If I had to pin myself down to a general conceptual area, I would say definitions. Synthesis might come in second—but I suppose that might just be another way of saying I like to steal a whole bunch of other people’s ideas, throw them together with a few alterations, and put my name on it. But hey, it seems to have worked for some people (cf. Alcaeus fr. 332 (Lobel-Page)). I’ve never really considered myself a particularly inventive or original person, though, and that’s frequently been a stumbling block for me when engaging in recreational mathematics.
Creativity tends to be associated with disciplines in the humanities, lately more and more with the sciences (cf. Einstein’s ubiquitous quote that “Imagination is more important than knowledge.”), but almost never with mathematics, probably on account of the fact that few people pursue the discipline further than the socially pragmatic function of numeracy would require. But even apart from that, the sheer invention upon which a fair amount of pure mathematical problem solving depends is difficult. When working proofs, success often hinges upon the ability literally to “build” a mathematical object with the proper attributes (Kudos to Prof. Burger for teaching me the basis of this). Some people find this pretty easy; I don’t. But I do find it incredibly fascinating to see it in action, esp. once the object has been built and the desired properties seem to come tumbling out of it like magic.
The examples I’m most acquainted with come from Number Theory, so here are two number theoretical proofs from this book.
-Prove that there are infinitely many prime numbers.
This elegant proof goes back to Euclid and can usually be found in any introductory book on Number Theory:
1) Assume that the set of prime numbers, P, is finite: P = {p1, …, px} for some integer x.
2) Build the integer M, the product of all primes plus one: M = (p1p2…px) + 1.
The integer M then has no prime factorization, since the division of M by any of our assumed finite primes leaves a remainder of one: For 1< n < x, M/pn is not an integer. This violates the Fundamental Theorem of Arithmetic, which states that every integer greater than one must have a unique prime factorization. Therefore, our initial assumption is false, and the number of primes in infinite.
As you can see, the proof hinges upon the properties of the integer M, which was built in such a way as to demonstrate whether or not our assumption was false.
A related proof that is much more interesting comes from the exercise section:
-Modify the proof of infinite primes in order to show that there exist an infinite number of primes congruent to five modulo six.
As before, we start off by assuming that the number of primes congruent to five modulo six is finite: Q = {q1, …, qr}, where for 1< n < r, qn ≡ 5 (mod 6).
Off the bat, I’ll admit that I had to look in the back to get a hint, and it gave me the proper object to investigate, namely, M = (6q1…qr) – 1. In hindsight, it seems pretty obvious. First, you need to account for all the primes modulo 6, which suggests dealing with their combined product as in the prior proof. Then, you need it to be congruent to 5 mod 6, so that suggests multiplication by 6 and subtraction of 1 for the purposes of factoring out the modulus. It all seems so simple! I guess I just need more practice at that kind of thinking.
Anyway, once you’ve got your object M, the rest just kind of falls out of it. We have built M specifically to be congruent to 5 (mod 6). Now, M > qr , which means that M cannot be prime, since we have assumed that qr is the largest prime congruent to 5 (mod 6). Our task is to prove that M cannot be composite, thereby violating the rule that an integer must be either prime or composite.
1) Start off by considering the hypothetical prime factorization of M:
M = p1a p2b … psg. Then, it’s clear that M = (6q1…qr) – 1 = p1a p2b … psg.
2) No element of Q can be a prime factor of M, since that division of M by any member of Q would leave a non-integer remainder. Therefore, for all pm, such that 1< m < s, pm is not congruent to 5 (mod 6).
3) But since M is odd, all of its prime factors must be odd primes, which means that for all pm, 1< m < s, either pm ≡ 1 (mod 6) or pm ≡ 3 (mod 6).
4) Now, for all integers z > 1, 3z ≡ 3 (mod 6).
5) Let ‘x’ be the sum of the exponents on the prime factors of M that are congruent to 1 (mod 6), and let ‘y’ be the sum of the exponents on the prime factors of M that are congruent to 3 (mod 6). Then,
M = p1a p2b … psg ≡ 1x3y = 3y ≡ 3 (mod 6).
This contradicts the previously established fact that M ≡ 5 (mod 6); therefore, we conclude that our assumption was false and that there is an infinite number of primes congruent to 5 (mod 6). Q.E.D.
Posted by evovae
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