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 May 4th, 2015, 01:35 PM #1 Newbie   Joined: May 2015 From: USA Posts: 2 Thanks: 0 Math Focus: Number Theory, Bayesian Probability Good explanation of Fundamental Theorem of Arithmetic? I'm going through Ireland and Rosen's "A Classical Introduction to Modern Number Theory", and I can't seem to get past the first chapter on the fundamental theorem of arithmetic. Do you guys have any suggestions on how to to understand that theorem better, with more explicit descriptions on each lemma?
 May 4th, 2015, 03:36 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra I find the Wikipedia page to be well written.
 May 5th, 2015, 11:03 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Admittedly a bit terse, but this provides good intuition: https://xkcd.com/5/
 May 5th, 2015, 02:25 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Perhaps it would help if you would state exactly what you think the "Fundamental Theorem of Algebra" is and what you feel is wrong with the proof given. Do you agree that if the original integer "n" is itself prime, we are done? Do you agree that if the original integer "n" is not prime then we can factor it into two integer, a and b, each smaller than n? Do you agree that if a and b are both prime, we are done? Do you agree that if either a, or b, or both, are not prime then that integer can be factored into smaller factors? Do you agree that, since the original integer, "n", is a finite number and each time we factor we have smaller factors, eventually this factoring must end?
 May 6th, 2015, 06:22 AM #5 Newbie   Joined: May 2015 From: USA Posts: 2 Thanks: 0 Math Focus: Number Theory, Bayesian Probability I like the idea of stating exactly where I'm going wrong. I can't get past Lemma 2: If a,b is in the ring of integers and b > 0, there exist q, r in the ring of integers such that a = qb + r with 0 <= r < b. This is in the part where they are proving preliminary results before giving the proof of the main theorem that for every nonzero integer n there is a prime factorization n = ((-1) ^E(n)) * multiplication( p^a(p)). EDIT: I apologize for not showing in a clearer format since I have no idea how to show this in a TeX format. Last edited by drshrey; May 6th, 2015 at 06:23 AM. Reason: Apologies for not show in TeX format
 May 6th, 2015, 06:42 AM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms This is called the division lemma. You use it every time you do long division. Imagine you have 256 divided by 6. You know that 25 = 4*6 + 1 and so you write 4 over 256 and subtract 24 from the 25, leaving 16; then you use that 16 = 2*6 + 4 and so you write a 2 after the 4 giving 42 + 4/6.
May 6th, 2015, 08:28 AM   #7
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The following is a proof for $a \ge 0$, taken from Stillwell's Elements of Algebra.
Quote:
 A proof of the division property, which also gives a way to find r, goes as follows. If b > a then r = a (and q = 0). If a > b then subtract b from a repeatedly until a number r < b is obtained. We must reach such a number because otherwise $\{a - b, a - 2b, a - 3b, ...\}$ form a set of natural numbers without least member.
It can easily be modified for $a \lt 0$ if required.

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