Talk:Quadratic irrational number

message: I simply cannot follow the application of the Fundamental theorem of arithmetic:
:For any rational non-integer in lowest terms there must be a prime in the denominator which does not divide into the numerator. When the numerator is squared that prime will still not divide into it because of the unique factorization. Therefore if an integer is not an exact square of another integer its square root must be irrational.
I'm OK up till "therefore".  
Could someone try to make this more explicit?
Dan kirshner (talk) 05:07, 3 September 2009 (UTC)
replies: text: That implies the square of a rational non-integer cannot be reduced to an integer so the square root of a non-square integer cannot be a rational non-integer. Dmcq (talk) 08:08, 3 September 2009 (UTC)
thread_title:  Clarification, please? 

message: What was Dedekind thinking when he claimed to prove this without appealing to the Fundamental Theorem of Arithmetic?  He's simply expanded a proof of the FTA at the point where one would appeal to it.  How is substituting a proof of a desired theorem different in any nontrivial sense from appealing to the theorem itself?  --Vaughan Pratt (talk) 21:14, 21 December 2008 (UTC)
replies: text: The proof doesn't even mention primes or anything similar to a prime number, why do you say it is a reformulation of the fundamental theorem of arithmetic? Dmcq (talk) 00:29, 22 December 2008 (UTC)
replies: text: Because the techniques used in Dedekind's proof are exactly those usable for a quick proof of unique factorization.  --Vaughan Pratt (talk) 11:12, 7 January 2009 (UTC)
replies: text: The proof was in relation to Dedekind cuts and he wasn't interested in the fundamental theorem or arithmetic, I believe he just wanted something self-contained, I don't think he was claiming anything. I put in the bit about the proof just depending on ordering and not making use of prime numbers. I'm not going to make a point of it so I'll just put in something about the proof being self contained instead, I only really put it in originally because there were so many proofs assuming the fundamental theorem without appealing to it and it seemed a good way of emphasizing that it wasn't altogether obvious. You can delete it if you think it is just taking up space. Dmcq (talk) 13:58, 7 January 2009 (UTC)
replies: text: If you want a short, self-contained proof, though, I'd say Estermann's is shorter and neater and assumes no more knowledge than Dedekind's.  There's a version of it on page 11 of his obituary for root two, and it's a simple generalisation to roots of non-squares in general. Chenxlee (talk) 15:14, 23 January 2009 (UTC)
replies: text: I didn't know that proof. Unfortunately it is written only for the square root of 2. It can be generalized easily but then there wouldn't be a citation. I would be better with a shorter proof and it certainly looks worthwhile searching for one. Dmcq (talk) 16:19, 23 January 2009 (UTC)
replies: text: There's a short note in The Mathematical Gazette, Vol. 83, No. 498 (1999), pp.502-503 by Colin Richard Hughes titled Irrational roots that extends Estermann's proof to cover any non-square natural number.  It's no longer than the root two proof and has a citation.  I can add it unless you know a neater or nicer proof. Chenxlee (talk) 17:11, 23 January 2009 (UTC)
replies: text: My access to things is a bit limited. Yes please, thanks. Dmcq (talk) 18:58, 23 January 2009 (UTC)
replies: text: All these proofs are essentially the same proof, namely the usual inductive proof showing that the integers form a principal ideal domain.  Every principal ideal domain is a unique factorization domain.  All that's happening in these proofs is the substitution of a proof of the FTA (unique factorizability of the integers) for a direct appeal to the FTA.  One does not need to use the word "prime" in the statement that the integers form a PID and hence a UFD.  --Vaughan Pratt (talk) 16:44, 6 May 2011 (UTC)
thread_title: Why Dedekind's proof?

message: The article claims "The quadratic irrationals with a given c form a field" but defines the concept in such a way (b nonzero) as to preclude rationals, in particular 0 and 1.  So how can they form a field?
Since there is no such thing as a "quadratic rational" what is the point of requiring b to be nonzero?   Granted every ham sandwich is a sandwich, but why must every quadratic irrational be irrational?  The concept becomes cleaner when b can be zero.  --Vaughan Pratt (talk) 17:30, 5 May 2011 (UTC)
replies: text: Well that's all very rational(!) but for instance irrationals don't include rational numbers and quadratic equations don't include where the coefficient of x squared is zero. It's just how they were defined and it has stuck. Dmcq (talk) 17:39, 5 May 2011 (UTC)
replies: text: So you don't mind the article claiming that the quadratic irrationals form a field when obviously they don't? --Vaughan Pratt (talk) 02:22, 6 May 2011 (UTC)
replies: text: Sorry I see what you're on about, it's just language, feel free to improve that bit if you can do it without making it sound too clumsy. Dmcq (talk) 21:34, 6 May 2011 (UTC)
thread_title: Not a field?

message: This is an excellent article. I am having trouble following this sentence:
The rational numbers together with all quadratic irrationals with a given c form a field, called a real quadratic field. In particular, their inverses are of the same form...
Which inverses are of the same form?Dratman (talk) 22:06, 10 June 2015 (UTC)
thread_title:  Which inverses are of the same form? 

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