Talk:Quadratic irrational

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[edit] Expanding this article

I'm interested in expanding this article. I see three possible avenues for expansion.

Analogies with the field of complex numbers. I think that the early development of algebraic number theory was driven (in part) by the realization that adjoining an irrational square root to the rational numbers is very similar to adjoining the imaginary unit i to the rationals. Think of similarities between say the Gaussian integers and more general algebraic integers.
Discussion of the history of algebraic numbers. I don't know as much about this as I should. But the first irrational numbers known to exist were quadratic irrationals (like the square root of two, and the golden ratio, and other irrationalities involving the square root of 5, all of which were understood by the Greeks). And there are close interrelationships between quadratic irrationals and interesting sequences of natural numbers, such as the Pell numbers (square root of two) and the Fibonacci numbers (square root of five). This could be tied in with, for instance, Brahmagupta's methods for deriving some of the solutions to Pell's equation.
Expansion of the discussion about quadratic surds and regular continued fractions. There are several interesting theorems about this, and I do have access to some historical data about those theorems (Euler, Legendre, and Galois all obtained interesting results).

Anyway, if anyone else is interested in this subject, please chime in. I'll keep an eye on this page. DavidCBryant 16:26, 23 April 2007 (UTC)

Just an expression of support. 04:57, 30 April 2007 (UTC)

Yes, really interesting article. According to current definition numbers of the form (listed at Continued fraction calculator) (I'm using denominations as in the article and adding d):

${a+d\sqrt{b} \over c} \qquad \ldots d \in \mathbb{Z}; \, d \ne 0 \; (\mathrm{or})\; d > 0 \!\,$

are not "listed in" this set, but they also have periodic continued fraction form. For example:

$2\sqrt{42} = 12,9614 \ldots = [12;\overline{1,24}] \!\, ,$

$1+2\sqrt{2} = 3,8284 \ldots = [3;\overline{1,4}] \!\, ,$

$1+42\sqrt{42} = 273,1911 \ldots = [273;\overline{5,4,3,2,1,10,2,2,2,1,10,1,7,10,1,59,1,1,2,1,3,10,1,5,3,1,1,1,3,5,1,10,3,1,2,1,1,59,1,10,7,1,10,1,2,2,2,10,1, 2,3,4,5,544}] \!\, ,$ (hm, quite a long period (length 54))

$(42+42\sqrt{42})/42 = 7,4807 \ldots = [7;\overline{2,12}] \!\, ,$

$(40+41\sqrt{42})/43 = 7,1095 \ldots = [7;\overline{9,7,1,3,4,5,1,11,1,21,5,38,1,399,1,11,1,3,1,1}\ldots] \!\, ,$ (a period with length 556!)

and such ... --xJaM (talk) 21:24, 9 January 2008 (UTC)

At the article for vinculum is said that: "quadratic irrational numbers are the only numbers that have" periodic continued fraction representation and these numbers are obviously also quadratic irrationals, so can be included in the article. --xJaM (talk) 17:15, 11 January 2008 (UTC)

Also strange is a sentence at periodic continued fraction "... $D > 0$ is not a perfect square, and $Q$ divides the quantity $P 2 - D$ ". There integers a, b and c are denoted P, D, Q respectively. Is the last condition really necessary, since in these examples, the case is not true?

$(42+\sqrt{5})/42 = 1,0532 \ldots = [1;18,\overline{1,3,1,1,1,1,4,1,1,1,1,3,1,36}] \!\, ,$ and $42 \not\vert \ 42^{2}-5 \!\, ,$

$(13+\sqrt{11})/79 = 0,2065 \ldots = [0;4,1,5,\overline{3,6}] \!\, ,$ and $79 \ \vert \ 13^{2}-11 \!\, .$ --xJaM (talk) 01:35, 12 January 2008 (UTC)

Quadratic surds are really a particular kind of quadratic irrational, those of the form

${a+\sqrt{b} \over c}$

with b square-free. If we have a general quadratic irrational then we can always take the coefficient of the square root inside the root to get the number in the form the article originally stated, but I think the definition now given avoids worrying about signs and agrees with what most number theory texts would call a quadratic irrational. I'd like to help expand this page when I find my notes on this stuff, but I'm not sure whether the continued fraction stuff would belong here or in the periodic continued fraction article. Chenxlee (talk) 19:06, 13 February 2008 (UTC)

And the condition Q divides P²-D is a consequence of the quadratic formula. If the original equation is ax²+bx+c=0 then Q is 2a and P²-D is just 4ac Chenxlee (talk) 19:26, 13 February 2008 (UTC)

[edit] My editing

I am not a specialist in this area, but certain problems in the article were apparent even to me.

The article lacks references, yet makes substantive claims and points of usage that are not in accord with all sources. The line taken here needs to be supported from the literature. Where, for example, is the supposed distinction between quadratic irrational and quadratic surd to be found in printed works of reference?

The definition, though recently changed and improved, I think, was factually and expositorily inadequate. Here is the way I have put it:

The quadratic irrationals, therefore, are all those numbers that can be expressed in this form:

${a+b\sqrt{c} \over d}$

for integers a, b, c, d; with b and d non-zero, and with c positive and not a perfect square.

First, it needs to be clear that all numbers satisfying the stated conditions are quadratic irrationals, and vice versa. Second, the restrictions on b and d are obviously essential. Third, there are two restrictions on c. Obviously it must be positive, since we are dealing with real numbers only (aren't we?). But it is constrained to be non-square, in the formula as we have it. It is not constrained to be square-free. ~~An alternative formula is available in which that component must indeed be square-free. Using our symbols the same way as much as possible:~~

~~${a\pm\sqrt{c} \over d}$~~

~~for integers a, c, d; with d non-zero, and with c positive and square-free.~~

This formula is equivalent, as reflection on the implications of being square-free under the square-root operator will reveal. This alternative formula appears to be used here and there in web sources. [NO! Retracted below.–– Noetica^♬♩ Talk 02:10, 14 February 2008 (UTC)]

Am I right? Comments?

– Noetica^♬♩ Talk 23:52, 13 February 2008 (UTC)

The non-zero restrictions are definitely needed, well spotted. As for those two definitions I think you've got the conditions on c the wrong way around. If c is not a perfect square then we can write √c as ±b√c′ for a square-free c′ just by "factoring out" the squares. But if c is square free then √c can only ever be written as √c, so the b in front is needed.

As to whether c must be positive, I'm really not sure. I looked in Hua's book on number theory today and he defines quadratic irrationals as algebraic numbers of degree 2, so that would include stuff like i=(0+√(-1))/1. I'd have to look through more texts to see if there's any general concensus about what constitutes a quadratic irrational. As for the difference between quadratic irrationals and surds, the only place I've seen the latter mentioned is on the linked Mathworld article. But again I'd need to flick through some books to see if there's any decent reference to them. Chenxlee (talk) 00:10, 14 February 2008 (UTC)

Chenxlee, I think you are right about the alternative formulation that I gave. Something is wrong with what I wrote above. But the formula I have put in the article is correct, and it makes the definition rigorous. Related to what you say, a quadratic irrational (QI) is always expressible with a non-square-free term under the square-root operator:

${1+6\sqrt{2} \over 1}$

is a QI and is equivalent to

${1+3\sqrt{8} \over 1}$

So the definition is more accurate the way I have done it, I think. We have to distinguish between numbers and expressions for numbers, don't we?

Many sources on the web make a mess of this. Let's not be among them!

– Noetica^♬♩ Talk 02:10, 14 February 2008 (UTC)

Yep, the definition does seem more rigorous as you've put it, good job. I removed the remark about Pell's equation since as it stood it didn't mean anything. I know periodic continued fractions can be used to solve Pell's equation but that deserves a little more explanation than the article gave, and I'll try to add something this weekend.

A quick flick through some text books seemed to show that when continued fractions were being discussed the term quadratic irrational was taken to mean a real irrational solution to a quadratic equation with integer coefficients. Which makes sense because then we have the result that a continued fraction is periodic precisely when the number is a quadratic irrational. In more general algebraic number theory texts, though, quadratic irrational seems to be used for any irrational root of a quadratic equation with integer coefficients, which again makes sense because the results about the quadratic field you then get don't rely on c being positive or negative, but only what it reduces to modulo 4. I'll try to get some decent references for all this and make the distinction in the article. Chenxlee (talk) 08:51, 14 February 2008 (UTC)

Fine, Chenxlee. I think the article is better now, without that reference to Pell's equation. I am not really a mathematician, myself. I am interested most in foundations and definitions, and that's hard enough! I did some snooping on the web, and was amazed at how badly quadratic irrationals were presented. Even at Mathworld, where the definition is very poor indeed. I'll keep watching here, and I'll do what I can to keep the article clear, and to discuss things with a view to improvement. After all, this article is the first find on a Google search for either "quadratic irrational" or "quadratic surd" (which I really think must be the same, in most "dialects"). Perhaps that gives us a special responsibility.

– Noetica^♬♩ Talk 01:38, 15 February 2008 (UTC)