My motivation for asking this question stems from Euclid's elegantly simple proof of the infinitude of prime numbers. I am not suggesting an alternative proof, since my method, even if is valid, requires assumption of the fundamental theorem of arithmetic, which Euclid's proof does not require. I just thought that this was a fun way of finding an additional prime number.
Given a set S of prime numbers, find an additional one as follows. Express S as the union of two non-overlapping subsets P and Q. Let m equal the product of the members of P and n equal the product of the members of Q. Then m+n is a number that is not divisible by a number in S.
Proof: For m+n to be divisible by a number in S, it must be divisible by some number in P or Q. It can't be divisible by some value p in P, since m is divisible by p but n is not. Similarly, m+n is not divisible by a number in Q. Therefore m+n is not divisible by a number in P or Q and therefore not by a number in S.
Asked
Active
Viewed 94 times
5
user1153980
- 1,188
-
1Yep that works and doesn’t use anything fancy (all you need is that $p|ab$ implies $p|a$ or $p|b$) – Eric Apr 06 '24 at 18:54
-
1What if the set S consists of the primes 3 and 5? 8 is relatively prime to the members of S, but not prime. – Andy Walls Apr 06 '24 at 18:54
-
4@AndyWalls I think the assumption is that we get the next prime number by taking some prime factor of $m+n$. In your case we get $3+5 = 8$ and then the prime $2 | 8$ is indeed a new prime not in $S$. – Jair Taylor Apr 06 '24 at 18:56
-
Example : $S={2,3,5,7}, P={2,3}, Q={5,7}, m=2.3=6, n=5.7=35, m+n=41$. Moreover, $41<7^2$ so $41$ is prime – Stéphane Jaouen Apr 06 '24 at 18:56
-
1Note that your proof actually becomes Euclid's proof in the case $Q = \emptyset$, taking the empty product to be $1$. – Jair Taylor Apr 06 '24 at 19:00
-
$m+n$ is a number not divisible by a number in $S$, but may still be composite. For example in the question using $Q=\emptyset$: Is Euclid's proof on the infinitude of primes flawed because it yields some composites? – peterwhy Apr 06 '24 at 19:02
-
2(+1) for the idea , but there is no question , and this idea is not really new. But you successfully redetected it. – Peter Apr 06 '24 at 21:42
-
1This is Stieltjes’s proof given in 1890. – jjagmath Apr 07 '24 at 03:52
-
This variation on Euclid's proof is attributed to Stieltjes (by Ribenboim), cf. here in the linked dupe. – Bill Dubuque Apr 08 '24 at 01:26