Talk:Fundamental theorem of arithmetic
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| Mathematics grading: | B Class | Top Importance | Field: Number theory | ||
| Needs information on history and more on applications. Tompw 18:39, 7 October 2006 (UTC) | |||||
This may be a simplification of the proof: Assume N=p1.p2...pn = q1.q2...qm is the smallest number with two prime factorizations. Assume q1 < p1. (p1-q1).p2...pn = q1.(q2...qm - p2...pn) is smaller than N, hence has unique factorization. q1 must then be a factor of p1-q1. So p1-q1=k.q1 for some integer k. Thus p1=(k+1)q1, which contradicts p1 being prime.
I think using · as a multiplication operator in this article is a bad idea. It's not recognisable to lay people. CGS 16:45, 21 Oct 2003 (UTC).
Except that the use of dot is uncluttered, consistent, and used by mathematicians as it is easier to write than cross. If you are worried by inconsistency with rest of wiki, then think of it as elegant variation, cf. Fowler Stan 23:49, 24 Feb 2004 (UTC)
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[edit] Applications
The 5th and 6th lines of this paragraph are absolutely obscure
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I second. Except I would venture to say that this entire section is poorly worded. I was unable to understand what was being said until I read it over and over and over again AND read some other references elsewhere.
- I replaced now at some places in this paragraph 'factor' by 'divisor' (prime and non-prime) to perhaps diminish the confusions between a divisor and a prime factor. Hopefully this helps. Bob.v.R 17:07, 15 September 2005 (UTC)
[edit] changeing the wording
How about changing the wording from
and there are no other such factorizations of 6936 or 1200 into prime numbers, except for reorderings of the above factors.
to
and there are no other possible factorizations of 6936 or 1200 into prime numbers, if we ignore the ordering of the factors.
[edit] Paragraph sounds funny
Suppose there were a positive integer which can not be written as a product of primes. Then there must be a smallest such number: let's call it n. This number n cannot be 1, because of our convention above. It cannot be a prime number either, since any prime number is a product of a single prime, itself. So it must be a composite number. Thus n = ab where both a and b are positive integers smaller than n. >>>Since n was the smallest number for which the theorem fails, both a and b can be written as products of primes.<<< But then n = ab can be written as a product of primes as well, a contradiction.
There is something wrong with the paragraph above.
Since n is the smallest number which can not be written as a product of primes then both a nd b cannot be written as a product of primes. BUT didn't we say n is the smallest number which can not be written as a product of primes and the number n , a and b are positive integers. This means a is smaller than n and b is smaller than n. If a cannot be written as a product of primes and a is smaller than n then n CANNOT be the smallest number which can not be written as a product of primes. Hence a contradiction.
- I prefer the first version. The first sentence of the second version jumps a step.
Charles Matthews 10:19, 29 Nov 2004 (UTC)
- I think the original poster misunderstands the nature of the proof. It is simply not true that "n is smallest number which cannot be written as product of primes" implies "a and b cannot be written as a product of primes". In fact, it is just the opposite, by our choice of n. We are using well-ordering, and then showing that the existence of such a least n always leads to a contradiction. Revolver 20:33, 29 Nov 2004 (UTC)
[edit] Proof by infinite descent
The unicity proof given as "by infinite descent" uses the division algorithm but not the Euclidean algorithm (much less its generalisation Bézout's lemma). In this way, it's more elementary (you need fewer facts about division, or if you prefer, you need not consider the consequences of the division algorithm in as much depth).
Who first came up with that argument?
—Toby Bartels 09:35, 26 July 2006 (UTC)
[edit] One
Might one say the the set of prime factors of one is the empty set? Thus, one is not prime because prime numbers are those the set of factors of which has a cardinality of 1.
