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I was once told that one must have a notion of the reals to take limits of functions. I don't see how this is true since it can be written for all functions from the rationals to the rationals, which I will denote $f$, that $$\forall L,a,x:(L,a,x\in\mathbb{Q})\forall \epsilon:(\epsilon>0)\space \exists\delta:(\delta>0)$$ $$\lim_{x\rightarrow a}f(x)=L\leftrightarrow(\mid x-a\mid<\epsilon\leftrightarrow\mid f(x)-L\mid<\delta) $$ Since, as far as I know, functions like $\mid x\mid$ and relations like $<$ can be defined on the rationals. Is it true you couldn't do calculus on just the rational numbers? At the moment I can't think of any rational functions that differentiate to real functions. If it's true that it isn't formally constructible on the rationals, what about the algebraic numbers?


Edit

Thanks for all the help, but I haven't seen anyone explicitly address whether or not we could construct integrals with only algebraic numbers. Thanks in advance to anyone who explains why or why not this is possible.

Praise Existence
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    You are right that you can treat $\Bbb Q$ as a metric space, in which case derivatives of rational functions using the limit definition are well-defined. However: you talk about derivatives, but what about integrals? And, rational functions are a very particular class of functions - things like radical, exponential and trigonometric functions are considered in courses even before calculus, and they require real numbers. – arctic tern Aug 04 '16 at 00:52
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    Rational number systems have holes. It is incomplete. – P Vanchinathan Aug 04 '16 at 00:53
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    Classic differential equations like $f'=f$ become impossible to solve. – GEdgar Aug 04 '16 at 00:57
  • @arctictern If you can define infinite sums and limits over the rationals, can't you define integrals? err..can the integral of a rational function be algebraic or transcendental? sorry if I'm just not thinking of the obvious examples – Praise Existence Aug 04 '16 at 00:59
  • @GEdgar yes of course, the calculus over rationals would study only rational functions. sorry I might've been unclear but I already had in mind the fact that it would exclude a great deal of useful functions. I was questioning only if a calculus over rational numbers considering strictly rational functions could be formally constructed. – Praise Existence Aug 04 '16 at 01:07
  • @Shalop Oh yeah thanks – Praise Existence Aug 04 '16 at 01:08
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    Re: your last comment to me: Who said you can define infinite sums and limits over the rationals? The idea that limits or infinite sums of rationals are themselves rationals is clearly false. – arctic tern Aug 04 '16 at 01:08
  • @arctictern yeah you're right, thanks – Praise Existence Aug 04 '16 at 01:09
  • Yes! Definitions and rules of calculus work in any ordered field. – GEdgar Aug 04 '16 at 01:18
  • @GEdgar wait what? you mean differential calculus? – Praise Existence Aug 04 '16 at 01:23
  • Definition of limit, definition of continuous, definition of derivative, chain rule, product rule, etc., etc. – GEdgar Aug 04 '16 at 01:25
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    The big thing about limits in the rationals is sequences that "should" converge (are cauchy, or bounded and increasing, or whatever) often don't. Indeed if $q_n \rightarrow v \not \in \mathbb Q$ is not convergent in Q. – fleablood Aug 04 '16 at 01:31
  • @fleablood good point, should make that an answer – Brevan Ellefsen Aug 04 '16 at 01:39
  • http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/real.pdf – Count Iblis Aug 04 '16 at 01:46
  • There is hardly anything in the universe simpler than an isoceles right triangle with legs of length $1.$ There's the hypotenuse staring at you, plain as day. And yet you are proposing that this simple triangle can't be admitted into the objects of study, simply because our mathematical ancestors, in a moment of almost unreal brilliance, deduced that the length of that hypotenuse cannot be rational? – zhw. Aug 04 '16 at 01:48
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    Re: your edit about algebraic numbers. The counterexamples given in answers can be adjusted accordingly. There are still only countably many algebraic numbers, and the algebraic numbers certainly aren't complete (the sequence of truncated decimal expansions of $e$, each term of which is algebraic, converges to $e,$ which is not algebraic, and the sequence is Cauchy), so, as far as I can see, the same problems hold with algebraic numbers as for rational numbers. The problem is essentially one of topology, and the set of algebraic numbers is not much different from $\mathbb{Q}$ in this setting. – Will R Aug 04 '16 at 03:05
  • Aside: you have the logical formula wrong. e.g. the equation involving the limit shouldn't be inside the quantifiers for $\epsilon$ and $\delta$ –  Aug 04 '16 at 04:13
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    @Will R: A much larger fragment of calculus is still good for the algebraic numbers -- see the theory of real closed fields. For example, every nonempty, bounded, *algebraically defined* subset of the real algebraic numbers has a least upper bound. –  Aug 04 '16 at 04:38
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    [A related question.](http://math.stackexchange.com/questions/387234) – J. M. ain't a mathematician Aug 04 '16 at 04:54
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    You could rephrase many calculus limits as eventual inequalities of rational numbers. This would basically be the same as using real numbers, though, just without the abstraction. – Owen Aug 04 '16 at 11:42
  • @PVanchinathan everything has holes, it's just a matter of defining something denser. ℚ is an ordered field, so this is a valid question. – Tobia Aug 04 '16 at 12:54
  • @CountIblis Is that a serious piece by Zeilberger that you linked to? It seems rather confused, for example the claim that continuous derivative is a degenerate case of continuous derivative by setting $h=0$ in the Newton quotient (without taking a limit) is contrary to the mathematical orthodoxy that such is not defined, being 0/0. – ziggurism Aug 04 '16 at 13:31
  • @ziggurism The point is not per se that you can't take limits but that the interpretation of such a limit is then different. Consider the following thought experiment. We run a cellular automaton in which mathematicians evolve who invent calculus, real numbers etc. etc. But everything they do must have a finite interpretation because everything inside the CA is finite. – Count Iblis Aug 04 '16 at 17:26
  • Can I do calculus with only the irrationals then? (as in real calculus, with the most of the known theorems and stuff.) – Simply Beautiful Art Aug 05 '16 at 00:53
  • You should totally read [this](https://www.dpmms.cam.ac.uk/~wtg10/reals.html) apropos. – Vandermonde Aug 05 '16 at 05:06
  • Would a geometer please come in? – IAmNoOne Aug 05 '16 at 12:12
  • Although rationals have "holes" between them, the holes are as small as you want them to be. – Kaz Aug 07 '16 at 15:25
  • @ziggurism While 0/0 is not defin_able_ in general (even if you tried), we might be able to create a sensible way that $h/h$, where $h=0$, is 1. One way to do that is to talk about the limit, but there may be other ways (albeit not necessarily compatible with mathematical orthodoxy). Maybe Zeilberger is proposing symbolic computation as another way, where we require the substitution to happen after symbolic simplification. – leewz Nov 02 '17 at 20:50
  • @PraiseExistence You may be looking for the [computable reals](https://en.wikipedia.org/wiki/Computable_number). They are not complete, but there's an analogous sense for which they are complete. – leewz Nov 02 '17 at 21:11

12 Answers12

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$\newcommand{\QQ}{\mathbb{Q}}$ Derivatives don't really go wrong, it's antiderivatives. (EDIT: Actually, the more I think about it, this is just a symptom. The underlying cause is that continuity on the rationals is a much weaker notion than continuity on the reals.)

Consider the function $f : \QQ \to \QQ$ given by $$f(x) = \begin{cases} 0 & x < \pi \\ 1 & x > \pi \end{cases}$$

This function is continuous and differentiable everywhere in its domain. If $x < \pi$, then there's a neighborhood of $x$ in which $f$ is a constant $0$, and so it's continuous there, and $f'(x) = 0$. But if $x > \pi$, there's a neighborhood of $x$ in which $f$ is a constant $1$, so it's continuous there too, and $f'(x) = 0$ again.

So the antiderivatives of $0$ can look rather messy. By adding functions like this, you can construct arbitrarily "jagged" functions with zero derivative. As you can imagine, this completely destroys the Fundamental Theorem of Calculus, and any results that follow from it.


This can happen in the real line to some extent, but it's not nearly as bad. The traditional antiderivative of $1/x$ is $\ln|x| + C$. But so is the following function: $$ g(x) = \begin{cases} \ln x + C_1 & x > 0 \\ \ln(-x) + C_2 & x < 0 \end{cases} $$

By changing $C_1$ and $C_2$, we can push the two halves of the real line around completely independently. This is only possible because $1/x$ isn't defined at $0$, and so we've "broken" the real line at that point.


If you like dumb physical metaphors, here's one:

The real line is kind of like an infinite stick. If you wiggle a section of it, the whole thing must move.

With the $1/x$ example, you've made a cut at $x = 0$, and now you have two half-sticks. They can be wiggled independently, but each half must still move as a unit.

The rational numbers are more like a line of sawdust. You can't really move one grain by itself, but you can certainly take an interval and move it around independent of its neighbors.

By completing the rationals, you're adding all the glue between the grains to form a stick again. (I hope no one from diy.stackexchange is reading this...)

Henry Swanson
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    I'd personally use tape and say "good enough." Gluing saw dust together is not for the faint of heart. – Axoren Aug 04 '16 at 01:47
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    Nice answer, though from another point of view you could say there is an issue with differentiability because the notion of what it means to be differentiable at a point becomes fuzzy. – Kimball Aug 04 '16 at 02:48
  • The option of having a different $C$ is not the only difference between the two $g(x)$ functions. The only way to take the logarithm of a negative number is to assume that all the values will be imaginary products - the coefficient being $\ln(−x)/i=n$. The absolute values will therefore be $x=\sqrt{e^{2ni}}=e^{ni}$. –  Aug 04 '16 at 06:56
  • @user1551 the reason I think of derivatives as behaving okay is that if I have a real function, its derivative is the same whether or not I restrict it to the rationals. But that's not the case for antiderivatives. And actually, f is continuous as a function on the rationals. The definitions for continuous and derivative are the same as for the reals, just with different domain. – Henry Swanson Aug 04 '16 at 08:30
  • @HenrySwanson Sorry, I misread your answer. I get what you mean now. – user1551 Aug 04 '16 at 08:36
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    I love your dumb physical metaphor! – Matthew Leingang Aug 04 '16 at 18:43
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    @Martin there is no way to extend $f$ to a continuous function on the reals, much less a differentiable one. – Henry Swanson Aug 04 '16 at 18:58
  • @Henry: yes, my bad. – Martin Argerami Aug 04 '16 at 23:33
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    “adding all the glue between the grains to form a stick again” — this is exactly what's called particle board or fiberboard. – Kevin Reid Aug 06 '16 at 01:03
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    A clean kill for the #giantswanman. – Spooky Aug 06 '16 at 03:20
  • It's probably worth noting that the same effect arises in the reals with, for example, the function $f(x)=0$ for $x\neq0$ and $f(0)=1$, which has $f'(x)=0$ for any finite $x$. One could argue that the two cases are equivalent, in that one cannot insert a value for which you get a nonzero derivative, yet the derivative isn't *really* zero everywhere. – Glen O Aug 06 '16 at 14:39
  • If you define Cauchy sequence equivalence as, "Stay close to each other," and define a stronger version of continuity, "Preserves equivalence of Cauchy sequences," it should throw out that piecewise function, because the resulting equivalence classes would be the real numbers. This is not compatible with topological continuity for the rationals (though topology may have turned out very differently if the earlier mathematicians had dropped the irrationals!). – leewz Nov 02 '17 at 20:37
  • We can also redefine continuity as something that happens for nested open intervals (of rationals), rather than around points. "A function is continuous if, for every sequence of nested open intervals such that the diameters approach 0, the diameters of the images of those intervals approach 0." I believe this is equivalent to the redefinition I gave above, and still isn't compatible with topological continuity. – leewz Nov 02 '17 at 21:19
91

This is a slightly softer answer.

You can 'do calculus' in-so-far as you can define the derivative and perhaps compute some things. But you'll get no theorems out: the main interval theorems (the Intermediate Value Theorem and the Extreme Value Theorem) rely heavily on the fact that the real numbers are complete, which the rationals aren't. In fact, the 'Intermediate Value Property' is equivalent to the completeness of the reals, and I'm pretty sure the 'Extreme Value Property' is as well.

Going on from there, Rolle's Theorem depends on the Extreme Value Theorem, the Mean Value Theorem depends on Rolle's Theorem, and Taylor's Theorem depends on the Mean Value Theorem.

Going in a different direction, L'Hopital's Rule is typically proven using the Cauchy Mean Value Theorem, which of course depends on the Mean Value Theorem. I don't know if there's some way to prove L'Hopital's Rule without this dependence, but I expect that if it's possible then the proof will depend crucially on completeness.

Above all, much of the usefulness (and beauty) of calculus comes from the theorems mentioned above. So while you can set up the usual definitions in non-complete spaces, and you may even be able to get some partial results, you eventually reach the question: is this really worth calling 'calculus'? It certainly doesn't compare to the real-variable theory.

This is all not-to-mention the lack of a meaningful theory of integrals, which is detailed in other answers.


Addendum: Consider the 'rational complex numbers' $\mathbb{Q}[i]$. If you like, you can extend your rational calculus to $\mathbb{Q}[i],$ but I don't think it will bear much resemblance to complex analysis. As a first example, the fact that satisfaction of the Cauchy-Riemann equations, along with continuity of partial derivatives of real and imaginary parts, implies complex differentiability of a complex function at a point depends, at least in the proof I've seen, on the Mean Value Theorem.

Will R
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    [This pdf](http://faculty.uml.edu/jpropp/reverse.pdf) is related. ​ ​ –  Aug 04 '16 at 04:58
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    @WillR Propp's paper, linked above, proves that the Extreme Value Theorem is equivalent to completeness. (It's a fun paper.) – Patrick Stevens Aug 04 '16 at 13:09
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    L'Hopital's rule couldn't possibly work - for instance, the functions defined by the two properties $$x\in (\pi/n,\pi/(n+1)) \Longrightarrow f_{c,\alpha}(x)=c\pi/n + \alpha x$$ $$f(x)=-f(-x)$$ have that $f_{c,\alpha}'(x) = \alpha$ and $\lim_{x\rightarrow 0}\frac{f_{c,\alpha}(x)}x=c$. Being able to shift intervals with irrational endpoints around at will just creates too much looseness. – Milo Brandt Aug 04 '16 at 14:26
  • @WillR Actually the extreme value theorem is equivalent to the *compactness* of a closed interval. Of course $[a,b]\cap\Bbb Q$ is neither compact nor connected. Connectedness immediately implies that the reals are complete, but I don't think the same can be said of compactness. – Mario Carneiro Aug 05 '16 at 21:20
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    +1 because relaxing hypothesis is cool but it doesn't mean you'll get something useful out of it – seldon Aug 07 '16 at 11:43
  • This answer is not fully satisfying: sure, the theorems of the smooth Calculus will not apply, if you try to find-replace $\mathbb{R}$ with $\mathbb{Q}$ blindly. But it's not clear *why* a meaningful analogue of continuity, Taylor's theorem, etc couldn't be defined for rational-valued functions of $\mathbb{Q}$. – user7530 Aug 09 '16 at 21:04
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It seems like most people are talking about integrals; let me answer with the first thing that popped into my head about derivatives.

$$\lim_{x\rightarrow a}f(x)=L\leftrightarrow(\mid x-a\mid<\epsilon\leftrightarrow\mid f(x)-L\mid<\delta) $$

While this may be a fine definition for rational-restricted limits, it winds up saying nothing about non-rational limits. Which are almost all of them. For example, the sequence {1, 1.4, 1.41...} approaching $\sqrt{2}$ has no limit according to this (rational-restricted) definition.

Likewise if we seek to use the definite integral as the area under some curve, then we find that the rationals are a set of measure zero and therefore the area over that set is automatically zero.

In short, because rationals are a set of measure zero, they account for almost none of the limits on rational functions or sequences, and no area under any curve of interest. I do find that a tiny bit of knowledge of Lebesgue measure theory makes a lot of this stuff immediately clear.

Daniel R. Collins
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    well, that's assuming that one would define integrals in the exact same way as lebesgue did even when working in $\mathbb Q$, which probably won't be true – Ant Aug 04 '16 at 18:31
  • @Ant: I'm skeptical that such a hypothetical, alternative construction would be coherent. – Daniel R. Collins Aug 07 '16 at 14:56
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    The area under a curve is the limit of the area of some trapezoids. Surely, rational numbers can define trapezoids. A 1 x 1 square (integer sides) has area 1, not zero. – Kaz Aug 07 '16 at 15:28
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    All numerical integration done with machines uses rational numbers. That is to say, floating-point. – Kaz Aug 07 '16 at 15:28
  • "The area under a curve is the limit of the area of some trapezoids." But most such limits would not be rational, i.e., would not exist under the proposed alternate definition of limit. – Daniel R. Collins Aug 10 '16 at 04:57
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Based on your comments, I think you might be particularly interested in the theory of real closed fields, or more generally, real algebraic geometry.

There is a formal, logical sense in which all real closed fields are the "same", and two prominent examples of real closed fields are the real numbers and the real algebraic numbers.

By leveraging this "sameness", it turns out that a large fragment of calculus still works the same way if you stick to algebraic numbers and algebraically defined functions.

It's mainly notions like continuity, completeness, or differentiation that carry over well; other techniques can be developed too, but I believe they tend to follow more along the lines of algebraic geometry than that of the calculus of real numbers.

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The simple answer here is that the Riemann Integral doesn't allow for you to have infinite discontinuities like that; this is a strong motivator for some uses of Lebesgue Integration. However, even if we consider the function $$f(x)=\begin{cases}0,&\text{if }x\in \mathbb{Q}\\ 1,&\text{if }x\notin \mathbb{Q}\;. \end{cases}$$ and we try to integrate over the domain $x \in [0,1]$ we find that $$\int_{[0,1]} f(t) dt = \int_{[0,1] - \mathbb Q} f(t) dt + \int_{\mathbb Q} f(t) dt = \int_{[0,1]- \mathbb Q} dt= 1$$
With measure $0$ the rationals simply aren't strong enough for many uses even in this context... the rationals are countable, and you can imagine an integral over the rationals as being an integral over a real function with infinite holes. This breaks many of the requirements for simple integral and derivative properties, fails to satisfy basic differential equations, etc. Now, you can define your own integral over the rationals and find what properties it satisfies, but you'll soon hit your limitations. Note that the concept of the derivative also gets a little tougher to define at any rational point due to the countability of the set, as a result of having to mess with the standard definition of continuity (again, you could simply take the inverse functional of the integral I said you could design earlier, but you will get a fundamentally different derivative)

Brevan Ellefsen
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    This example, though with Q's =1, is one that most easily satisfies the easy reader. We know that the area will be width times height. And as this is 'rectangular', it should also be height time width, so we turn the 'graph' on its side, slide all the strips down to create a rectangle and see apparent contradictions because, seen on its side the graph has zero area. (mind you, the catch 22 is that we start by assuming there are non rationals that matter, and that they are implicitly more numerous!) – Philip Oakley Aug 07 '16 at 19:39
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You can define limits of functions from $\mathbb{Q}$ to $\mathbb{Q}$ and also define limits of sequences with rational values and thus in general any concept based on limit of functions and sequences can be developed easily (almost in line with the way usual calculus is developed with real numbers). In particular it is possible to define derivatives and Riemann integrals.

However such a system of calculus over rationals will be almost totally uninteresting because it will lack all the significant theorems of calculus which are based on completeness property of real numbers. In essence the "calculus over rationals" will be nothing more than just fancy algebra. In general most of the interesting limits will not exist in this system.

It is a rather deep misconception that the power of calculus is because of ideas like derivatives and integrals (differential and integral calculus). The power of calculus comes entirely out of the structure of real numbers and by a sheer act of smartness/shrewdness the importance of real numbers is never emphasized in introductory calculus textbooks and students are left with the feeling that the strange new techniques of derivatives / integrals and their applications are the backbone of calculus.

Paramanand Singh
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  • Your last paragraph really could use justification. – djechlin Aug 05 '16 at 21:50
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    @djechlin: Which particular statement you wish justification for? – Paramanand Singh Aug 06 '16 at 08:06
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    Calculus with rationals is immensely interesting; it is called numerical analysis, and is of tremendous significance in, oh, engineering and science. – Kaz Aug 07 '16 at 15:29
  • @Kaz, a bit exaggerated, but there is a kernel of truth: in most models of computer arithmetic you are further restricted to work only with a specific set, namely the dyadic rationals. – J. M. ain't a mathematician Aug 07 '16 at 15:40
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    @Kaz : calculus **with** rationnal is very different from calculus **on** rationnal. If numerical analysis was about calculus **on** rationnal, you wouldn't try to compute an approximation of $\sqrt{2}$, because such thing doesn't exist. But it makes perfect sense to compute an approximation of $\sqrt{2}$ (a real number) with rationnal numbers. – Tryss Aug 07 '16 at 17:23
  • @Kaz: Numerical analysis is not specifically related to calculus, rather it deals with approximation techniques of various kinds. For example there are techniques to approximately solve a linear system of equations. However the basis (and justification) of most such approximation techniques lies in calculus (the proof that approximations will get better and better as number of iterations increases). I would therefore never think of numerical analysis as calculus over rationals. – Paramanand Singh Aug 08 '16 at 08:25
  • @Tryss I don't follow the reasoning. Suppose we have the function $y = x$, but we remove the $x = 0$ from the domain. Thus the range value $0$ doesn't exist in that function. Yet, the limit of that function as $x$ approaches $0$ is $0$. Similarly, if we are restricted to rationals, $\sqrt 2$ doesn't exist among them; yet we can approximate it as finely as we want. Why not? Moreover, we can still regard the real number $\sqrt 2$ as the (unattainable) limit value. We can find rationals $a$ and $b$ such that $a < \sqrt 2 < b$, such that the $[a, b]$ interval is arbitrarily small. – Kaz Aug 08 '16 at 20:19
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    @Kaz : And if you do that, you do "analysis **with** rationnal numbers", not "analysis **on** $\Bbb Q$". Your sequence has no limit in $\Bbb Q$, and if you consider it's limit in $\Bbb R$, it cannot be considered analysis *on* $\Bbb Q$ no? – Tryss Aug 08 '16 at 20:50
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    @Tryss I'm not sure. Philosphically, we can ask the question, is there a rational number which, if multiplied by itself, equals two? Whether it exists or not, you can give it the notation $\sqrt 2$ and use that to express the limit. I.e. there is a "hole" in the rationals where this $\sqrt 2$ would otherwise be, and we can approach that hole. – Kaz Aug 08 '16 at 20:58
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    @Kaz And if you ask that question you just went outside $\mathbb Q$ and you are not doing Calculus on $\mathbb Q$ anymore. For the same reason, when we build a number $i$ such that $i^2=-1$, that is called complex analysis, it is not real analysis anymore. – N. S. Feb 12 '17 at 02:30
  • @N.S. but we can do algebra over reals despite the need of $ i $, so why can we not do calculus without completing the rationals? – user73236 Nov 18 '19 at 00:02
  • @user73236: completeness is important. Without it you lose all the significant theorems. – Paramanand Singh Nov 18 '19 at 01:41
  • @ParamanandSingh But we could define some $ \epsilon $ calculus, masking the real values with some value arbitrarily close that is rational for IVT and such. It sounds messy, but none of the significant theorems are not approximately true for rationals. – user73236 Nov 19 '19 at 04:17
  • @user73236: here is another catch. All functions have to have range and domain as subsets of $\mathbb {Q} $. That restricts us mostly to rational functions with rational coefficients. No trig, no log and no exponential. Then what do we do with these rational functions which are most adequately handled by algebra alone? – Paramanand Singh Nov 19 '19 at 07:26
  • @user73236: Real numbers are essential for IVT. Consider the sets $A, B$ defined as $A=\{x\mid x\in\mathbb {Q}, x\geq 0,x^2<2\}, B=\{x\mid x\in\mathbb {Q}, 0\leq x\leq 3,x^2>2\} $ and define $f:A\cup B\to\mathbb{Q} $ as $f(x) =1$ if $x\in A$ and $f(x) =2$ if $x\in B$. The set $A\cup B$ is the set of all rationals in $[0,3]$ and we might think it of as a rational interval. Then $f(0)=1,f(3)=2$ and there is no rational $x$ in $A\cup B$ for which $f(x) =3/2$ even though $f$ is continuous on rational interval $A\cup B$ under any sensible definition of continuity. – Paramanand Singh Nov 19 '19 at 07:35
  • @user73236 : If you start to deal with things that can be approximated via rationals to an arbitrary degree of precision then what you get is the system of rationals. – Paramanand Singh Nov 19 '19 at 07:40
  • @ParamanandSingh, the issue isn't cuts or real numbers, the issue is uncountability. Instead of completing rationals from the get-go (uncomputable! I hate that), we should use real numbers approximations only as we need them for a theorem and countably many times in the proof. Anyways what use is a real number without approximation (its absurd). I think IVT requires only one real number to be found. Deep controversial mathematical issues of infinities since Kronecker/Cantor are potentially avoided. Imagine if IVT worked but isn't approximatable over rationals; it becomes useless. – user73236 Aug 06 '21 at 18:07
  • @ParamanandSingh no result of material value (i.e. constructible value) in real analysis (or almost any other branch of math) requires the uncountable, that is my position. The uncountable is only useful for expressing a lack of results (incomputability). – user73236 Aug 06 '21 at 18:17
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Something the other answers haven't really brought up is that it is profitable to study rational functions defined over the rationals using (among other things) some techniques from differential calculus. This is done in arithmetic algebraic geometry (also known simply as "arithmetic geometry.")

Daniel McLaury
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    Can you give a basic example? What kind of problem might this come up in, and what calculation would need to be done? Obviously you don't need to give full details, but your answer could be made much more helpful with just a few extra pieces of information. – Will R Aug 04 '16 at 01:55
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    I'm not sure if I can give a really basic example, but for instance you can take the product rule together with linearity as the *definition* of the derivative for a polynomial, even in contexts where limits don't make sense. If you want to understand what the set of rational points on a cubic curve looks like, this depends crucially on whether the curve is a elliptic curve or not, or in other words if there are singularities. These singularities can be detected with these limit-free derivatives. – Daniel McLaury Aug 04 '16 at 03:08
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    This isn't a representative example, but it takes more than calculus alone to explain the more "normal" applications. – Daniel McLaury Aug 04 '16 at 03:20
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Just some warning this answer may contain traces of engineering math.

You clearly can do calculus on rational numbers since

  1. Computers and electronics are not able to / focused on representing irrational numbers.

  2. There are plenty of useful applications utilizing differential calculus in basically all fields of science and engineering by now.

But one needs to consider at least two facts:

  1. You can approximate real numbers as well as you want or need to with rational numbers.
  2. Differentiation in practice is never purely infinitesimal. Anything that is measurable in our world is when measured (a.k.a. sampled) in some sense a sum or an integral, so whatever you would hope to calculate a differential on is in fact a differential onto an integral or approximation by finite difference of such sums.
mathreadler
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    ...and it is here where the concepts of *truncation error* and *discretization error* become relevant. – J. M. ain't a mathematician Aug 04 '16 at 20:30
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    Yes, absolutely very important concepts. I also had to truncate my answer somewhere or it would easily turn into half a course or more and that would not be very discrete. – mathreadler Aug 04 '16 at 21:10
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    I think it's wrong to suggest that computers can't handle irrational numbers, e.g. [here is a Haskell library for dealing precisely with square roots and sines and cosines of rationals](http://hackage.haskell.org/package/cyclotomic). Moreover I'm not really sure what you mean by "differentiation in practice is never purely infintesimal". Once I model some physical process by a mathematical function, I can differentiate that precisely to model its derivative. I don't need to literally divide rise by run. There's some imprecision, sure, but there already is with any model. – Ben Millwood Aug 06 '16 at 14:43
  • Yes, any irrational number that we are able to express can be stored as the string that expresses it. There are many symbolic algebra packages which can represent and express numbers that are not rational. I should probably have been more specific and mention commonly used examples like the floating point standards. You can do symbolic algebra on the model with a computer and derive expressions how to choose or measure parameters. But once you need to deal with measurements of some sort in an application you will often need to define some way to approximate the derivative or integral. – mathreadler Aug 06 '16 at 16:46
  • @J.M. Truncation error is relevant because of calculations being done with a fixed number of bits. The game changes if you have arbitrary-precision numbers. Man fundamental operations are then exact; addition and multiplication and so on. There are still errors in measurement; but once the inputs are obtained, the machine does exact math. – Kaz Aug 07 '16 at 15:32
  • @Kaz, that looks a bit confused. If you work in a computing system that supports *exact* arithmetic, truncation error is of no concern. With arbitrary precision, you still have truncation error, unless you can tell me about a system that knows every digit of $\pi$ and $e$. That the computer is finite is precisely why there is truncation error in computer arithmetic. – J. M. ain't a mathematician Aug 07 '16 at 15:37
  • @J.M. The inability to represent π isn't a truncation error. That's a representational error. Truncation error means that the machine adds error of its own when performing a computation. For instance, the sum of two numbers isn't exact, because it requires, say 65 bits, but must fit into 64 due to the type being used. The fact that you don't have an exact π in your system is a representational limitation, not the result of a calculation truncation. The machine didn't start with π and then truncate it. – Kaz Aug 07 '16 at 15:47
  • That's different from the definition of "truncation error" that I am accustomed to @Kaz, but we can agree to disagree on that notion. What I am certain is that arbitrary precision arithmetic doesn't totally banish truncation error; you can set up numbers to have as many digits as needed, but you can't pull out any more digits from those. – J. M. ain't a mathematician Aug 07 '16 at 16:08
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By the way, I just recently found that the completeness of the field we're working over is equivalent to the Intermediate Value Theorem for functions over that field. Since the rationals are incomplete, IVT does not apply. I know IVT is necessary with the Extreme Value Theorem to prove the Mean Value Theorem for Integrals, so it would certainly be impossible to prove FTC for the rationals although I don't know that the absence of IVT would have any effect on differential calculus.

Praise Existence
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  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - [From Review](/review/low-quality-posts/665495) – Math1000 Aug 05 '16 at 07:47
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    @Math1000 Disagree, this is an okay answer by the OP, even if it is not perfectly complete, and contains points mentioned in the other answers, it is certainly an answer. – Mario Carneiro Aug 05 '16 at 23:13
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    I have a hunch that the IVT does apply to a rational function that is approximated by straight line segments, connecting points defined at rational domain values (to as fine a resolution as we care to make the approximation). – Kaz Aug 11 '16 at 00:36
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If you handle real-valued functions defined on the rationals, then these functions are obviously not Lebesgue-integrable with respect to the $\sigma$-algebra of $\mathbb{Q}$ inherited from $\mathbb{R}$, except when they vanish outside finite sets. This is because the only sensible measures on the rationals are those that place a weight on each rational number. However, this will guarantee that the traditional derivative and the Lebesgue-antiderivative of a function have completely nothing to do with one another.

On the other hand, there may be some sense in finding the Riemann-integral of such a function, given that it is sufficiently nice. For example, continuous functions (with respect to the topology on the rationals inherited from the reals) may work well. One can simply lend the usual theory and demand that the partitions are done at rational points. Anyhow, I would expect many odd things to happen. Brevan Ellfsen's function would indeed be continuous relative to the rationals, i.e., the zero function, with constant functions as Riemann-antiderivatives, but as Henry Swanson's shows, there are also non-constant functions with this zero derivative.

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On the rational numbers you can not really do Algebra, much less Calculus.

Since the rational numbers are not "complete", you can not for instance define the square root function. Similarly you can not define many other functions, e.g. power functions with non-integer exponent, ln(x), log(x), exp(x), sin(x), cos(x) etc.

Even if you restrict yourself only to functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$, you can not always take limits, derivatives, antiderivatives etc. as pointed out in other answers.

So, the problem is a lot more fundamental than "you'll get no theorems out".

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Can I give my view? For me, there are many huge distinctions between a "rational analysis" and the real analysis, and few of them are dependent on the concept of integration.

Derivatives and integrals are indeed defined on sets which are very different from the real numbers; the Cantor sets for instance. This means that derivatives and integrals are not what makes real analysis real analysis. The monotone convergence theorem and, equivalently, the fact that every bounded sequence has a convergent subsequence is a point of real calculus which I think is prior to integrals. Riemman integrals, themselves, are defined via supremes and infs, or via limits, which exist thanks to the monotone convergence theorem. So, the difference between rationals and reals comes first from facts relating to sequences and series, and then from antiderivatives.

Before learning about integrals, we learn things like mean value theorem or rolle theorem. Both come from topology of the real number set, which is very different from the topology of reals, thanks to the existence of supremes and infs. Mean value theorem would not hold in the rationals set. For example, the square of x would have values above and bellow 2, without passing through 2.

Of course, all of that also makes us think whether real numbers are something fundamentally necessary, or else a matter of convenience. In fact, thinking about all these examples, one can have the impression that real numbers are a construct which allows us to talk about some limits, even when we can't calculate them. This would, in my opition, be a precise observation, and which takes us to the old criticism of Dedekind to Cantor's theory. In my opinion, Cantor's set theory is much more important from being a huge and ingenious notational aid, than from being a fundamentally and filosofically consistent description of mathematics. The same observation applies to real numbers: they make analysis much easier, as we can talk about limiting values (as pi or e) using a single name, symbol, and thinking of them as numbers like 1 or 0, while without completeness theorem we would have to talk about "aproximate behaviours" in each and every place in mathematics were these numbers appear.

The values of many and many integrals are some of these limiting numbers, about which we can talk naturally in real analysis, but not in "rational analysis". Without completeness, we would not be able to talk about the integral of any continuous function on a closed set; we would have to put the huge restriction that the integration results in a rational number.

But, as I said, calculus is all about limits, and integrals are only one of the examples of limits which are only defined thanks to supremes and infs.