text: edited Aug 10, 2016 at 21:11
  asked Aug 4, 2016 at 0:49
  $\begingroup$ 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. $\endgroup$
  Aug 4, 2016 at 0:52
  $\begingroup$ Rational number systems have holes. It is incomplete. $\endgroup$
  Aug 4, 2016 at 0:53
  $\begingroup$ Classic differential equations like $f'=f$ become impossible to solve. $\endgroup$
  Aug 4, 2016 at 0:57
  $\begingroup$ 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. $\endgroup$
  Aug 4, 2016 at 1:31
  edited Aug 6, 2016 at 21:47
  $\begingroup$ I'd personally use tape and say "good enough." Gluing saw dust together is not for the faint of heart. $\endgroup$
  Aug 4, 2016 at 1:47
  $\begingroup$ 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. $\endgroup$
  Aug 4, 2016 at 2:48
  $\begingroup$ I love your dumb physical metaphor! $\endgroup$
  Aug 4, 2016 at 18:43
  $\begingroup$ @Martin there is no way to extend $f$ to a continuous function on the reals, much less a differentiable one. $\endgroup$
  $\begingroup$ "adding all the glue between the grains to form a stick again" — this is exactly what's called particle board or fiberboard. $\endgroup$
  Aug 6, 2016 at 1:03
  edited Aug 4, 2016 at 2:43
  answered Aug 4, 2016 at 1:39
  $\begingroup$ This pdf is related. ​ ​ $\endgroup$
  Aug 4, 2016 at 4:58
  $\begingroup$ @WillR Propp's paper, linked above, proves that the Extreme Value Theorem is equivalent to completeness. (It's a fun paper.) $\endgroup$
  $\begingroup$ 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. $\endgroup$
  $\begingroup$ @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. $\endgroup$
  $\begingroup$ +1 because relaxing hypothesis is cool but it doesn't mean you'll get something useful out of it $\endgroup$
  Aug 7, 2016 at 11:43
  edited Aug 4, 2016 at 16:20
  answered Aug 4, 2016 at 3:42
  $\begingroup$ 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 $\endgroup$
  Aug 4, 2016 at 18:31
  $\begingroup$ @Ant: I'm skeptical that such a hypothetical, alternative construction would be coherent. $\endgroup$
  $\begingroup$ 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. $\endgroup$
  Aug 7, 2016 at 15:28
  $\begingroup$ All numerical integration done with machines uses rational numbers. That is to say, floating-point. $\endgroup$
  $\begingroup$ "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. $\endgroup$
  answered Aug 4, 2016 at 4:49
  answered Aug 4, 2016 at 1:12
  $\begingroup$ 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!) $\endgroup$
  $\begingroup$ @djechlin: Which particular statement you wish justification for? $\endgroup$
  Aug 6, 2016 at 8:06
  $\begingroup$ Calculus with rationals is immensely interesting; it is called numerical analysis, and is of tremendous significance in, oh, engineering and science. $\endgroup$
  $\begingroup$ @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. $\endgroup$
  $\begingroup$ @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? $\endgroup$
  Aug 8, 2016 at 20:50
  $\begingroup$ @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. $\endgroup$
  Aug 8, 2016 at 20:58
  answered Aug 4, 2016 at 1:50
  $\begingroup$ 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. $\endgroup$
  Aug 4, 2016 at 1:55
  $\begingroup$ 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. $\endgroup$
  $\begingroup$ This isn't a representative example, but it takes more than calculus alone to explain the more "normal" applications. $\endgroup$
  Aug 4, 2016 at 3:20
  $\begingroup$ ...and it is here where the concepts of truncation error and discretization error become relevant. $\endgroup$
  Aug 4, 2016 at 20:30
  $\begingroup$ 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. $\endgroup$
  Aug 4, 2016 at 21:10
  $\begingroup$ 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. 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. $\endgroup$
  Aug 6, 2016 at 14:43
  $\begingroup$ 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. $\endgroup$
  Aug 6, 2016 at 16:46
  $\begingroup$ @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. $\endgroup$
  answered Aug 5, 2016 at 3:39
  $\begingroup$ @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. $\endgroup$
  $\begingroup$ 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). $\endgroup$