Question about Euclid's infinite prime proof

Question

Suppose that $p_1=2 < p_2 = 3 < \cdots < p_r$ are all of the primes. Let $P = p_1p_2...p_r+1$ and let $p_s$ be a prime dividing $P$ where $p_s$ is not in our original list $p_1, p_2, \cdots, p_r$. If $p_s$ can divide $p$, wouldn't it be able to divide 1?

i.e.

$7 = 2*3 + 1$ meaning $p_s | 2*3$ and $p_s | 1$?

How can a prime divide $1$?This would mean that $\frac{1}{p_s}$ is integer which is only possible for $p_s=1$ in natural numbers — kingW3, Jan 12 '15 at 19:11

score 2 · Accepted Answer · answered Jan 12 '15 at 19:06

2

"Let $p_s$ be a prime dividing $P$ where $p_s$ is not in our original list." - This doesn't make sense. You assumed you have a list of all primes $\{p_i\}$. So $p_s$ must belong to $\{p_i\}$. This is what results in the contradiction.

answered Jan 12 '15 at 19:06

David P

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Question about Euclid's infinite prime proof

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