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In my opinion, it doesn't make sense because I can't think of any number that meet that criteria for the number 4, which is not prime.

a must be different from b, right?

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No, not necessarily different. $4=2\cdot 2$. And $4|4$ but $4\not|2$.

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How is 4 not divisible by 2? – Dong Min Son Sep 30 '12 at 19:27
Ah, I see. n can't divide a. I thought the notation meant a can't divide n. – Dong Min Son Sep 30 '12 at 19:36

It proves: the integer $\rm\,p>1\,$ is prime (i.e. irreducible) if it satisfies PDP = Prime Divisor Property (i.e. $\rm\,p\,$ divides a product $\rm\Rightarrow p$ divides some factor). Below is a proof that clarifies this inference.

Theorem $\ $ In the following, $\rm\ (1)\:\Rightarrow\:(2)\iff (3)$

$\rm(1)\ \ \ a\ \ |\ \ bc\ \Rightarrow\ a\:|\:b\ \ or\ \ a\:|\:c\quad$ (Definition of $\rm\:a\:$ satisfies PDP = Prime Divisor Property)

$\rm(2)\ \ \ a=bc\ \Rightarrow\ a\:|\:b\ \ or\ \ a\:|\:c\quad$ (Definition of $\rm\:a\:$ is prime, i.e. irreducible) $\ $ (alternative)

$\rm(3)\ \ \ a=bc\ \Rightarrow\ b\:|\:1\ \ or\ \ c\:|\:1\quad$ (Definition of $\rm\:a\:$ is prime, i.e. irreducible) $\ $ (classical)

Proof $\ \ \ (1\Rightarrow 2)\ \ $ If $\rm\: a = bc\:$ then $\rm\:a\:|\:bc\:$ so $\rm\:a\:|\:b\:$ or $\rm\:a\:|\:c\:$ by $(1).\:$ Thus PDP $\Rightarrow$ irreducible.

$(2\!\!\iff\!\! 3)\ \ \ $ If $\rm\:a = bc\:$ then $\rm\:b/a = 1/c\:$ so $\rm\:a\:|\:b\iff c\:|\:1.\:$ Similarly $\rm\:a\:|\:c\iff b\:|\:1.$

Remark $\, $ In general domains, invertibles, i.e. divisors of $1,$ are called units, and nonunits satisfying the Prime Divisor Property $(1)$ are called primes, and elements satisfying $(2)$ or $(3)$ are called atoms or irreducibles. So the inference $(1\Rightarrow 2)$ says that a prime is an atom = irreducible. The converse, atoms are prime, holds iff factorizations into atoms are unique (up to order and unit factors), i.e. when the domain is a Unique Factorization Domain (UFD), e.g. the integers $\,\Bbb Z.$

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Hey, Math Gems is back! – The Chaz 2.0 Feb 10 '13 at 4:52

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