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Proof by contradiction
It seems that Euclid also does not claim that 'there are infinitely many prime numbers'. Euclids statement seems much closer to 'for every finite list of prime numbers, there is a prime number not on the list'. What does infinite mean here really? If it means not finite, then the argument by contradiction is hidden in 'therefore there must be infinitely many prime numbers' as stated in this article (Euclid does not need this since he never makes this step it seems). I don't think the comments about argument by contradiction in the article at the moment are helpful, at least not in the context that they are made. When a contradiction is invoked to prove this theorem along Euclids lines, the statement of the theorem is typically a negation of something (the set of primes is not finite). In this case, argument by contradiction is the only way to prove the statement. — Preceding unsigned comment added by David Eklund (talk • contribs) 04:46, 8 March 2012 (UTC)
The main point is that whether or not the argument follows Euclid exactly or some variant is used where the whole statement is a negation (and proven by contradiction) is of rather minor importance. One might note though that there is a significant difference between constructing a prime number not on Euclids finite list, compared to simply assuming that there is no such thing and deriving a contradiction.
- What Euclid actually proved is that given any set of exactly THREE primes, there is a fourth prime not in the list. He says that he's going to prove that the primes are greater than "any given multitude" of primes, but he only addresses the case that the "multitude" contains three primes. Mnudelman (talk) 04:10, 17 October 2013 (UTC)
"If q is not prime then some prime factor p divides q." Doesn't this hold as well if q is prime? I mean, q divides q, right? So why the distinction whether q is prime or not? 188.8.131.52 (talk) 14:12, 25 April 2009 (UTC)
- What Euclid, in the 3rd century BC, meant by "divides" (the translators say "measures") is not clear from the article. But even if you're right, the way it's presented is how Euclid wrote it. Michael Hardy (talk) 17:28, 19 March 2010 (UTC)
I should point out that that q cannot possibly be prime, because P is the product of odd numbers and, therefore, an odd number itself; and any odd number plus one is an even number, and therefore not a prime number. -184.108.40.206 (talk) 15:04, 19 January 2010 (UTC)
- P is not necessarily a product of odd primes; the number 2 can be included in the initial list, and in some cases in which 2 is included, q is prime. Michael Hardy (talk) 17:26, 19 March 2010 (UTC)
I made a deletion of some text from the first paragraph. Perhaps this was a strange bit of defacing. More likely, it was an attempt at adding some formatted text that went awry. Assuming this latter case, perhaps this deletion will prompt the author to return and try to again.--220.127.116.11 (talk) 19:16, 10 February 2010 (UTC)
It would be great if there were some references that attest to the significance of the recent proofs. I'm sure there are tons of proofs of Euclid's theorem, what makes these proofs so special that they need to be included in this article?--345Kai (talk) 03:44, 12 September 2011 (UTC)
Hey... just been looking at the proof using the irrationality of Pi. ... I'm not sure if that series as given is a legitimate thing- tried it out using the first hundred or so primes and got Pi=3.15... There doesn't appear to be any citations in that part of the page, so it's possible that that formula is a hoax. If it is a legitmate formula, can someone please give a citation, and if it isn't, then it should probably be take down. — Preceding unsigned comment added by 18.104.22.168 (talk) 02:13, 28 May 2012 (UTC)
- I agree with 345Kai's statement. The reason you get π=3.15 is because, as stated in the article, "If there were finitely many primes, π would be rational." You mentioned trying the first hundred or so primes; this does not work because as long as you have finitely many numbers inside the formula, you approach but never reach pi. However, your number may not be displayed accurately because there is a limited amount of space on a calculator or computer screen. There is not enough room for an irrational or even a rational number with many digits to be displayed properly, and the number is automatically rounded to a more appropriate size. This is Madness300 04:47, 30 June 2012 (UTC)
New"?" Easier Argument For the Infinitude of Primes?
Providing this argument isn't false (because it definitely is not a proof) it may be the most easy-to-understand argument for the infinitude of the primes.
Rather than ordering numbers by quantity, order them in a different way. Order them like this, where all the letters are the prime numbers (which we know not the quantitative value of yet):
A, B, 2A, 2B, C, 3A, 3B, 3C, 2C, D, 4A, 4B, 4C, 4D, 3D, 2D, E, 5A, 5B, 5C, 5D, 5E, 4E, 3E, 2E, F, 6A, 6B, 6C, 6D, 6E, 6F, 5F, 4F, 3F, 2F, G, 7A, 7B, 7C, 7D, 7E, 7F, 7G, 6G, 5G, 4G, 3G, 2G, H,.......
I think the pattern can be seen by now, provided there are no typos. It's kind of organizing numbers in such a way that they are put in order by groupings of primes rather than by quantity. A is the prime of lowest value, B is the prime of second lowest value, etc.
Using this system, every prime and composite number should be able to be generated eventually, even though they'll be generated out of quantitative order, and some composite numbers will be repeated. (If this were a proof I'd probably have to reference the fundamental theorem of arithmetic here.) Provided all primes and composites are generated by the pattern above, it's pretty easy to see that by following the pattern, you'll have to generate a prime (represented by the letter characters) in order to keep counting once you run out of multiples of previous primes. Therefore the primes are infinite.