---
title: "prime"
type: concept
related: [Divide, Number, Integer, Prime]
source: https://www.jemoka.com/posts/kbhprime/
confidence: high
status: active
---

An integer \(p &gt; 1\) is prime if it has no positive divisors other than \(1\) and itself.
No even number, except \(2\), is prime. Because 2
additional information There are infinitely many primes Credit: Euler.
Proof:
Assume to the contrary that there are finitely many primes. \(p_1, &hellip;, p_{n}\). We desire to make a new prime to reach contradiction.
Consider:
\begin{equation} N = p_1 \times \dots \times p_{n} + 1 \end{equation}
Note that \(p_1 \times &hellip; \times p_{n}\) is divisible by each of the \(p_{j}\). If some \(p_i |N\), \(p_{i}|1\), which is impossible as \(1\) is not divisible by anything. So, no \(p_{i}\) divides \(N\).
If \(N\) is now prime, we are done as it is not in the list of \(p_{j}\). If not, pick any prime divisor \(p\) of \(N\). We will note that given no \(p_{j}\) divides \(N\), therefore any prime divisor is a new prime.
Having made a new prime, we reach contradiction. \(\blacksquare\)
coprime Two integers \(a, b\) is considered coprime if \(\gcd (a,b) = 1\). Therefore, because greatest common divisor is a linear combination
