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anonymous

  • one year ago

Let m = p1^e1 * p2^e2 * ... * ps^es, where pi are distinct primes. How does the Fundamental Theorem of Arithmetic proves m|n if and only if pi^ei | n for all i?

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  1. anonymous
    • one year ago
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    I only need to prove the backward direction.

  2. ganeshie8
    • one year ago
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    Fundamental theorem of arithmetic guarantees you that there are no other prime factorization representations for \(m\)

  3. anonymous
    • one year ago
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    But how does that prove that m|n?

  4. ganeshie8
    • one year ago
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    According to fundamental theorem of arithmetic, the prime factorization representation is unique up to the order of factors.

  5. ganeshie8
    • one year ago
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    so if each of the prime powers divide \(n\), then by euclid lemma it follows that the product of the prime powers divide \(n\) since \(\gcd({p_i}^{e_i},~{p_j}^{e_j})=1\) for all \(i\ne j\)

  6. ganeshie8
    • one year ago
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    since fundamental theorem of arithmetic guarantees that there are no other prime factorization representations for \(m\), the proof ends.

  7. ganeshie8
    • one year ago
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    for example : \(m = 2^35^4\) suppose \(2^3\mid n\) and \(5^4\mid n\) since \(\gcd(2^3,5^4)=1\), by euclid lemma we have \((2^3*5^4)\mid n\) by fundamental theorem of arithmetic, we know that there are no other prime factorization representations for \(m\), so it follows that \(m\mid n\).

  8. anonymous
    • one year ago
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    so the Euclid lemma states that if p is a prime, and p | ab, then p|a or p|b correct?

  9. ganeshie8
    • one year ago
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    that is one version of euclid lemma

  10. ganeshie8
    • one year ago
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    here is another version https://i.gyazo.com/9bef4aa867f4a65e0da4c77295d83bdc.png

  11. ganeshie8
    • one year ago
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    there are couple more variants, but all of them are same.. you can derive one from the other

  12. anonymous
    • one year ago
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    I'm trying to see that pattern of which letters in Euclid's lemma corresponds to which letters in the question. So if a|bc, and gcd(a,b) = 1, then a | c gcd(a,b) is like gcd(p1^e1, p2^e2, ..., ps^es) right?

  13. anonymous
    • one year ago
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    see *the*

  14. ganeshie8
    • one year ago
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    we need to prove this : If \(a\mid c\) and \(b\mid c\), with \(\gcd(a,b)=1\), then \(ab\mid c\)

  15. anonymous
    • one year ago
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    right. That's what I just noticed and somehow it doesn't look like Euclid lemma

  16. ganeshie8
    • one year ago
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    above is another variant of euclid's lemma

  17. anonymous
    • one year ago
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    O: it's another variation of Euclid's lemma???

  18. ganeshie8
    • one year ago
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    Yes, its proof is trivial, try it

  19. ganeshie8
    • one year ago
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    what is euclid's lemma according to ur book ?

  20. anonymous
    • one year ago
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    Ok. Let me try if you wouldn't mind. Had I known this variation, I wouldn't have asked this question to begin with. gcd(a,b) = 1 implies there exists x,y such that ax + by = 1

  21. anonymous
    • one year ago
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    since a|b and b|c, by definition, ak = b and bs = c

  22. anonymous
    • one year ago
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    oops a|c implies ak = c

  23. anonymous
    • one year ago
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    multiply c for both sides gives acx + bcy = c. and subsitute a(bs)x + b(ak)y = c

  24. ganeshie8
    • one year ago
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    that'll do

  25. anonymous
    • one year ago
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    ab (sx + ky) = c implies ab|c whoah I did it!! :D

  26. anonymous
    • one year ago
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    ok, now I can apply this veriation of Euclid lemma to prove the question each pi^ei divides n and since gdc(p1^e1, .., ps^es) = 1, by Euclid lemma m|n

  27. ganeshie8
    • one year ago
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    sry actually this is not euclid's lemma, this is corollary of bezout's identity

  28. anonymous
    • one year ago
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    oh ... lol

  29. anonymous
    • one year ago
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    but anyways, I understood it now. I only need this corollary of bezout's identity to prove. Makes much more sense!

  30. anonymous
    • one year ago
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    also, I noticed that m * gcd(k1,k2,...ks) = n where p1^e1 * k1 = n, ..., ps^es * ks = n. Do you think it's true?

  31. ganeshie8
    • one year ago
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    i don't get you

  32. ganeshie8
    • one year ago
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    what are k1, k2, ... ks ?

  33. anonymous
    • one year ago
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    by givens, p1^e1 | n, ... ps^es | n . The k1, ...., ks are the integers in the definition of divisibility

  34. anonymous
    • one year ago
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    I was just trying to prove by definition. I need to find an integer k such that m k = n. and what I noticed is k = gcd(k1, k2, ..., ks)

  35. anonymous
    • one year ago
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    let's use the example above. 2^3 | 10,000 5^4 | 10,000 so, 2^3 * 1250 = 10,000, where k1 = 1250 5^4 * 16 = 10,000, where k2 = 16 gcd(1250,16) = 2 and coincidentally, 2^3 * 4^2 * 2 = 10,000

  36. anonymous
    • one year ago
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    I think this only works if gcd(2^3, 5^4) = 1, which it is

  37. anonymous
    • one year ago
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    typo :O 2^3 * 5^4 * 2 = 10,000

  38. anonymous
    • one year ago
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    so here is the conjecture. If a | n and b|n and gcd(a,b) = 1, then ab * gcd(x,y) = n, where ax = n and by = n. If you have any comments to this conjecture, let me know. What matters is that I understood how to prove the original question.

  39. anonymous
    • one year ago
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    btw @ganeshie8 Thank you for helping me with the original question :)

  40. ganeshie8
    • one year ago
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    looks good to me, try proving it :)

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