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

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SUMMARY

The discussion clarifies the definition of prime numbers, emphasizing that a prime number is defined by its divisibility properties rather than solely by being divisible by 1 and itself. Specifically, a prime number divides a product, necessitating that it divides at least one of the factors. The conversation also highlights the distinction between primality and irreducibility, particularly in the context of certain mathematical domains like ##\mathbb{Z}[\sqrt{-5}##, where irreducible integers may not be prime. This nuanced understanding corrects common misconceptions taught in schools.

PREREQUISITES
  • Understanding of basic number theory concepts
  • Familiarity with divisibility rules
  • Knowledge of irreducibility in mathematics
  • Basic understanding of algebraic structures, particularly rings
NEXT STEPS
  • Research the properties of prime numbers in number theory
  • Study the concept of irreducibility in different mathematical domains
  • Explore the structure of the ring ##\mathbb{Z}[\sqrt{-5}## and its implications
  • Learn about the Fundamental Theorem of Arithmetic and its exceptions
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Mathematicians, educators, and students seeking a deeper understanding of number theory, particularly those interested in the properties of prime and irreducible numbers.

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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.
 
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