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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 mn if and only if pi^ei  n for all i?
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 mn if and only if pi^ei  n for all i?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I only need to prove the backward direction.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Fundamental theorem of arithmetic guarantees you that there are no other prime factorization representations for \(m\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But how does that prove that mn?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0According to fundamental theorem of arithmetic, the prime factorization representation is unique up to the order of factors.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so 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 \(i\ne j\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0since fundamental theorem of arithmetic guarantees that there are no other prime factorization representations for \(m\), the proof ends.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0for 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 \(m\mid n\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the Euclid lemma states that if p is a prime, and p  ab, then pa or pb correct?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0that is one version of euclid lemma

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0here is another version https://i.gyazo.com/9bef4aa867f4a65e0da4c77295d83bdc.png

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0there are couple more variants, but all of them are same.. you can derive one from the other

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to see that pattern of which letters in Euclid's lemma corresponds to which letters in the question. So if abc, and gcd(a,b) = 1, then a  c gcd(a,b) is like gcd(p1^e1, p2^e2, ..., ps^es) right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0we need to prove this : If \(a\mid c\) and \(b\mid c\), with \(\gcd(a,b)=1\), then \(ab\mid c\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right. That's what I just noticed and somehow it doesn't look like Euclid lemma

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0above is another variant of euclid's lemma

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0O: it's another variation of Euclid's lemma???

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes, its proof is trivial, try it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what is euclid's lemma according to ur book ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since ab and bc, by definition, ak = b and bs = c

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops ac implies ak = c

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0multiply c for both sides gives acx + bcy = c. and subsitute a(bs)x + b(ak)y = c

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ab (sx + ky) = c implies abc whoah I did it!! :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, 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 mn

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0sry actually this is not euclid's lemma, this is corollary of bezout's identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but anyways, I understood it now. I only need this corollary of bezout's identity to prove. Makes much more sense!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also, I noticed that m * gcd(k1,k2,...ks) = n where p1^e1 * k1 = n, ..., ps^es * ks = n. Do you think it's true?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what are k1, k2, ... ks ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0by givens, p1^e1  n, ... ps^es  n . The k1, ...., ks are the integers in the definition of divisibility

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think this only works if gcd(2^3, 5^4) = 1, which it is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0typo :O 2^3 * 5^4 * 2 = 10,000

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so here is the conjecture. If a  n and bn 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw @ganeshie8 Thank you for helping me with the original question :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0looks good to me, try proving it :)
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