DecoratorFawn82

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?

skipjack

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.

SDK

If you have a sum of the form $A = B + C$ and a number $p$, divides both $A$ and $B$, then it must also divide $C$. In this case your sum looks like $P = p_1 \cdot ... \cdot p_r + 1$.

Maschke

Because dividing $p_1 p_2 \dots p_r + 1$ by any of the $p_i$'s leaves a remainder of 1.

DecoratorFawn82

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.

Maschke

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.

skipjack

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.

JeffM1

Another way of seeing what is going on is to use a more friendly notation.

$\mathbb L = \{p_1,\ ... \ p_r\} = \text { the FINITE LIST of primes.}$

$\text {Let } \displaystyle y = \prod_{j=1}^r p_j \implies y \in \mathbb Z,\ y > 1, \text { and }$

$\text {for any } p \in \mathbb L,\ y \ge p \text { and } \dfrac{y}{p} \in \mathbb Z \ \because p \text { is a factor of } y.$

$\text {Let } x = y + 1 \implies x \in \mathbb Z,\ x > y > 1 \text { and, for any } p \in \mathbb L,\ x > p .$

$x \text { is either a prime or else not a prime because } x \text { is an integer } > 1.$

$\text {Case I: } x \text { is a prime.}$

$p \in \mathbb L \implies x > p \implies x \not \in \mathbb L \implies$

$\text {there is at least one prime not in the finite list.}$

$\text {Case II: } x \text { is not a prime.}$

$\therefore \exists \text { a non-empty set } \mathbb F \text { of prime factors of } x.$

$\text {ASSUME } \exists \ u \text { such that } u \in \mathbb L \text { and } u \in \mathbb F.$

$v = \dfrac{x}{u} \implies v \in \mathbb Z \ \because \ u \in \mathbb F.$

$w = \dfrac{y}{u} \implies w \in \mathbb Z \ \because \ u \in \mathbb L.$

$v = \dfrac{x}{u} = \dfrac{y + 1}{u} = w + \dfrac{1}{u}.$

$\text {By hypothesis, } u \text { is a prime } \implies u > 1 \implies 0 < \dfrac{1}{u} < 1.$

$\therefore w < w + \dfrac{1}{u} < w + 1 \implies \left (w + \dfrac{1}{u} \right ) \not \in \mathbb Z \ \because \ w \in \mathbb Z.$

$\therefore v \not \in \mathbb Z, \text {a CONTRADICTION.}$

$\therefore \not \exists \ u \text { such that } u \in \mathbb L \text { and } u \in \mathbb F.$

$\text {But } \mathbb F \text { is not empty } \implies$

$\text {there is at least one prime not in the finite list.}$