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A Ridiculously Advanced Proof of a Simple Theorem
- henning 18y agoLet's see if I still remember second year math. Suppose the number of primes is finite, call them p1, p2, ..., pn. Let P be their product, P = p1...pn. Since P > pi for all i, P+1 is not prime. Therefore some pk divides P+1. but pk also divides P since it is in the list of primes multiplied. Sp pk|((P+1)-P), i.e. pk|1, a contradiction. Therefore the number of primes is infinite. That proof depends on the fundamental theorem of arithmetic, and a theorem that says that if p|X and p|Y then p|(X-Y).
- sysop073 18y agoI'm not sure where "Since P > pi for all i, P+1 is not prime" comes from, but maybe I'm just misunderstanding you. I think you're trying to state this proof: http://en.wikipedia.org/wiki/Prime_number#There_are_infinitely_many_prime_numbers http://en.wikipedia.org/wiki/Prime_number#There_are_infinite...
- PieSquared 18y agoActually, I'm pretty sure that that proof doesn't use the Fundamental Theorem of Arithmetic. FTA states that every natural number has exactly one prime factorization, up to sign, and this proof doesn't use that.
- sh1mmer 18y agoI was interested in the comment on the blog post, that you should be able to validate any proofs you use or at least have a good excuse. I think this is a general good practice for life. Extolling the work of others that you don't understand, will not often end well. Particularly if someone questions you about that work and it's implications.