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The mismatch between the sensitivity of 'mathematical calculus' and the flexibility of 'real world calculus' has been bothering me a bit recently. What I mean is this: in the real world, I can trust that calculus will work whether $\mathbb{R}$ is "actually real" or not. "Continuity" can be considered as relative to a certain scale, and this is good enough for many purposes. Mathematically of course, this is not the case. Weakening even to $\mathbb{Q}$ will cause, say, the intermediate value theorem to fail. But I guess what I'm really wondering is whether or not this is just a problem with the chosen definitions involved, and if a clean formalism exists that more accurately captures how/why calculus is so widely applicable in real-world problems. And so I come to my main questions:

  1. If we replace the idea of exactly hitting a number (as used in the IVT, EVT, and MVT) with "getting arbitrarily close to it", can we still cleanly and/or consistently develop calculus? It seems like $\mathbb{Q}$ could support some form of calculus with this, since as far as I can see the whole '$\forall\epsilon\exists\delta$' paradigm is left intact. The normal counterexample to the IVT in $\mathbb{Q}$ is given by asking for roots of $y=x^2-2$, but this would be avoided because $y$ gets arbitrarily close to $0$. Moreover, restricting to smaller and smaller intervals where $y$ gets arbitrarily close to $0$ can also show that it doesn't get arbitrarily close to any other number "at the same time". Could this be considered continuous?
  2. Is there a form of calculus that deals with finite error without actually lugging explicit error terms around (i.e. a 'fuzzy calculus')? As in "any differences less than the tolerance $\epsilon_0$ are ignored"? I suppose this is somewhat related to nilpotent infinitesimals/dual numbers and big O notation, but since these still consider numbers of infinite precision they aren't quite what I'm looking for. I'd imagine that there's no way of doing this without getting into the same ugly issues of floating point numbers, but I figure it doesn't hurt to ask.
Robert Mastragostino
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  • Related: http://math.stackexchange.com/questions/387234/how-far-can-one-get-in-analysis-without-leaving-mathbbq – Asaf Karagila Jul 24 '13 at 12:46
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    The graph of $y=e^x$ will only have one point on it over the rationals, $(0,1)$. How will you solve $y'=y$? – Gerry Myerson Jul 24 '13 at 12:50
  • perhaps there again could be some sort of an approximate solution. Anyway, if you restrict youself with rationals only, the whole concept of a derivative has to be reworked somehow since this is a limit, but i've not figured out yet how to do so ... i mean, well, if the limit turns out to be irrational, should we then conclude it doesn't exist? – W_D Jul 24 '13 at 13:08
  • There's also $p$-adic calculus. – Michael Hardy Jul 24 '13 at 13:09
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    Your point one is precisely how $\mathbb{R}$ is constructed in most text books. $\mathbb{R}$ is just convergent sequences of rational numbers, i.e. approximations. For example: The solution of $x^2=2$ is the approximation $3/2, 17/12, 577/408, 665857/470832, ...$ and its equivalent ones. – OR. Jul 24 '13 at 13:23
  • @RGB Wow, I've been using Cauchy sequences for a long time and that never really clicked before for some reason. I worry though that this simply means that I've asked my question poorly. While saying "point 1 is equivalent to calculus in $\mathbb{R}$" is certainly very enlightening, it somewhat serves to reinforce the point that mathematical calculus depends on $\mathbb{R}$ (more aggressively than I thought!) while "real-world calculus" functionally does not. It is this apparent gap that I really wanted to get into. – Robert Mastragostino Jul 24 '13 at 13:34
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    Yes, it can be done. And then, to express things more cleanly, we would have to invent the reals. – André Nicolas Jul 24 '13 at 13:49
  • I think the point in which things become non-trivial is not being able to approximate with arbitrarily good precision, or certain fixed precision, but rather when one says: Let's take all approximations (all convergent power series), and we mod out the equivalent ones. What does it mean "all approximations"? In real world it seems to be we only have access to some approximations, those we construct by some procedure. – OR. Jul 24 '13 at 22:12
  • @GerryMyerson The solution would be y(x)=0. – Christopher King Nov 16 '13 at 23:26
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    @PyRulez, that "solution" misses the one and only rational point, $(0,1)$. – Gerry Myerson Nov 17 '13 at 04:27
  • @GerryMyerson It actually goes through a lot of rational points. Just not that one. – Christopher King Nov 17 '13 at 13:42
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    @PyRulez, it misses the one and only rational point on the graph of $y=e^x$. – Gerry Myerson Nov 17 '13 at 23:15
  • $e$ isn't a rational number. The solution to $y'=y$ is $y=0$. – Christopher King Nov 18 '13 at 23:43
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    You should check out Non-Standard Analysis. The idea that there exist "hyperreal" numbers that get "infinitely close to, but not equal to" real numbers is an intriguing one, even at young age. Just ask high schoolers whether or not $.9999999\dots$ equals $1$. – chharvey Dec 09 '13 at 01:23

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In general, this is more of an question on the epistemology of mathematics. Einstein answered this with "as far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality."

But more on the question itself: The key difference is the completeness axiom. That is, any set of real numbers has a lowest upper bound. This doesn't hold for the rationals. For example, the set $\{x \in Q : x^2 < 2\}$ doesn't have a lowest upper bound in Q. This axiom underlies (almost?) everything in real analysis.

Tyler
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  • I think it'd be cool if you described something that actually fell apart without completeness. –  Dec 09 '13 at 01:24
  • To what extent? I don't think I can prove that not completeness axiom implies not X, where X is some important theorem. But, in IVT, for example, given $f : R \to [u,v]$ and some $w \in (u,v)$, we pick $c:f(c)=w$ by defining $c=sup{x:f(x)<w}$. And the reason the supremum exists is this case is the completeness axiom. – Tyler Dec 09 '13 at 01:31
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    Suppose the completeness axiom fails; let $S$ be a non-empty bounded above set without a supremum. Define $f(x)$ to be $1$ if $x$ is an upper bound of $S$, and $0$ otherwise. Then $f$ will be a continuous function for which the intermediate value property fails. – Oskari Virtanen Dec 09 '13 at 16:14