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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 8 months |
| seen | Apr 27 at 21:27 | |
| stats | profile views | 33 |
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Apr 12 |
accepted | Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. |
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Apr 11 |
comment |
Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Sorry if it was heard rude, but I didn't intend to. I know what you used is an elementary fact but even the theorem I proposed is also so elementary as not to be proved. What I was trying to do was proving such elementary stuffs not with rough intuitions but purely by logical steps. Sorry also that I didn't clearly state what I was doing and about Euclid's lemma nor the definition of prime; at the time when I was posting the question, my mind was quite confused about this. Anyway, I've roughly written what I was doing on the comments on my question. |
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Apr 10 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Any rough intuition can be doubted before proved, and I was doubting this because I didn't know the proof of it though I've used it for quite long time anywhere that I need to use it |
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Apr 10 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. "If p is a prime in the factorization of a, it must be in the factorization of b. Why? Well, you can use the property that if a prime number divides a product, it must divide one of the factors." What you said is exactly the Euclid's Lemma, which must be proved before being used. And my essential speculation was on this very matter. Please read the above posts if you want to understand what I was curious about. The very intuition that 'if a prime number divides a product, it must divide one of the factors' must be mathematically proved if mathematically used. |
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Apr 9 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. So even though the above theorem that I proposed may be a quite direct corollary of The Fundamental Theorem, it must take more logical steps. |
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Apr 9 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. I'm not sure which Fundamental Theorem you are looking at. But according to wiki's one, the theorem states that any integer can be decomposed into purely by prime factors and any kind of this process will actually end up in the same factors. But you must recognize that the theorem does not state these are 'all the possible divisors' of the given integer! They are only all the prime divisors, nothing more. Purely in logical manner, prime divisor is never the same as general integer divisor. |
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Apr 8 |
comment |
Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. What I feel weird about this is that Euclid's lemma looks stronger than the above theorem that I formulated (because the above one deals only with primes). And this is exactly why I was trying to use this theorem as a lemma to prove Euclid's lemma, not the other way. (I mean, I felt it'd be easier to prove the above theorem than Euclid's.) But maybe my feeling is wrong. Anyway thanks for the link and you guys' comments. |
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Apr 8 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Maybe for you but not for me. But Fundamental Theorem of Arithmetics does not really deal with divisors in general; it focuses on prime factorization 'structures' and possibly only them. |
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Apr 8 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Okay so then without using Bezout's technique and Euclid's Lemma, is it not possible to prove the above theorem? |
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Apr 8 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Hm... the thing is I was actually trying to prove Euclid's Lemma but then I need the above formula as a lemma for proving Euclid's... |
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Apr 8 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. @copper.hat Not much actually... The thing is even though I assume $d$ is not one of $1, p_1 , p_2 , p_1 p_2$ I cannot derive a contradiction. |
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Apr 8 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. The former one. I was actually trying to prove the former and the latter are equivalent, and in the course of it, I needed the above theorem as a lemma. |
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Apr 8 |
revised |
Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. added 42 characters in body |
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Apr 8 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Prime number is a number that is divided by 1 and itself. I generally consider only positive integers actually.. |
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Apr 8 |
asked | Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. |
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Mar 29 |
accepted | (Baby Rudin) To show the set of all condensation points of a set in Euclidean space is perfect |
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Mar 29 |
comment |
(Baby Rudin) To show the set of all condensation points of a set in Euclidean space is perfect not sure... I've tried to derive a contradiction from this point for 3 hours but still stuck.. |
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Mar 29 |
asked | (Baby Rudin) To show the set of all condensation points of a set in Euclidean space is perfect |
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Feb 14 |
revised |
Is this function $(\partial / \partial x)f$ discontinuous at $(0,0)$? added 18 characters in body; edited title |
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Feb 12 |
revised |
Is this function $(\partial / \partial x)f$ discontinuous at $(0,0)$? added 54 characters in body |
