Euclid's Proof of Infinite Number of Primes: Suppose that $\displaystyle p_1=2 < p_2 = 3 < \dots < p_r $ are all of the primes. Let $\displaystyle P = p_1p_2...p_r+1 $ and let $\displaystyle p $ be a prime dividing $\displaystyle P $; then $\displaystyle p $ can not be any of $\displaystyle p_1 $, $\displaystyle p_2 $, ..., $\displaystyle p_r $, otherwise $\displaystyle p $ would divide the difference $\displaystyle P-p_1p_2\dots p_r=1 $, which is impossible. So this prime $\displaystyle p $ is still another prime, and $\displaystyle p_1, p_2, ..., p_r $ would not be all of the primes. $\displaystyle \square $

(a) Why can't $\displaystyle p $ divide $\displaystyle 1 $, where $\displaystyle p $ is one of $\displaystyle \{p_{1}, p_{2}, p_{3}, \ldots, p_{r} \} $?

Proof: We want to show that $\displaystyle P \equiv P-1 (\mod p) $ is false. To prove this, choose $\displaystyle a,b \in \mathbb{Z} $ with $\displaystyle a \neq 0 $. Then $\displaystyle a $ divides $\displaystyle b $ if there is an element $\displaystyle c \in \mathbb{Z} $ such that $\displaystyle b = ac $. Suppose for contradiction that $\displaystyle p $ divides $\displaystyle 1 $. Then $\displaystyle 1 = pc $ where $\displaystyle c \in \mathbb{Z} $. Since $\displaystyle p > 1 $ by definition, $\displaystyle c \not \in \mathbb{Z} $. Contradiction. Thus $\displaystyle p $ cannot divide $\displaystyle 1 $. $\displaystyle \square $

Is this correct?