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Engineer's honest question to a mathematician: is this proof more simple than the one the multiplies all the supposedly finite primes and shows that their produ
by buzzdenver 10y ago
Engineer's honest question to a mathematician: is this proof more simple than the one the multiplies all the supposedly finite primes and shows that their product plus or minus one is not divisible by any of those primes, ergo there are always more primes ?
- quantumtremor 10y agoI'm not a mathematician but I'd say no since trig functions inherently require a lot more math to formulate, whereas Euclid's proof follows from much simpler facts.
- schoen 10y agoI see that proof, due to Euclid, as simpler because it doesn't require any facts about trigonometric functions. It might be harder to write out in mathematical notation in one line, but maybe I should try (using set-builder notation or something).
- quantumtremor 10y agoWe know it can be written out as a haiku: https://xkcd.com/622/ https://xkcd.com/622/!
- schoen 10y agoWhile that's very clever, it doesn't accurately present Euclid's argument because the haiku is Top prime's divisors' product (plus one)'s factors are...? Q.E.D., bitches! This doesn't include Euclid's argument about multiplying all of the primes, mistakenly referring instead to "top prime's divisors". The "top prime's divisors' product" would be equal to the top prime itself, so Randall's haiku asks "if there is a largest prime p, what are the divisors of (p+1)?" which doesn't create any contradiction (it could simply be divisible by various smaller primes!). Maybe we should amend it to Take factorial of top prime, then add one: what are the divisors?
- adrianratnapala 10y agoIt's not the factorial either, that would include composites in the product too.
- schoen 10y agoEuclid didn't use the factorial in his original proof, but it still produces a logically correct argument and it has fewer syllables. Metri causa. :-)
- JadeNB 10y agoUsing the factorial is better, I think, even if ahistorical; it avoids the slight unpleasantness of having to show that every non-unit integer is divisible by a prime.
- wangarific 10y agoBut then you can't exclaim Q.E.D., bitches! and it would be a terrible cartoon. :)
- schoen 10y agoIt's restored in a version further down in the thread.
- JadeNB 10y agoCouldn't you just change "divisors' product" to "factorial's"? I guess that it does some violence to the metre.
- schoen 10y agoYou could work with it and get the original last line back: Factorial of top prime, plus one: factor that! Q.E.D., bitches!
- mikeash 10y agoDepending on how much you're willing to assume and what exactly counts as "one line," it could be as simple as: > One plus the product of all primes is itself prime, and larger than all primes. This omits a bunch of stuff, but it seems like less than the linked paper.
- dllthomas 10y ago> One plus the product of all primes is itself prime One plus the product of the first N primes is not necessarily prime. Consider 2 * 3 * 5 * 7 * 11 * 13 + 1 = 30031 = 59 * 509
- mikeash 10y agoHow about: one plus the product of a list of primes has a prime factor not in the list.
- jwatte 10y agoYes! Because primes are infinite! The statement assumes limited primes. And, because it falsifies, it proves primes must be unlimited. I find this argument based on prime factors much simpler.
- dllthomas 10y agoI would have been fine with "would have to be prime"; I like the edit Mike suggested even better.
- deleted 10y ago
- yequalsx 10y agoPeople tend to think of the proof you are thinking of as easier because they are more familiar with it. There are some subtle things that the proof you mentioned glosses over. Namely, if p|A and p|B then p|(A-B). Now this is not a deep concept but it is something which people who are not accustomed to thinking of proofs tend to gloss over. Namely, they will see the part that if P = p1 p2 p3...pn + 1 and p|P then p|1 and not think that this does itself require justification. Personally, I really like the proof using sines. I won't claim it is easier or simpler but it is very nice. I've never thought to prove the infinitude of primes this way and it uses facts from trigonometry. I will now incorporate this into my trig classes. I think it is quite straightforward and easy for one to understand. Students in trigonometry and calculus generally do not have experience in proving statements about arithmetic. Here is an excellent way to show a connection between concepts that seemingly have nothing to do with each other. It is this that mathematicians really like.
- JadeNB 10y ago> Namely, they will see the part that if P = p1 p2 p3...pn + 1 and p|P then p|1 and not think that this does itself require justification. While your point that every step of a proof requires justification is a good one, this is really, really easy to verify. That doesn't excuse one from actually giving that justification, in the course of a formal proof; but there's at least as much implicit knowledge bound up in the linked proof. (For example, one needs the existence of sine, its 2π-periodicity, the fact that it is positive from 0 to π/2 ….)
- yequalsx 10y agoYes, definitely there is more information implicitly wrapped into the sine proof. That's why it's not simpler. However, it is not a worse proof either. It's just a different one that is clever and can help to reinforce concepts from trigonometry.
- bonoboTP 10y agoThere's not much trigonometry in this proof, just the fact that sine has period 2pi is zero at pi. Any function that has these two properties could directly replace sine in this proof. And actually any periodic function that is zero somewhere could be used in a tiny bit modified version of this proof.
- adrianratnapala 10y agoAs others have said, the proof relies on some trig facts (Though maybe you can replace `sin` with any zero-crossing periodic function). But it's worse than that. The only proof I can see for the second equality there depends on showing the numerator is composite. And that in turn depends on something very like the argument you mention. So unless there is a much simpler proof of that equality, the proof is only a "one liner" because it leaves Euclid's argument as an exercise to the reader.
- deleted 10y ago[deleted]
- blevinstein 10y agoThis proof is nearly equivalent to that proof, but uses the sine function to generate a contradiction. The crucial step involves the fact that: 1 + 2 * product(p', p') must be divisible by some prime number, where product(p', p') is the product of all primes you were talking about.