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September 2nd, 2018, 07:54 AM  #1 
Member Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0  Euclid's Proof of the Infinitude of Primes
So I'm going to refer to this link that the explanation is from: https://primes.utm.edu/notes/proofs/...e/euclids.html Theorem. "There are more primes than found in any finite list of primes." (Chris K. Caldwell) Proof. "Call the primes in our finite list p1, p2, ..., pr. Let P be any common multiple of these primes plus one (for example, P = p1p2...pr+1). Now P is either prime or it is not. If it is prime, then P is a prime that was not in our list. If P is not prime, then it is divisible by some prime, call it p. Notice p can not be any of p1, p2, ..., pr, otherwise p would divide 1, which is impossible. So this prime p is some prime that was not in our original list. Either way, the original list was incomplete." (Chris K. Caldwell) So all the parts of the proof is clear to me until he says: Notice p can not be any of p1, p2, ..., pr, otherwise p would divide 1, which is impossible. So p is a prime number that divide P in this case, but I do not understand why p can't be any of p1, p2, ..., pr? So using the definition of a prime number this tells me that I get 1 divided by p (1 / p) which is impossible? Why is that impossible? 
September 2nd, 2018, 08:33 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,882 Thanks: 1835 
He means that 1/p can't be a positive integer, as it's a fraction lying strictly between 0 and 1. The number 1 isn't considered to be a prime.

September 2nd, 2018, 08:58 AM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 504 Thanks: 283 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
 
September 2nd, 2018, 10:11 AM  #4 
Senior Member Joined: Aug 2012 Posts: 2,082 Thanks: 595  
September 2nd, 2018, 10:23 AM  #5 
Member Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 
So p is a prime number that divide P in this case, but I do not understand why p can't be any of p1, p2, ..., pr? Is the reason that we have assumed that any of p1, p2, ..., pr are prime numbers and so are p. Definition: a prime number is only divisible by 1 and itself and is greater than 1. So a prime is not divisible by another prime number (so that the result is a positive integer). So if p where any of p1, p2, ..., pr then: if p1 = p > p / p1 = 1 > p = 1 * p1 > 1 / p = p1 if p2 = p > p / p2 = 1 > p = 1 * p2 > 1 / p = p2 So this shows me that it is impossible because 1 / p can't become a positive integer since p is a prime number. 
September 2nd, 2018, 11:23 AM  #6  
Senior Member Joined: Aug 2012 Posts: 2,082 Thanks: 595  Quote:
Look at an example. Consider $31 = 2 \times 3 \times 5 + 1$. If you divide it by 2, you get remainder 1. If you divide it by 3, you get remainder 1. If you divide it by 5, you get remainder 1. So 31 must be either be prime (which it is, in this case) or else divisible by some prime other than 2, 3, or 5.  
September 2nd, 2018, 11:35 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,882 Thanks: 1835 
Put simply, if p is one of the first r primes, p1p2p3p4...pr and p1p2p3p4...pr + p are consecutive integer multiples of p. As P must be strictly inbetween those two integers, it can't be an integer multiple of p, which means that p doesn't divide P, contrary to the assumption that it does.

September 3rd, 2018, 05:38 AM  #8 
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489 
Another way of seeing what is going on is to use a more friendly notation. $\mathbb L = \{p_1,\ ... \ p_r\} = \text { the FINITE LIST of primes.}$ $\text {Let } \displaystyle y = \prod_{j=1}^r p_j \implies y \in \mathbb Z,\ y > 1, \text { and }$ $\text {for any } p \in \mathbb L,\ y \ge p \text { and } \dfrac{y}{p} \in \mathbb Z \ \because p \text { is a factor of } y.$ $\text {Let } x = y + 1 \implies x \in \mathbb Z,\ x > y > 1 \text { and, for any } p \in \mathbb L,\ x > p .$ $x \text { is either a prime or else not a prime because } x \text { is an integer } > 1.$ $\text {Case I: } x \text { is a prime.}$ $p \in \mathbb L \implies x > p \implies x \not \in \mathbb L \implies$ $\text {there is at least one prime not in the finite list.}$ $\text {Case II: } x \text { is not a prime.}$ $\therefore \exists \text { a nonempty set } \mathbb F \text { of prime factors of } x.$ $\text {ASSUME } \exists \ u \text { such that } u \in \mathbb L \text { and } u \in \mathbb F.$ $v = \dfrac{x}{u} \implies v \in \mathbb Z \ \because \ u \in \mathbb F.$ $w = \dfrac{y}{u} \implies w \in \mathbb Z \ \because \ u \in \mathbb L.$ $v = \dfrac{x}{u} = \dfrac{y + 1}{u} = w + \dfrac{1}{u}.$ $\text {By hypothesis, } u \text { is a prime } \implies u > 1 \implies 0 < \dfrac{1}{u} < 1.$ $\therefore w < w + \dfrac{1}{u} < w + 1 \implies \left (w + \dfrac{1}{u} \right ) \not \in \mathbb Z \ \because \ w \in \mathbb Z.$ $\therefore v \not \in \mathbb Z, \text {a CONTRADICTION.}$ $\therefore \not \exists \ u \text { such that } u \in \mathbb L \text { and } u \in \mathbb F.$ $\text {But } \mathbb F \text { is not empty } \implies$ $\text {there is at least one prime not in the finite list.}$ 

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euclid, infinitude, primes, proof 
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