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Suppose a picked the positive integer 7. Of course there are more than 7 prime numbers in Z. I just don't know how to show that. How do you show that?

nyorkr23
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    This is Euclid's classical proof: consider the first $n$ primes, multiply them, and add one. Any prime dividing that number must be different than the primes you started with. – Ittay Weiss May 07 '16 at 20:39
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    The primes are infinite, as was proven by Euclid in his very beautifull proof. What you are asking derives directly from that. – MathematicianByMistake May 07 '16 at 20:39
  • You can try to show any finite list of primes is incomplete by making out of such a list a number with prime factors which can't be in the list. – Bernard May 07 '16 at 20:43
  • If there were less than $n$ primes, say exactly $k$ primes, then it is possible to create a new prime like how Euclid did. Which is a contradiction. – peterwhy May 07 '16 at 20:48
  • I see Euclid's proof, and they said when there are $k$ primes and I make a new number $p$ with $k$ primes plus $1$, and divide $p$ by one of the $k$ primes, let's say $km$, I get a remainder of 1. The proof states it's a contradiction, but why is it a contradiction? – nyorkr23 May 07 '16 at 20:49
  • The contradiction arises because N = 1+product is > 1 so it has a prime factor p, but p isn't in your original list of primes since N leaves remainder 1 when divided by any of them. It's better not to use contradiction. Instead, it shows that given any finite set of primes you can construct a new prime by choosing any prime factor of N = 1+product. – Bill Dubuque May 07 '16 at 20:55
  • Because it contradicts the fundamental theorem of arithmetic that every positive integer can be expressed as a product of primes in a unique way – joy May 07 '16 at 20:56
  • @joy No, it has nothing to do with FTA. – Bill Dubuque May 07 '16 at 20:58
  • Yes, it does because if we have only finitely many primes to begin with then every positive integer is a product of those primes only. So, if we consider the number $p_1...p_k + 1$ in particular, then none of the $p_i$ divides that number and hence it must have a prime factor other than those $p_i$ – joy May 07 '16 at 21:02
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    @joy No it doesn't, because Euclid's idea works much more generally in rings that are not UFDs, e.g. see here. Using FTA here is extreme overkill.. All that is needed is that every integer > 1 has a prime factor, which is much more elementary. Moreover, this result is usually proved long before FTA in many textbooks. – Bill Dubuque May 07 '16 at 21:14
  • Ok. I got it. Only the existence of a prime factor is what is required here. We don't need to bother about the uniqueness of the representations. – joy May 07 '16 at 21:17
  • @joy Right, and that's trivial since the least nontrivial factor of any integer > 1 is necessarily prime. – Bill Dubuque May 07 '16 at 21:21
  • Maybe, what the OP bothers is : Can we assume that we have exactly $n$ primes for every $n$ ? But since the number of primes is not finite, it is trivial that we can do that (we can just choose the smallest $n$ primes). – Peter May 07 '16 at 21:30
  • The easiest way to prove that there are infinite many primes is : First, show that every number $n>1$ is divisible by a prime. This is true because the set of divisors $d>1$ with $d|n$ is non-exmpty because of $n|n$. The smallest element of this set must be a prime. Then, assuming $p_1,...,p_n$ were all the primes, the number $p_1\cdot ...\cdot p_n+1$ is not divisible by any prime, contradicting that every number $n>1$ is divisble by a prime. – Peter May 07 '16 at 21:33
  • This argument only uses the fact that every non-empty set containing natural numbers has a smallest element. Here, the set is even finite, so it is obvious that it has a smallest element. With this argument, we do not even need FTA. – Peter May 07 '16 at 21:37
  • @Peter Yes, that's the standard proof, but contradiction is unneeded and it greatly complicates the proof for beginners (and it spoils the beautiful constructivity of Euclid's original proof). There's a simpler proof in my link above. – Bill Dubuque May 07 '16 at 22:33

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The Wikipedia article on "Euclid's Theorem" (https://en.wikipedia.org/wiki/Euclid%27s_theorem) gives several proofs of the theorem which are very accessible. Some of them were entirely new to me. (Does everyone but me know of the beautiful proof by Furstenberg?)

P.S. I can't comment on this site, but i just saw your question about the contradiction at the heart of Euclid's proof. It's very simple: at the beginning you assume that there are k primes -- what is meant is that there are EXACTLY k primes (k is some particular number). But then Euclid produces a number which has a non-zero remainder when divided by every one of the k primes. If there were only exactly k primes, then that new number would have to be divisible by one of them. But it's not -- so either there is some OTHER number, not among the list of k primes, that is a prime divider, or that number itself is prime. Either way, it's just not true that there are exactly k primes.

P.P.S. Not that I particularly care, but I notice that within seconds of posting this answer, there were two down votes. I found that strange: I actually answered his actual question, which was "how do you show that?" There are many many ways to show the result, only one of which is ascribed to Euclid. I gave a link to a wide variety of ways to "show" that. And then I answered his particular question about how Euclid showed that. I can only assume that the down votes were because the original hyperlink led to some error page. Well, that error page was erroneous (you can't believe everything you read in Wikipedia :)) -- there IS an article in Wikipedia with the exact name I gave. In any case, the link is now correct. Perhaps the hater-voters can unhate their previous votes, or explain the hate. Thx!

Scott Petrack
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