[psykotic]

> Many people feel a blend of slight obscurity with succinctness are the core ingredients for an elegant solution.

My exemplar for mathematical elegance is page 3 of Serre's A Course in Arithmetic. In a tidy page of text, it proves what takes many textbooks a long and tedious chapter, introducing important concepts like the Frobenius. It's so short and neat I can reproduce it here in full:

Let K be a field. The image of Z in K is an integral domain, hence isomorphic to Z or to Z/pZ, where p is a prime; its field of fractions is isomorphic to Q or to Z/pZ = F_p. In the first case, one says that K is of characteristic zero; in the second case, that K is of characteristic p.

The characteristic of K is denoted by char(K). If char(K) = p != 0, p is also the smallest integer n > 0 such that n 1 = 0.

Lemma. If char(K) = p, the map sigma : x -> x^p is an isomorphism of K onto one of its subfields K^p.

We have sigma(xy) = sigma(x) sigma(y). Moreover, the binomial coefficient (p choose k) is congruent to 0 (mod p) if 0 < k < p. From this it follows that sigma(x + y) = sigma(x) + sigma(y); hence sigma is a homomorphism. Furthermore, sigma is clearly injective.

Theorem 1. (i) The characteristic of a finite field K is a prime number p != 0; if f = [K:F_p], the number of elements of K is q = p^f. (ii) Let p be a prime number and let q = p^f (f >= 1) be a power of p. Let Omega be an algebraically closed field of characteristic p. There exists a unique subfield F_q of Omega which has q elements. It is the set of roots of the polynomial X^q - X. (iii) All finite fields with q = p^f elements are isomorphic to F_q.

If K is finite, it does not contain the field Q. Hence its characteristic is a prime number p. If f is the degree of the extension K/F_p, it is clear that Card(K) = p^f, and (i) follows.

On the other hand, if Omega is algebraically closed of characteristic p, the above lemma shows that the map x -> x^q (where q = p^f, f >= 1) is an automorphism of Omega; indeed, this map is the f-th iterate of the automorphism sigma : x -> x^p (note that sigma is surjective since Omega is algebraically closed). Therefore, the elements in Omega invariant under x -> x^q form a subfield F_q of Omega. The derivative of the polynomial X^q - X is q X^(q-1) - 1 = p p^(f-1) X^(q-1) - 1 = -1, and is not zero. This implies (since Omega is algebraically closed) that X^q - X has q distinct roots, hence Card(F_q) = q. Conversely, if K is a subfield of Omega with q elements, the multiplicative group K* of nonzero elements in K has q-1 elements. Then x^(q-1) = 1 if x in K* and x^q = x if x in K. This proves that K is contained in F_q. Since Card(K) = Card(F_q) we have K = F_q which completes the proof of (ii).

Assertion (iii) follows from (ii) and from the fact that all fields with p^f elements can be embedded in Omega since Omega is algebraically closed.

> proof courses

I can't think of a single course I took, starting with the first semester, that wasn't a "proof course".

[RegEx]

> I can't think of a single course I took, starting with the first semester, that wasn't a "proof course".

ODE, Cal 1-3 and their labs, Multivariate, Matrix methods. Congrats to you for being able to skip 8 courses, not all of us are that talented.

[kragen]

When I studied the equivalents of calc 1 and 2 (from a textbook), the textbook proved everything, although somewhat informally. I mean, with epsilons and deltas and everything, but with a lot more prose than you see in a math paper on arXiv. The different textbook I later used for calc 3, which I actually did take a class in, also proved everything. As did the professor, in class, on the blackboard. I basically never read the book or did any of the homework for that class; I just rederived things from first principles during the exams, based on my memories of the lectures. I always finished the exams last, but I got an A in the class. This was all in the US. My father's calculus textbook, from which I'd learned calculus to start with, was also from his calculus courses in the US.

I take it my experience was atypical?

[bradleyjg]

No in many US universities rigorous calculus proofs are saved for a class called something like Real Analysis which is typically take as the 4th or 5th class for a math major.

Compare e.g.

Transcendental functions, techniques and applications of integration, indeterminate forms, improper integrals, infinite series.

with

Algebraic and topological structure of the real number system; rigorous development of one-variable calculus including continuous, differentiable, and Riemann integrable functions and the Fundamental Theorem of Calculus; uniform convergence of a sequence of functions; contributions of Newton, Leibniz, Cauchy, Riemann, and Weierstrass.

[RegEx]

Professors at my university do prove things in the classes I mentioned above, but I was attempting to make a distinction between courses where the emphasis is the proof(Abstract/Contemporary algebra) and courses where the emphasis is the process (Matrix methods). Sorry for any confusion.

[psykotic]

That must be a US thing. It seems strange to first do a course without proofs and then later redo it with proofs. My point was that where I studied, the courses for mathematicians were all "proof courses"; there were separate courses for engineers and other scientists who are happy to accept mathematics on faith.

[RegEx]

I'm not arguing which education system is better, but having a degree of familiarity in the subject matter you plan on proving things in helps out a lot. Proving things about continuity/discontinuity would be a bit difficult if you've never had to identify continuous functions with calculus.

[psykotic]

Well, you're never going to deeply understand everything in a subject in only one or even several passes over the material, but there's no reason you can't concurrently treat analysis both formally and informally, with lots of concrete examples and intuition to go along with abstract definitions and proofs.