger Sep 01 '15 at 13:17

10 Answers10

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What kind of algebraist "refuses to acknowledge the existence of any characteristic 0 field other than $\mathbb{Q}$"?? But there is a good question in here nevertheless: the basic definitions of limit, continuity, and differentiability all make sense for functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$. The real numbers are in many ways a much more complicated structure than $\mathbb{Q}$ (and in many other ways are much simpler, but never mind that here!), so it is natural to ask whether they are really necessary for calculus.

Strangely, this question has gotten serious attention only relatively recently. For instance:

$\bullet$ Tom Korner's real analysis text takes this question seriously and gives several examples of pathological behavior over $\mathbb{Q}$.

$\bullet$ Michael Schramm's real analysis text is unusually thorough and lucid in making logical connections between the main theorems of calculus (though there is one mistaken implication there). I found it to be a very appealing text because of this.

$\bullet$ My honors calculus notes often explain what goes wrong if you use an ordered field other than $\mathbb{R}$.

$\bullet$ $\mathbb{R}$ is the unique ordered field in which real induction is possible.

$\bullet$ The most comprehensive answers to your question can be found in two recent Monthly articles, by Jim Propp and by Holger Teismann.

But as the title of Teismann's article suggests, even the latter two articles do not complete the story.

$\bullet$ Here is a short note whose genesis was on this site which explains a further pathology of $\mathbb{Q}$: there are absolutely convergent series which are not convergent.

$\bullet$ Only a few weeks ago Jim Propp wrote to tell me that Tarski's Fixed Point Theorem characterizes completeness in ordered fields and admits a nice proof using Real Induction. (I put it in my honors calculus notes.) So the fun continues...

Martin Sleziak
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Pete L. Clark
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    James Propp's article is also on the arXiv: http://arxiv.org/abs/1204.4483 – kahen May 10 '13 at 19:20
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    If I may comment in this slightly inappropriate place on your real induction instruction linked to above: the statement and proof of the Real Induction Theorem 2 could be made a bit cleaner. Drop RI1, but allow $a=x$ in RI3; in the proof distinguish only the two cases $\inf S'\notin S'$ and $\inf S'\in S'$. Now $\inf S'=a$ is handled in the former case without needing explicit mention, and as $\inf S'=b$ can only occur when $S'=\{b\}$ it falls in the latter case, so in the other case $\inf S'\notin S'$ it suffices to say that $\inf S'=b$ cannot happen. – Marc van Leeuwen Jun 07 '13 at 11:42
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    Did not see this anywhere, but $\mathbb{Q}$ isn't a complete space which causes some issues with uniform convergence and the such – DanZimm Jun 09 '13 at 00:30
  • When you say "absolutely convergent series which are not convergent" are you talking about Cauchy sequences which are not convergent (because $\mathbb{Q}$ is not complete), or do you mean something else? – robjohn Feb 05 '14 at 19:09
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    @robjohn: An "absolutely convergent series which is not convergent" means that $\sum_n |a_n|$ converges while $\sum_n a_n$ diverges. Yes, this is permitted to happen because not all Cauchy sequences in $\mathbb{Q}$ converge. I invite you to read the note! – Pete L. Clark Feb 05 '14 at 19:30
  • "$\mathbb{R}$ is the unique ordered field in which real induction is possible." I would think any complete real closed field should allow real induction. Am I missing something? – Pavel Čoupek Dec 15 '14 at 19:45
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    @PavelC: In that real induction implies the least upper bound axiom: yes. – Pete L. Clark Dec 16 '14 at 02:00
  • @PeteL.Clark What is the mistaken implication? – Red Banana May 11 '17 at 16:29
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    @AskYourself: Theorem 10.17 of Schramm's book claims that in an ordered field in which all Cauchy sequences converge, the Bolzano-Weierstrass Theorem holds. This is false: the counterexamples are precisely the Cauchy-complete non-Archimedean ordered fields, e.g. the Laurent series field $\mathbb{R}((t))$. The mistake in the proof occurs when an interval is repeatedly bisected and it is claimed that the lengths approach zero. – Pete L. Clark May 11 '17 at 17:38
  • @PeteL.Clark Thanks. Have you informed him? I guess it would be a good thing. – Red Banana May 11 '17 at 18:37
  • @AskYourself: No, I haven't. Please feel free to do so if you wish. – Pete L. Clark May 12 '17 at 02:23
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This situation strikes me as about as worthwhile a use of your time as trying to reason with a student in a foreign language class who refuses to accept a grammatical construction or vocabulary word that is used every day by native speakers. Without the real numbers as a background against which constructions are made, most fundamental constructions in analysis break down, e.g., no coherent theory of power series as functions. And even algebraic functions have non-algebraic antiderivatives: if you are a fan of the function $1/x$ and you want to integrate it then you'd better be ready to accept logarithms. The theorem that a continuous function on a closed and bounded interval is uniformly continuous breaks down if you only work over the rational numbers: try $f(x) = 1/(x^2-2)$ on the rational closed interval $[1,2]$.

Putting aside the issue of analysis, such a math student who continues with this attitude isn't going to get far even in algebra, considering the importance of algebraic numbers that are not rational numbers, even for solving problems that are posed only in the setting of the rational numbers. This student must have a very limited understanding of algebra. After all, what would this person say about constructions like ${\mathbf Q}[x]/(x^2-2)$?

Back to analysis, if this person is a die hard algebraist then provide a definition of the real numbers that feels largely algebraic: the real numbers are the quotient ring $A/M$ where $A$ is the ring of Cauchy sequences in ${\mathbf Q}$ and $M$ is the ideal of sequences in ${\mathbf Q}$ that tend to $0$. This is a maximal ideal, so $A/M$ is a field, and by any of several ways one can show this is more than just the rational numbers in disguise (e.g., it contains a solution of $t^2 = 2$, or it's not countable). If this student refuses to believe $A/M$ is a new field of characteristic $0$ (though there are much easier ways to construct fields of characteristic $0$ besides the rationals), direct him to books that explain what fields are.

KCd
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    Of course, in constructive mathematics, $A/M$ is not provably a field, since checking the convergence of a sequence to $0$ is equivalent to the halting problem even if the sequence is proven to be Cauchy. Then again, there are quite few fields in constructive mathematics. – darij grinberg May 10 '13 at 07:03
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    @darijgrinberg: for constructivists, there's always [Abstract Stone Duality](http://www.paultaylor.eu/ASD/). – leftaroundabout May 10 '13 at 08:00
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I did my undergrad project on "Analysis in a rational world", because it seemed like fun. It turns out that using definitions over $\mathbb{Q}$ means you can solve $d^n F/d x^n=G(x,F,dF/dx,\ldots,d^{n-1}F/dx^{n-1})$ with each derivative bijective on almost whatever interval you like. Fun, but not very like real analysis.

Stupid Questions Inc
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Jessica B
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The Bolzano-Weierstrass, Cantor's theorem, and other theorems which essentially talk about compactness of closed intervals will become messy.

Whereas every bounded sequence of real numbers have a convergent subsequence, if you take a sequences of rational numbers which converges to $\sqrt 2$ it does not have a convergent subsequence in $\Bbb Q$.

Similarly, the intersection of decreasing sequences of closed intervals is no longer guaranteed to be non-empty. And I am quite certain that we can somehow find a way to construct a continuous function on a bounded interval which is not uniformly continuous.

Asaf Karagila
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    Yes - compactness is at the heart of analysis! – Hendrik Vogt May 10 '13 at 20:31
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    Heart of Darkness, if one is so inclined. Mista Kurtz, he dead. – Asaf Karagila May 10 '13 at 21:30
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    I think that $f:\mathbb{Q}\cap [0,2]\rightarrow \mathbb{R}$ given by $f(q)=\frac{1}{q-\sqrt{2}}$ is a continuous function on a closed bounded interval in $\mathbb{Q}$ that is not uniformly continuous, also unbounded. The problem now is that continuous functions on $\mathbb{Q}$ aren/t as nice as continuous functions on $\mathbb{R}$ – Amr Jan 22 '14 at 13:13
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What counts as “without leaving $\mathbb{Q}$”? Suppose our framework for theorising about the rationals is the axiomatizable fragment of second-order logic which only allows predicative instances of comprehension. That framework arguably is logic, not set theory in disguise, and the resulting second-order theory is arguably talking about nothing over and above the rationals. But such a theory will be equivalent in strength to the weak second-order arithmetic $\mathsf{ACA_0}$, in which it is well known that we can reconstruct classical analysis and a lot more besides.

Turn the OP's question around a bit: how much beyond a (first-order) theory of the rationals is actually required to reconstruct classical analysis? This question has been intensively investigated in the project of Reverse Mathematics (which I'm surprised others haven't mentioned yet): in headline terms, the answer is “very much less than you would have imagined”. If the project is not familiar to you, the place to start is with the freely available first chapter of Stephen Simpson's now classic book.

Peter Smith
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    The title of the article of Jim Propp linked to in my answer is "Real analysis in reverse". Although (I think) he doesn't say so explicitly, I have always taken this to be an allusion to Simpson's "reverse mathematics". The game that Propp/Teismann/myself and others like to play here is not the same as the reverse mathematical one but it has some family resemblance. Anyway, I think you are right to call attention to Simpson's work: +1. – Pete L. Clark May 10 '13 at 13:42
  • Added: I just checked the introduction of the final version of Propp's paper: in fact *he does* make explicit reference to Simpson and reverse mathematics. (But no one here had done so explicitly, as you say...) – Pete L. Clark May 10 '13 at 13:44
  • @PeteL.Clark And thanks for all the references in your answer, which I'll be interested to follow up. – Peter Smith May 10 '13 at 13:48
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I think he's in trouble when you ask him what the length of the diagonal of a unit square is.

Quinn Culver
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    For that algebraist guy, I guess, unit square means $1^2$, and 'its diagonal' is an undefined term. – Berci May 10 '13 at 07:56
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    @Berci The point is that the distance function is something that requires irrationals. Otherwise there is no distance between $(0,0)$ and $(1,1)$. – Quinn Culver May 10 '13 at 16:17
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Even in the OP you've already went past the point where your stubborn algebraist runs into trouble: you started talking about functions. But if your algebraist doesn't believe in characteristic zero fields like $\mathbb{Q}(x)$, then he can't believe in functions either!

A less stubborn algebraist can get pretty far without giving in and accepting the reals though. Introducing number fields as needed to deal with algebraic irrationals, or dealing with all of them by using the algebraic closure of the rationals. Better yet, there is the real closure of the rationals -- the set of algebraic irrational numbers with imaginary part zero -- which is a real closed field, meaning that it has exactly the same properties as the real numbers do, so long as one sticks to things that can be described in terms of polynomials and inequalities. (in particular, nothing that requires induction is allowed, since you can't describe the integers with polynomials and inequalities)

There are purely algebraic formulations of notions like differentiation and manifolds. Antiderivatives would exist less frequently, but they could be added formally as needed, and (co)homology can play many of the purposes one uses integration for anyways. Many things one can do with trig functions can be done with suitable rational functions instead (e.g. see the Chebyshev polynomials). There are exponential rings and modules related to differentiation and lots of other stuff I have little familiarity with (and many more I've probably never even heard of)

The main trouble the less stubborn algebraist will run into is that things will become progressively more complicated the more he tries to include. Or maybe some of the purely algebraic facts whose only known proofs are analytic.

Either that, or formal power series rings would give him trouble, since despite their algebraic use, allowing them would make it more difficult to justify avoiding the construction of $\mathbb{R}$.

  • I find the last sentence rather cryptic. Should I read it as "avoiding the construction of $\Bbb R$ is motivated by the desire to avoid any uncountable sets, and therefore elements that are fundamentally impossible to describe in finite terms; however formal power series also form an uncountable set, with the same drawbacks"? Or is there some more direct connection between formal power series and real numbers you are hinting at? – Marc van Leeuwen Jun 07 '13 at 10:05
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    @Marc: Mainly, I have power series rings, p-adic numbers, I-adic completions, and the real and complex numbers all fitting into a similar place in my head. One similarity is they were uncountable. Another (and the one that first sprung to my mind) is that they are all topological completions with respect to a valuation. –  Jun 07 '13 at 20:07
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The fact that the only functions on an interval with derivative zero everywhere are constants is at the heart of applications to science. This fact is equivalent to the completeness axiom. Continuity, differentiation and intervals make sense in any ordered field but if there exists a nonempty set with an upper bound but no least upper bound, then there exists a non-constant function with derivative zero everywhere.

Satish
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I think it is worth point out that limiting oneself to the rational numbers means that certain analytical number theory results that can only be established with real or complex numbers cannot work out merely by rational numbers. The huge problem we had with $\mathbb{Q}$ is not merely analytical, but also algebraic. Any elementary theory methods which only employ addition, substraction, division cannot distinguish $\mathbb{Q}$ and other fields if we start with $\mathbb{Z}$. So in some sense $\mathbb{Q}$ gives the limit of elementary number theory.

Bombyx mori
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Real numbers are needed to establish new theorems. For example $\exists L \neq 0$ s.t. $\sin(x) = \sin(x+L)$ isn't true if we restrict the definition of domain to rational numbers. You may restrict the definition of domain even more to obtain a function s.t. $f: D \rightarrow \mathbb{Q} \ , \ \ f(x) = \sin(x)$, but it restricts the possibilities of choosing suitable numbers $L\in \mathbb{Q}$. Hence there will not be such a number.

Real numbers were needed to express the solutions of $x^2 = 2$. Of course we can restrict the range of applications, but it discloses for example finding the side length of a square, when area is known. This has also geomatrical significance. It is known, that there is a way to construct $\sqrt{x}$ from $x$ using pencil and compasses. So there is a way to see the solution. So there is left to make a sufficiently general and simple construction for numbers (real numbers).

Juha-Matti Vihtanen
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