# 6 is a prime number? Misunderstanding something in the definition of a prime number

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

• 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. – Amitesh Datta Jun 20 at 20:49
• $a | bc \implies a |b \text{ or } a|c$ 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. – Jair Taylor Jun 20 at 20:51
• $6 \mid 2\cdot 3$, but $6$ doesn't divide $2$ and $3$. – rtybase Jun 20 at 20:52
• \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.) – Michael Hardy Jun 21 at 0:37

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 $$6\nmid4$$.