# Proof

(related to Proposition: Existence of Prime Divisors)

• Trivially, every prime divides the integer $0.$ Also, because with every divisor $$d\mid n$$ it is also true that $$d\mid -1$$, it is sufficient to show the existence of prime divisors for natural numbers $$n > 1$$, instead of integers $$n\neq\pm 1$$.
• Let $$D(n):=\{d\in\mathbb N:b>1,~d\mid n\}$$ be the set of all divisors of $$n$$.
• $$D(n)$$ is not empty, since $$n\in D(n)$$.
• Therefore, according to the well-ordering principle, $$D(n)$$ must have a smallest element, which we denote by $$p$$.
• By construction, $$p > 1$$.
• We will now show that $$p$$ is a prime number.
• For if it was composite, then it would have a non-trivial divisor $$q\mid p$$, i.e. a divisor with $$1 < q < p$$.
• Because $$q\mid p$$ and $$p\mid n$$, it would follow from the divisibility laws that $$q\mid n$$ and so $$q\in D(n)$$.
• This is a contradiction to the minimality of $$p$$ in $$D(n)$$.
• Therefore, $$p$$ must be prime.

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### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013