2

I will post my own answer below. That should not deter others from answering. There are many ways to prove this.

Michael Hardy
  • 1
  • 1
    Most of your question looks like a meta question. It should be a comment. – Git Gud Jul 22 '14 at 23:13
  • See this. – David Mitra Jul 22 '14 at 23:14
  • 1
    Someone posted this question within the past hour, and several people instantly responded in comments, saying "Don't you know Euclid's proof?", etc. Within minutes, the original poster deleted the question. I think those who posted comments ought to have posted answers instead, and the poster should have left the question intact. It appears that a newbie might have been made to feel that the question was inappropriate. But it's a far better question than many that appear here. – Michael Hardy Jul 22 '14 at 23:21
  • @GitGud : I have now edited the question. – Michael Hardy Jul 22 '14 at 23:21
  • 3
    There are really seven prime numbers, they just run around very fast. – Will Jagy Jul 22 '14 at 23:22
  • If you accept the assumption that there is a largest natural number, then there are only finitely many prime numbers. If you don't, you have a bone to pick with Zeilberger. – Asaf Karagila Jul 22 '14 at 23:27
  • I see that at least two people have down-voted the question. Is there a reason for that? – Michael Hardy Jul 22 '14 at 23:32
  • @Michael Maybe the question would have been better received if you had given a compelling reason for posting yet another duplicate of one of the most frequently asked questions on the site. I don't see here any compelling reason to do this (what you write in your answer is already mentioned in many other answers) – Bill Dubuque Jul 22 '14 at 23:37
  • @BillDubuque : I didn't recall seeing the question posted here before. I'm not convinced it's asked frequently. However, if we cannot create redirect pages such as those on Wikipedia, maybe we should encourage the following: Someone deliberately posts a duplicate question, with the title phrased differently from the original, then notes in a comment that it's intended as a redirect to help people find a question phrased differently from the way they'd have have thought to phrase it. Those who agree could up-vote the question and vote to close it as a duplicate. – Michael Hardy Jul 22 '14 at 23:41
  • @BillDubuque : Your comment seems to suggest that a question cannot be well received by those who vote to close it as a duplicate. I don't see that. Is that actually what you're suggesting? If so, why? – Michael Hardy Jul 22 '14 at 23:43
  • Good question.! Vote up! – johannesvalks Jul 22 '14 at 23:46

1 Answers1

5

The number $1+\Big(2\times17\times61\times97\Big)$ does not have $2$, $17$, $61$, or $97$ among its prime factors, because when divided by any of them, it leaves a remainder of $1$. Either it is prime, or it has prime factors not among those listed. Either way, there is at least one more prime than those listed.

With any finite set of primes, you can find at least one more prime not among them, by proceding as above.

This argument was written by Euclid of Alexandria (after whom Euclidean geometry and the Euclidean algorithm are named) in the third century BC.

Many authors rearrange it into a proof by contradiction, thus: suppose only finitely many prime numbers exist; multiply them and then add $1$, getting a number not divisible by any prime numbers. But every positive integer (except $1$) is divisible by some primes, so our assumption that only finitely many exist has led to a contradiction. However, making it a proof by contradiction merely adds an extra complication that doesn't help. Many authors who write it that way mistakenly say that Euclid wrote it as a proof by contradiction. Mathematicians are better at mathematics than at history.

In many published accounts, instead of an arbitrary set of prime numbers, one uses the smallest $n$ prime numbers, for some $n$. The conclusion is then given thus: there is always at least one prime number larger than the one used at the outset. But Euclid merely said there is always at least one other than the ones used at the outset. I now see that the number mentioned above, $1+\Big(2\times17\times61\times97\Big)$, is divisible by $11$. In some cases, some or all of the primes other than those in the initial set are smaller than all of the primes in the inital set. That happens, for example, if the initial set is $\{5,7\}$.

Michael Hardy
  • 1
  • I see that the number I started with here is divisible by $11$. – Michael Hardy Jul 22 '14 at 23:26
  • 3
    You know, I'm embarrassed to say, but not too proud to say, that I never really understood exactly why $1+p_1p_2\cdots p_n$ isn't divisible by any of the $p_k$ ($1\leq k\leq n$). Now, thanks to your answer to this elementary question, I understand it perfectly. Thank you for that very enlightening answer. +1. – MPW Jul 22 '14 at 23:26