I have two full questions on some number theory questions I've been working on, I guess my best bet would be to post them separately.(adsbygoogle = window.adsbygoogle || []).push({});

1) Suppose that n is in N (natural numbers), p_{1},....,p_{n}are distinct primes, and l_{1},....l_{n}are nonnegative integers. Let m = p_{1}^{l1}p_{2}^{l2}....p_{n}^{ln}. Let d be in N such that d ≥ 2 and d divides m.

a) Using the fundamental theorem of arithemetic prove that p is a prime that divides d, then p_{i}for some i in {1,...,n}.

Attempt: Since d is in N and d ≥ 2, this means that d itself is a product of primes (prime factorization) or a prime itself. This means d = p_{1}p_{2}...p_{n}(product of primes). Then there exists a p that divides d. (Since we are starting from 1 in the natural numbers)

Is this the right idea and what would be a clearer way of writing this as a proof?

b) Suppose the i is in {1,...,n}. Prove that p_{i}^{k}does not divide d if

k > l_{i}

Attempt: if k > l_{i}then d ≠ p_{i}^{k}, but d may equal any prime in m = p_{1}^{l1}p_{2}^{l2}....p_{n}^{ln}since we know d | m.

this would imply d has a cononical factorization : p_{1}^{l1}p_{2}^{l2}....p_{n}^{ln}and we know k > l_{i}therefore p_{i}^{k}does not divide d.

Right idea?

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# Homework Help: Number Theory fundamental theorem of arithemetic

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