Talk:Casus irreducibilis

It should be noted that a solution given as Cbrt (2+3i) is just as valid as a solution given as Cbrt (5). Both can be numerically evaluated to any desired precision using various approximation methods including Newton's derivative method.

Hence, a solution to a cubic equation such as Cbrt (2+3i) + Cbrt (2-3i) can be numerically approximated to the correct real value.

The only reason for using trigonometry is because not all calculators include complex operations. — Preceding unsigned comment added by 173.164.116.225 (talk) 17:28, 19 January 2019 (UTC)Reply

Take the cubic polynomial x³ − 3x² + 2x. Its discriminant is equal to 4. According to the article, this mean we are in casus irreducibilis. However, the polynomial factors as x(x − 1)(x − 2).

Furthermore, what is the meaning of real field in the section Formal statement and proof?

• If the meaning is that F is the field of real numbers, and the three real roots are r₁, r₂, and r₃, the polynomial can be written in the form a(x − r₁)(x − r₂)(x − r₃) and so is by definition not irreducible over F. Also, then why not simply use ℝ instead of F?
• If F is a formally real field, then what does this have to do with polynomials having real roots?

 --Lambiam 14:37, 1 May 2008 (UTC)Reply

Hi Lambiam, let me respond to your points.

• The polynomial has to be irreducible over F (or, in the case of the lead, Q). I see that this condition had been left out of the discriminant condition, even though it was mentioned in the previous paragraph. I have added it to the second paragraph as well, since that is an obvious source of confusion.
• By "real field" I meant "formally real field". Thanks for catching the fact that real field is a disambiguation page. I think the terminology "real field" is quite established for what we are calling a "formally real field", and that the only reason there is a disambiguation page at all is because of a certain laziness in linking. At any rate, any real field is a subfield of the real numbers (although the real numbers are not needed to define a real field). In this context, "real root" means "root contained in the real closure". I will try to clarify this in the article. silly rabbit (talk) 14:55, 1 May 2008 (UTC)Reply

With your recent edits all my points have been satisfactorily addressed. Thanks.  --Lambiam 15:07, 1 May 2008 (UTC)Reply

The first part of the article defines the discriminant D such that D>0 is the case of two complex roots, but the second part of the article says that D is positive in the three-real-roots case. I think the former notation is common in the literature on cubics and the latter is common in the literature on polynomials in general, but I think the article should adopt a common notation throughout. 174.98.109.4 (talk) 14:55, 4 October 2009 (UTC)Reply

I agree, it's totally confusing. The Cardano page directly contradicts the paragraph in here that links to it. It would be great if someone put in a definition of D. I'd just copy it from the Cardano page, but I don't understand the F business so I don't want to make an even bigger mess. Kallog (talk) 08:18, 23 July 2010 (UTC)Reply

I don't believe that the claim about casus irreducibilis being Cardano and Tartaglia's original motivation for introducing complex numbers is correct. First of all, Tartaglia never published on cubics or imaginaries; Cardano's (1545) book credits Tartaglia (along with Scipione del Ferro and himself) in the solution of cubics but does not mention him in connection with imaginaries. Tartaglia did share his solution with Cardano (after much pleading from Cardano, as Cardano tells the story, and after a specious promise not to publish, as Tartaglia tells it), but they did not have a working relationship. So it seems misleading to me to talk about Tartaglia as having (along with Cardano) introduced complex numbers. Secondly, I don't believe that the connection between cubics and imaginaries is mentioned in Cardano's book. The book includes many chapters devoted to cubics, but all the examples are real. It has one chapter (I believe it is called "On the Rule for Postulating a Negative" and I think it is Ch. 27) where he works with what we now recognize as imaginaries, but in the relevant section in this chapter he does not mention cubics; he works with a quadratic that has complex solutions ("split 10 into two parts whose product is 40"). According to the first chapter of Tristan Needham's book Visual Complex Analysis, the use of complex numbers in solving cubics was introduced by Bombelli, 25-30 years later. Thirdly, it is misleading to speak of Cardano introducing "the complex number system." In his day there wasn't even yet a "system of real numbers." All Cardano's discussion of negatives in the book is somewhat skeptical and speculative. He mentions the possibility of taking a square root of negatives, and then describes the idea as "as subtle as [it is] useless"; this is the extend of his "introduction of the complex number system." Benblumsmith (talk) 15:23, 21 December 2011 (UTC)Reply

The sentence following the one about Cardano and Tartaglia is also anachronistic. The word "imaginary" is not used in Cardano's book. Since it appears that the original article author is no longer active, I am taking the liberty of removing these two sentences.Benblumsmith (talk) 15:36, 21 December 2011 (UTC)Reply

The Generalization section says

Casus irreducibilis can be generalized to higher degree polynomials as follows. Let p ∈ F[x] be an irreducible polynomial which splits in a formally real extension R of F (i.e., p has only real roots). Assume that p has a root in ${\displaystyle K\subseteq R}$  which is an extension of F by radicals. Then the degree of p is a power of 2, and its splitting field is an iterated quadratic extension of F.

Does this mean that casus irreducibilis can only occur beyond the cubics when the degree of the polynomial is 2^k? The reason I ask is that the following source

• Solving Solvable Quintics

which appears in the external references section of the article quintic function, says on p. 17 (the last paragraph before the references)

....this is the Casus Irreducibilis for quintic polynomials, where the five real roots of the quintic are expressed by radicals of necessarily non-real complex numbers....

Can someone who understands this clarify and extend the Generalization section? Thanks. Duoduoduo (talk) 22:34, 9 January 2012 (UTC)Reply

As far back as Bombelli's example of solving the cubic x^3 - 15x - 4, with a root x=4 that the cubic formula derives as (2+i) + (2-i), the casus irreducibilis has been about the paradoxical but unavoidable (hence "irreducibilis") appearance of imaginaries in the cubic formula even to find 3 real roots, regardless of whether the roots can be found by other means. The stronger condition that the roots not be in a real-radical extension is of interest from a modern point of view but is not part of the historical definition of casus irreducibilis (which I think was stable and standard for centuries). I don't think it was mentioned at all before the relatively recent papers on solvability in real radicals raised this issue and resolved it. 73.89.25.252 (talk) 06:03, 14 June 2020 (UTC)Reply

As already partially discussed above and pointed out as a reason for misunderstanding, I have found literature which proves the very wide-spread MIS-understanding:

casus irreducibilis = modern irreducible case

Further remarks:

1. I have added an irreducible polynomial (in the modern sense) which is NOT casus irreducibilis.
2. I have added a reducible polynomial (in the modern sense) which for Cardano IS a casus irreducibilis.
3. I have added the product of differences of the roots which is closer to the things which happen than the discriminant.
4. It should be pointed out (also in other algebraic WP-articles) that the algebraic solutions are NOT (necessarily intended as) numeric solutions (although numeric solutions may be deduced from them). The algebraic solutions very frequently apply to the so-called general equations, i.e. equations with variable coefficients – and not with numbers, moreover, over non-real, non-complex fields. In that sense casus irreducibilis applies to formally real fields, as already mentioned in section Casus_irreducibilis#Formal statement and proof.

–Nomen4Omen (talk) 17:13, 17 November 2021 (UTC)Reply

With only a few changes, Cardano's formula and Vieta's trigonometric method for solving the cubic equation become suitable for solving equations of odd degree >=3. I have summarized these methods in an image that I have posted in the link below, accompanied by a caption with a brief description of how to determine the equations of this category.

https://imgur.com/a/7xvsaFu

Geoplous (talk) 23:32, 25 September 2024 (UTC)Reply