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anonymous
 one year ago
help! It said the Fundamental Theorem of Arithmetic is used to prove. Let m = p1^e1 * p2^e2 ... ps^es, where pi is a prime. mn if and only if pi^ei  n for all i.
anonymous
 one year ago
help! It said the Fundamental Theorem of Arithmetic is used to prove. Let m = p1^e1 * p2^e2 ... ps^es, where pi is a prime. mn if and only if pi^ei  n for all i.

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dan815
 one year ago
Best ResponseYou've already chosen the best response.0all numbers can be written as a product of primes

dan815
 one year ago
Best ResponseYou've already chosen the best response.0primes have factors of 1 and themself

dan815
 one year ago
Best ResponseYou've already chosen the best response.0now this means that if m divide n then every prime^exponent divides n

dan815
 one year ago
Best ResponseYou've already chosen the best response.0maybe u need to say this statement too if mn then n=k*m there fore k*(p1^e1*p2^e2....)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, m  n means mk = n for some integer k p1^e1 ( p2^e2 ... ps^es * k) = n implies p1^e1  n p2^e2 * (p1^e2 * p3^e3 ... ps^es k) = n implies p2^e2  n and so on to ps^es. How do you prove the other direction?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's an if and only if statement

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay since u saw that n=k*m then n=k*p1^e1*p2^e2... therefore p1^e1,p2^e2... all have to be factors

dan815
 one year ago
Best ResponseYou've already chosen the best response.0and u say primes cannot be decomposed into other primes so u are done

dan815
 one year ago
Best ResponseYou've already chosen the best response.0there is no other prime representation for m, so it goes both ways

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, we just proved that direction. The other direction is if pi^ei  n for all i, then mn

dan815
 one year ago
Best ResponseYou've already chosen the best response.0hmm to me its the same thing lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ok how about saying it like this

dan815
 one year ago
Best ResponseYou've already chosen the best response.0if pi^ei  n for all i then (p1^e1)(p2^e2)(p3^e3).....(pn^en)  n so mn

dan815
 one year ago
Best ResponseYou've already chosen the best response.0because if* p1n and p2n then p1*p2n if p1 and p2 are prime

dan815
 one year ago
Best ResponseYou've already chosen the best response.0this has to be true as a 2 different primes cannot share factors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so a lemma was used a long the way Given p1 and p2 are primes. If p1 and p2n, then p1*p2  n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think m * gcd(k1, k2, ... ks) = n will prove the result.
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