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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. m|n if and only if pi^ei | n for all i.

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  1. dan815
    • one year ago
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    all numbers can be written as a product of primes

  2. dan815
    • one year ago
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    primes have factors of 1 and themself

  3. dan815
    • one year ago
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    now this means that if m divide n then every prime^exponent divides n

  4. dan815
    • one year ago
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    does it make sense?

  5. dan815
    • one year ago
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    maybe u need to say this statement too if m|n then n=k*m there fore k*(p1^e1*p2^e2....)

  6. anonymous
    • one year ago
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    ok, 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?

  7. dan815
    • one year ago
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    what do u mean

  8. anonymous
    • one year ago
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    it's an if and only if statement

  9. dan815
    • one year ago
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    okay since u saw that n=k*m then n=k*p1^e1*p2^e2... therefore p1^e1,p2^e2... all have to be factors

  10. dan815
    • one year ago
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    and u say primes cannot be decomposed into other primes so u are done

  11. dan815
    • one year ago
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    there is no other prime representation for m, so it goes both ways

  12. anonymous
    • one year ago
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    well, we just proved that direction. The other direction is if pi^ei | n for all i, then m|n

  13. dan815
    • one year ago
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    hmm to me its the same thing lol

  14. dan815
    • one year ago
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    ok how about saying it like this

  15. dan815
    • one year ago
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    if pi^ei | n for all i then (p1^e1)(p2^e2)(p3^e3).....(pn^en) | n so m|n

  16. dan815
    • one year ago
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    because if* p1|n and p2|n then p1*p2|n if p1 and p2 are prime

  17. dan815
    • one year ago
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    this has to be true as a 2 different primes cannot share factors

  18. anonymous
    • one year ago
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    so a lemma was used a long the way Given p1 and p2 are primes. If p1| and p2|n, then p1*p2 | n

  19. anonymous
    • one year ago
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    I think m * gcd(k1, k2, ... ks) = n will prove the result.

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