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Hi I cant seem to find anything about this, I'm sure I am just missing something simple but it is driving me crazy!

From my textbook, the definition of a prime in $\mathbb{Z}$ is:

If $a \in \mathbb{Z}$ is neither $0$ nor a unit (only $1$ and $-1$ in $\mathbb{Z}$) we say that $a$ is prime iff, whenever $a$ divides a product, that is $a \mid bc$ where $b,c \in \mathbb{Z}$ it follows that $a\mid b$ or $a\mid c$ or both.

My question is do we first have to assume that $a$ is irreducible? the textbook doesn't imply so. We know later that prime $\implies$ irreducible in $\mathbb{Z}$.

The example I'm having trouble with:

$6\mid 24 = 2\cdot 12$, then $6\mid 12 \implies 6$ is prime?

We also see that $6\mid 12 = 3 \cdot 4$ and $6 \nmid 3$ and $6 \nmid 4 \implies 6$ is not prime.

What am I missing?

Edit: does it have to be true in all cases where $6 \mid bc$?

Thanks

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    $\begingroup$ Hi @liamod! $p$ is prime if "$p| ab$ $\implies$ $p|a$ or $p| b$" $\textit{for all}$ $a$ and $b$. In your case, $6$ is not prime because the condition fails for at least one pair of $a$ and $b$: $a = 3$ and $b = 4$ as you have shown. $\endgroup$ – Amitesh Datta Jun 20 at 20:49
  • $\begingroup$ $a | bc \implies a |b \text{ or } a|c$ has to hold for every possible choice of $b,c$ for $a$ to be prime. You've supplied one counterexample ($b=3$, $c=4$) and so $a$ is not prime. $\endgroup$ – Jair Taylor Jun 20 at 20:51
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    $\begingroup$ $6 \mid 2\cdot 3$, but $6$ doesn't divide $2$ and $3$. $\endgroup$ – rtybase Jun 20 at 20:52
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    $\begingroup$ $$ \begin{align} \text{right: } & a\nmid b \\ {} \\ \text{wrong: } & a\not|b \\ {} \\ \text{right: } & a\mid b \\ {} \\ \text{wrong: } & a|b \end{align} $$ The two labeled "right" above are coded as a\nmid b and a\mid b; the others are a\not|b and a|b. (The lack of ability of so many mathematicians to learn simple points like this amazes me. I don't know how they do it.) $\endgroup$ – Michael Hardy Jun 21 at 0:37
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You are missing the word “whenever”. Yes, $6\mid2\times12$ and $6\mid12$. But it doesn't follow from this that $6$ is prime because it doesn't follow from this that whenever $6\mid bc$, then $6\mid b$ or $6\mid c$. And, in fact, as you have noticed, $6\mid3\times4$, but $6\nmid3$ and $6\nmid4$.

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    $\begingroup$ Ok perfect thanks! So I just go until I find a counter example and then I know it isn't prime. $\endgroup$ – liamod Jun 20 at 20:51
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    $\begingroup$ @liamod Yes, that is correct. $\endgroup$ – José Carlos Santos Jun 20 at 20:51

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