Can we teach calculus without reals? 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.
 A: 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. 
A: 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.

UPDATE
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.
A: 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.
