(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that if a given natural number M has no factors less than or equal to M^{1/2}, then M is prime.

2. Relevant equations

None!

3. The attempt at a solution

So I was wondering, say I have two sets [itex]P[/itex] and [itex]Q[/itex]. Furthermore, say that [itex]P \bigcup Q = \mathbb{N}[/itex] and [itex]P \bigcap Q = \emptyset[/itex]. Now, can I assume that if all elements of set P obey a certain property, then any element of N that does not obey that property belongs to Q?

I ask because I'm trying to prove that if a natural number M has no factors less than or equal to M^{1/2}then M is prime. I'm hoping to use a proof that I almost have complete that if a natural number N is composite, then N has a factor less than or equal to N^{1/2}, but I don't know if the property can transfer this way. Help?

EDIT: Also, my proof for the latter statement, if it helps:

Assume that N is composite and that it has no factors less than or equal to N^{1/2}. Now, since N is composite, it has a factor A, which must be greater than N^{1/2}for the assumption to hold. Therefore, A>N^{1/2}. Now, by inverting the two sides of the equation, we can see that A^{-1}<N^{-1/2}. Multiplying both sides by N, we get that N/A < N^{1/2}. However, as A is a factor of N, N/A must be an integer B that is less than or equal to N^{1/2}. Therefore, N = A*B, and B is a factor of N. But B is less than N^{1/2}, which contradicts the original assumption. Therefore, if N is composite, N has a factor less than or equal to N^{1/2}

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# Homework Help: Opposite Sets

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