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Proof of prime numbers

Mathematics Asked by Lorenza Fuller on February 13, 2021

Prove that for every integer $m geq 2$, if there is no prime number $p$ such that $p leq sqrt{m}$ and $p mid m$ (evenly divides), then $m$ must be prime.

Don’t even know where to start with this one.

*Edit: In class we covered the Fundamental Theorem of Arithmetic, but again I’m not sure how to start applying it.

One Answer

This is basically the Sieve of Eratosthenes. The point is that once you pass $sqrt m$, you have already exhausted the possibilities. Take $11$. Then you only need to check up to $sqrt{11}$ for factors, because anything bigger than $sqrt{11}$ would need to be paired with something less than $sqrt{11}$. None of $2,3$ is a factor, so we have proved that $11$ is prime.

Answered by Chris Custer on February 13, 2021

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