We know that calculus does not take real values and takes variables like infinity and approaching zero. So can we conclude that even calculus cannot define reality and is different from the "real" value?

  • But "where is" the "real" value if we cannot define/calculate/specify ... it ? Jun 9, 2014 at 16:54
  • Good question - but why would you think that calculus would even define reality? Jun 9, 2014 at 21:35
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    What dou you mean? Can you elaborate?
    – Hector
    Jun 15, 2014 at 23:59
  • One should be careful about the word "real" in this context. In the context of calculus, "real" refers to the completion of the set of rational numbers (i.e., all decimal numbers). Sep 23, 2015 at 19:07

4 Answers 4


All physical measurements in the real world are approximate. To the extent that calculus is used to model physical phenomena, it is an approximation to reality [if one believes in an underlying reality at all ... a separate question.]

However, within the scope of mathematics, calculus is exact. That's because calculus is based on a logically rigorous theory of the real numbers, which are based on a logically rigorous account of set theory.

That is, we can start from the Zermelo-Fraenkel axioms of set theory, published in the early 1920s, and we can then develop the theory of the real numbers and calculus from those axioms. In that respect, calculus is an exact theory of some imaginary or fictional entities; since (as far as we know) there are no real numbers in the physical universe.

So, you are entirely correct that calculus is only an approximation to the real world, whatever that may be. But within the domain of pure mathematics, calculus is exact. That's because the real numbers and sets in general are only abstract mathematical entities. They do not necessarily have any analog in the physical world.

  • +1: nice answer. I think its useful to note that its also true that there are other theories of the calculus that are based on generalised set theories; historically speaking, the calculus wasn't exact even in a purely mathematical situation when Newton/Leibniz discovered, which was the crux of the argument by Berkeley; but it was effective which is what a lot, but not all physicists cared about. Jun 10, 2014 at 1:35
  • @MoziburUllah Thanks. I mentioned that ZF was published in the 1920s to indicated that it's a relatively recent, historically contingent development. Before ZF ... was calculus exact? One could argue not. Certainly not before Weierstrass and Cauchy. And if set theory is found to be inconsistent tomorrow morning ... what then? Hard to say.
    – user4894
    Jun 10, 2014 at 16:53
  • Sure, is anyone arguing though its inconsistent? I'm no expert - but I've just thought that there is one historical figure who did exactly that - Brouwer with his intuitionism; and calculus has been placed on firm foundations there - where luckily all functions are continuous - now wouldn't that have made real analysis much easier! Jun 10, 2014 at 17:03

A clarification of when you say 'calculus does not take real values and takes variables like infinity and approaching zero'. This is not correct, and might be leading to confusion.

The central notion in calculus, which underlies all of of the other notions is that of a limit, i.e. the limit of a function at a particular point. For this, we usually write something like, as x→a, f(x)→b, with important instances being f(x)→0 and f(x)→∞. I expect that this is where you get the idea that calculus 'takes variables like infinity and approaching zero'.

But these notions are precisesly defined, and the functions for which they are defined are precise, not approximate.

In particular, the idea that a function tends to infinity at a point means that, although the function itself does not take a value at that point, it takes values arbitrarily close to that point, and we can make the value arbitrarily large by getting sufficiently close to that point. This is spelled out rigorously by what is called the ε-δ definition. (Rather than explain it here, it's probably best to look at the Wikipedia article.)


Calculus is exact because derivatives approach a value indefinitely closely. That is to say, if they are seen as 'limits' then as the increment of the independent variable (i.e. the infinitesimal dx) is reduced any reversals in the approach of the derivative's value to its limiting value are inevitably themselves reversed.

Calculus can take real values - it results in finite difference calculus. Starting with the gradient formula you just leave the 'infinitesimals' and their powers in when you're doing the algebra. On whether conventional smooth calculus (and by implication physics) defines reality or not: not exactly, because there is no perfect circle, or perfect anything. But we do have good approximations, simplifying assumptions, finite element analysis and fuzzy logic to model reality.


When you talk about "real," you start from the implication that some thing can be distinguished as "real" as opposed to "modeled." That distinction is more important than the mentioning of calculus.

To use a Tarskiian truth: Calculus only defines reality if and only if Calculus defines reality. It would be completely valid to state "Calculus defines a reality, which is the reality defined by calculus."

I think the deeper question is whether the reality "you" live in is defined by calculus, and by FAR the hardest part of that question is defining what "you" are. From there, the rest is easy. However, that step has been a source of disagreement for philosophers since the invention of language.

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    Well that just completely went over my head. I am sorry. Jun 9, 2014 at 17:27
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    Perhaps you should clarify your question then. If you're on a philosophy forum, and assume everyone agrees on your meaning of things like "real" and "define," you're in a sink or swim type of domain.
    – Cort Ammon
    Jun 9, 2014 at 20:05
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    "Calculus only defines reality if and only if Calculus defines reality." The way I recently learned about Tarski's definition of truth is: "P" is true iff P. That format exactly. The first P, in quotes, represents some sentence P in a formal language. Syntax only, no meaning. The second P is a proposition, something that has the property of being either true or false once some interpretation of P is provided in some possible world. And "true" is a predicate in a meta-language. Because a formal language can't define its own truth. Just wanted to mention that.
    – user4894
    Jun 9, 2014 at 22:58
  • Good catch, I totally missed putting those in!
    – Cort Ammon
    Jun 10, 2014 at 1:54

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