Math Myth: A prime is only divisible by 1 and itself

[Greg Bernhardt]

From @fresh_42's Insight
https://www.physicsforums.com/insights/10-math-things-we-all-learnt-wrong-at-school/

This is wrong. Well, yes and no. Strictly speaking, this definition describes irreducibility. And irreducibility is different from primality. A prime number is actually a number, that if it divides a product, then it has to divide one of the factors. $$7\,|\,28=2\cdot 14 \;\Longrightarrow \;7\,|\,2\text{ or }7\,|\,14$$ $$4\,|\,12=2\cdot 6\text{ but }4\,\nmid \,2 \text{ and } 4\,\nmid \,6$$ It is a bit more complicated, but it is the correct definition. However, irreducible integers are prime integers and vice versa which is why the correct definition is replaced at school by the more handy one. However, there are domains in which this is not the case. The standard example is $\mathbb{Z}[\sqrt{-5}]$ where $$6=2\cdot 3=(1+\sqrt{-5})\cdot (1-\sqrt{-5})$$ is a decomposition into irreducible factors that are not prime.

 

[BvU]

What is going on here ?

 

[Greg Bernhardt]

What is going on here ?

I divided up parts of @fresh_42's latest Insight to facilitate discussion on each

 

[BvU]

That's 11 entries on the unanswered threads list !

 

[FactChecker]

I'm glad that my teacher in grade school didn't teach these facts.