This question is related to another question, Do we really need reals?, and could be considered a duplicate, so I would not be surprised if it will be put on hold. But I'm especially interested in the teaching aspects of the problem so I ask it in the following form.

An anecdote.

Years ago, when I was a high school teacher, I used to introduce real numbers showing first that $\sqrt{2}$ is irrational and that there are infinitely many algebraic numbers of the same type. Then I used to add (obviously without any proof) that there are other numbers, such as $\pi, e, 2^{\sqrt{2}}$ (said transcendental) that are not algebraic. All of these new numbers have a non periodic representation and, added to the rationals, form the set of real numbers.

To taste the beauty of mathematics, I was then used to sketch the Cantor's diagonal proof, to show that the real numbers are much more numerous than the rationals and form a set called continuous.

Once a student asked me if were the transcendental numbers (as $\pi, e ...$) that make the set of reals continuous. I was a bit uncomfortable and I thought about it for a while before I gave an answer; finally the answer was: NO, we don't really know the numbers that make the reals continuous because those numbers are not computable. The student was a bit astonished by that answer and he commented that mathematics was not such an exact knowledge as he hoped.

After that day I was convinced that students have to be exposed with caution to the mysteries of real numbers.

Now the question.

What is the minimal extension of the rational field that we need to teach (and learn) the calculus at a beginner level?

My guess is that is enough an exponential extension $\mathbb{E} / \mathbb{A}$ of the algebraic numbers field $\mathbb{A}$ considered as a subfield of the complex numbers $\mathbb{C}$ and constructed as in Exponential extension of $\mathbb{Q}$$.

As shown in that post, such a field is countable and all its elements are obviously computable. As far as I know, we don't know if $e$ is an element of that field, but however if we add it to $\mathbb{A}$ (possibly with some other helpful transcendent numbers) the field closure is anyway countable and its exponential estension is entirely computable.

  • 4
    $\begingroup$ You maybe should have asked the student what she or he meant by the reals being “continuous” (a property usually attributed rather to functions)? Maybe she or he meant “connected”, and in this regard, was sort of right as the reals are the metric/order completion of the rationals yielding them connected (as opposed to totally disconnected – the state in which the rationals are). $\endgroup$ – k.stm Dec 28 '14 at 20:09
  • 3
    $\begingroup$ On some profound level we know all the reals pretty well. It's just that giving some (even most) reals a name is impossible. It's like fish. We know a lot about fish, but most fish we will never ever see, but we still know fish. $\endgroup$ – Ittay Weiss Dec 28 '14 at 20:21
  • 1
    $\begingroup$ So rereading the question, I think you need to clarify a few things: (1) What do you mean by the reals being “continuous”? (2) What do you mean by a number being “computable” (having a finite $b$-adic representation for some base $b = 2, 3, …$)? (3) What kind of calculus are you trying to teach? (Is it high school level still?) The intermediate value theorem will be false if you take any proper subfield of the reals, so you should state what results/mathematics you want to impart. $\endgroup$ – k.stm Dec 28 '14 at 20:22
  • $\begingroup$ @k.stm well. for "continuous" see encyclopediaofmath.org/index.php/Continuous_set. Not much used but expressive. $\endgroup$ – Emilio Novati Dec 28 '14 at 20:27
  • 1
    $\begingroup$ Okay, I now get your question – interesting. It seems to me that you/your student were unpleased with that you cannot list, compute or even define all of the numbers which are responsible for the reals being connected (which are, of course, all real numbers). But I don’t think you can say we don’t know the numbers which make the reals continuous: It’s not like the question of whether a given real number is responsible for the reals being continuous is undecidable – the answer is always “yes”. $\endgroup$ – k.stm Dec 28 '14 at 21:04

I would suggest the following set:

All the numbers that can be calculated using a formula which contains a finite amount of:

  • Natural numbers
  • The basic arithmetic operations ($+,-,\times,\div$)
  • The infinite-repetition operator (e.g., $\sum\limits_{n=1}^{\infty}$ or $\prod\limits_{n=1}^{\infty}$)

In fact, you only need $\left[1,+,-,\sum\limits_{n=1}^{\infty}\right]$ but I wanted to keep the definition above simple.

In any case, this set contains all the algebraic numbers, as well an infinite amount of transcendental numbers (including $\pi$, $e$, etc).

I'm pretty sure that this amount is countable, since we are using a finite amount of symbols in order to represent every element in the set, but I'm not sure how to prove it.


After positing a related question, I have realized that such set has already been defined (the credit goes to a comment made by @PeterFranek).

It is the set of computable numbers, which contains many of the specific real numbers that appear in practice, including all real algebraic numbers, as well as $e$, $\pi$, and many other transcendental numbers.

You may want to focus on the section which refers the rather philosophical question of whether or not the computable numbers can be used instead of the real numbers.

| cite | improve this answer | |
  • $\begingroup$ Such a field/number system contains $ℚ$ and if it’s closed under $\sum_{n=1}^∞$ (which is the same as taking limits as far as I can tell), doesn’t this imply it contains all the reals? $\endgroup$ – k.stm Dec 28 '14 at 20:26
  • 1
    $\begingroup$ You can write every real as an infinite sum of rationals. $\endgroup$ – user98602 Dec 28 '14 at 20:28
  • $\begingroup$ @k.stm: My point is, that there are real numbers which we cannot represent in any known method (the uncountable "majority" of $\mathbb{R}$. I might be wrong in explaining that, and will be happy to know in what way (if you can explain). $\endgroup$ – barak manos Dec 28 '14 at 20:28
  • 1
    $\begingroup$ @EmilioNovati: First of all, you provided a post that has received no answers or voting, so I think that it is hardly something to rely upon at this point. Second, your $P_F$ there is by itself a set of real numbers if I understand correctly. I did not allow the use of those symbols in my definition. You can use only what you're able to compute (even $\pi$ and $e$), but you cannot use, for example, "for every real number $r\in[0,1]$". $\endgroup$ – barak manos Dec 28 '14 at 20:51
  • 1
    $\begingroup$ @barak It's an open question to see if the construction that you proposed build up all the field of computable numbers. The usual algorithmic definition of such numbers is not so easy for beginner students. But I accept your answer. $\endgroup$ – Emilio Novati Dec 29 '14 at 15:36

For the basic notions of the calculus, like continuity and limits, you don't need the reals if you are happy to substitute them for something abstract. There are two ways this can be done. One is topology, but this is almost certainly not going to appeal to someone who did not already know enough calculus. The axiomatics of topology allows you to speak rigorously of the basic notion of calculus without mentioning the reals. Another possibility is to generalize metric spaces. Classically a metric space takes values in the reals, but you can replace the reals by what is called a value quantale. This axiomatization is much more easy to digest, so it can be used to introduce metric spaces without the reals, and again introduce the common notion of the calculus.

You are what is the minimal extension of the rationals needed to speak of calculus. Well, it would seem that a crucial property to have is that whatever the extension is it must be a complete lattice. Any complete lattice extension of the rationals must contain the reals, so the minimal such would be the reals.

| cite | improve this answer | |

It is possible to teach a form of differential calculus entirely algebraically. For functions $f, g$ of one variable define as follows:

Let $x'=1$ (for $f(x)=x$)

and impose linearity so that $(af(x)+bg(x))'=af'(x)+bg'(x)$

And the product rule $(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$

The difficulty here is in making it useful, because there is no natural interpretation (gradiant of graph) to hand. [It is also important to ensure that the definition is consistent - for example if $h=fg=de$ then the two products give the same result.]

It is easy enough to show that this will detect double roots of a polynomial. It is also possible to show that $f(x)$ is monotonic near $x$ when $f'(x) \neq 0$, and that you can recover a polynomial from its derivatives (Taylor Series).

But this algebraic definition comes unmotivated, and generally appears rather later in mathematical development when the motivation is clearer.

| cite | improve this answer | |
  • $\begingroup$ While I’m a fan of differential algebra (which is the term I learned for what you described), it’s probably not a substitute for calculus in a school setting. You really want things like the mean value theorem, and without reals (or, if you take a rather complicated route, constructible reals), it’s just plain wrong. $\endgroup$ – Christopher Creutzig Dec 28 '14 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.