Euclid's Lemma for Prime Divisors/General Result/Proof 3

From ProofWiki
Jump to navigation Jump to search


Let $p$ be a prime number.

Let $\displaystyle n = \prod_{i \mathop = 1}^r a_i$.

Then if $p$ divides $n$, it follows that $p$ divides $a_i$ for some $i$ such that $1 \le i \le r$.

That is:

$p \divides a_1 a_2 \ldots a_n \implies p \divides a_1 \lor p \divides a_2 \lor \cdots \lor p \divides a_n$


Let $p \divides n$.

Aiming for a contradiction, suppose:

$\forall i \in \set {1, 2, \ldots, r}: p \nmid a_i$

By Prime not Divisor implies Coprime:

$\forall i \in \set {1, 2, \ldots, r}: p \perp a_i$

By Integer Coprime to all Factors is Coprime to Whole:

$p \perp n$

By definition of coprime:

$p \nmid n$

The result follows by Proof by Contradiction.


Source of Name

This entry was named for Euclid.