28
$\begingroup$

The current issue (May 2015) of the American Mathematical Monthly has a one-line proof that there are an infinite number of primes, and I don't see why it is correct.

Here is the proof:

If the set of primes is finite, then

$$0 < \prod\limits_{p} \sin\left(\frac{\pi}{p}\right) = \prod\limits_{p} \sin\left(\frac{\pi(1+2\prod_{p'}p')}{p}\right) =0 .$$

(That's the whole proof.)

I see why the first equality holds, since, if there are only a finite number of primes, $p \mid \prod_{p'}p'$ for all $p$.

But I do not see why the second equality ("$= 0$") holds. None of the terms in the product are zero, and, since there are only a finite number of them, the product is not zero.

So, do I not understand the proof, or is the proof incorrect?

Thank you.

$\endgroup$
  • 3
    $\begingroup$ I really like the comments to Strants' answer. $\endgroup$ – marty cohen Jun 15 '15 at 18:00
24
$\begingroup$

We must have that $1+2\prod_{p'}p'$ is divisible by some prime $q$, so $1+2\prod_{p'}p' = kq$ for some integer $k$. But then, $$\sin\left(\frac{\pi(1+2\prod_{p'}p')}{q}\right) = \sin \pi k = 0$$ which gives the right-hand equality.

$\endgroup$
  • 19
    $\begingroup$ This is a good answer. But...to me the proof reads like a joke, something like an answer to the question "What is the sine of Euclid's proof of the infinitude of primes?" Is there anything serious going on here? $\endgroup$ – Pete L. Clark Jun 15 '15 at 5:45
  • 8
    $\begingroup$ @PeteL.Clark agreed; this is Euclid's proof made abstruse. There may be no "royal road to geometry", but there is no reason to stick a needle in one's foot and hop towards understanding. $\endgroup$ – guest Jun 15 '15 at 6:30
  • 5
    $\begingroup$ @guest : This differs from "Euclid's proof made abstruse" in that Euclid's actual proof did not begin with an assumption that only finitely many primes exist: Euclid's proof was not by contradiction. Euclid showed that (rephrased into modern concepts) for every finite set $S$ of primes (which need not be the smallest $n$ primes for some $n$) the prime divisors of $1+\prod S$ are not in $S$. Thus $S$ can always be extended to a larger finite set of primes. Dirichlet and many later mathematicians erroneously wrote that Euclid's proof was by contradiction. That's a naked emperor. $\endgroup$ – Michael Hardy Jun 15 '15 at 11:56
  • 3
    $\begingroup$ @MichaelHardy Then again, Euclid's proof was only about three primes ... $\endgroup$ – Hagen von Eitzen Jun 15 '15 at 13:42
  • 3
    $\begingroup$ @MichaelHardy: shouldn't your naked emperor be a straw man? $\endgroup$ – TonyK Jun 15 '15 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.