Euclid
Proof. Suppose that p1=2 < p2 = 3 < … < pr are all of the primes. Let P = p1p2…pr+1 and let p be a prime dividing P; then p can not be any of p1, p2, …, pr, otherwise p would divide the difference P-p1p2…pr=1, which is impossible. So this prime p is still another prime, and p1, p2, …, pr would not be all of the primes.
(Ribenboim’s statement of Euclid’s proof)
June 28th, 2005 at 10:30 am
Over my head. You need to start a separate blog called,” My Small Ship” for famous mathematical quotes.
June 28th, 2005 at 7:56 pm
you don’t how much i missed you through my trip