2

In the case of cubic equations,

Casus irreducibilis occurs when none of the roots is rational and when all three roots are distinct and real (...) —Wikipedia's Casus irreducibilis article

So, $x^3-3x+1=0$ is definitely an example of casus irreducibilis.

Cardano's formula can express a rational root in terms of non-real radicals (yet it is unnecessary), as in this example: $x^3-15x-4=0$. Some (Working with casus irreducibilis) call this equation a casus irreducibilis, but this disagrees with the (supposed) Wikipedia definition (which is described below), as it has a rational solution, namely $x=\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$.

Does the question in the link just involve a misinterpretation of casus irreducibilis, or are there any trustworthy books or other sources which support the fact that equations like $x^3-15x-4=0$ (which yield a rational root through Cardano's formula, though unnecessarily, using roots of complex numbers) are casus irreducibilis?

I suppose that the Wikipedia definition should read

Casus irreducibilis occurs if and only if none of the roots is rational and if and only if all three roots are distinct and real (...)

instead, as this defines casus irreducibilis precisely.

Wane
  • 317

3 Answers3

2

The "irreducibilis" part of casus irreducibilis is irreducibility over the rationals. Hence $x^3-15x-4=0$ is not casus irreducibilis.

The linked question, however, has not really misused the term.

My question is, using Cardano's method for casus irreducibilis...

implies that it is about a situation where the rational root test isn't used beforehand, and the equation is assumed casus irreducibilis; we want to tell if the root obtained from Cardano's formula is really a rational in disguise.

Parcly Taxel
  • 103,344
  • In the linked paper, I found this sentence: "It might be added that there exist more complicated examples of the casus irreducibilis where there are no simple formulae for real roots of cubics, except by using imaginary numbers." just after solving $x^3-15x-4=0$, as if it suggested that $x^3-15x-4=0$ alone is a casus irreducibilis. Isn't that misleading? – Wane Apr 11 '20 at 18:16
  • 1
    @Wane Yes, it's misleading. – Parcly Taxel Apr 11 '20 at 18:17
1

Historically casus irreducibilis simply meant the case where the discriminant is negative regardless of the existence of rational roots.

For instance, Lagrange's lectures on elementary mathematics discusses the irreducible case in connection with Bombelli's example $x^3 = 15x + 4$, which was chosen because it has an integer solution and so can be used to test the cubic formula.

The idea that "irreducible case" means the cubic polynomial has to be irreducible seems to be a modern idea from field and Galois theory, projected backward. Before Galois the concern was with generic solution formulas where the coefficients of the polynomial are parameters. Solvability and unsolvability of particular equations not in a parametric family was not really subject to a general theory until Galois placed it in the context of field extensions, Galois group actions and so on.

In short, the Wikipedia is probably wrong, but for the subsequent modern analysis of whether casus irreducibilis can be circumvented, of course the first thing to require is that there be no rational solution.

Trivial Notions
  • 11
  • Casus irreducibilis actually implies the existence of polynomails equations not solvable in radicals. For

    $$\cos(tan^{-1}(x))$$ is algebraic for any algebraic $x$.

     – Michael Ejercito Feb 25 '24 at 21:55
  • If $r(\cos(x)+i\sin(x))$ is algebraic, $r\cos(x)$ and $r\sin(x)$ are algebraic.

    If $r\cos(x)$ and $r\sin(x)$ can not be expressed in radicals over the real numbers, neither can their minimal polynomials be solved by radicals!

     – Michael Ejercito Feb 25 '24 at 22:01
0

The real and imaginary parts of a cube root of an arbitrary complex number can not always be expressed as a finite sequence of arithmetic operations, square root extractions, and cube root extractions over the real numbers.

A proof is here.

https://en.wikipedia.org/wiki/Casus_irreducibilis

Implicit is that the real and imaginary parts of solutions to cubics and quartics can not always be expressed as a finite sequence of arithmetic operations and root extractions over the coefficients and constants.

This does not contradict the solvability of cubics and quartics. Each solution as a whole can be expressed as a finite sequence of arithmetic operations and root extractions over the coefficients and constants.

there is a pattern here.

For quadratics, the solutions' real and imaginary parts each can be expressed as a finite sequence of arithmetic operations and square root extractions over the coefficients and constants

For cubics and quartics, the solutions as a whole can be expressed as a finite sequence of arithmetic operations, square root extractions, and cube root extractions over the coefficients and constants, but their real and imaginary parts not necessarily so.

For quintics and higher, the solutions are not always expressible (even as a whole) as a finite sequence of arithmetic operations, nth root extractions, ($n$ being positive integers) over the coefficients and constants.

Another corollary.

Let $\arg(a,b)$ be the number of radians of an arc between ($a,b$) and ($\sqrt{a^2+b^2},0$)on a circle with center ($0,0$), and let $i^2=-1$

Then the cube roots of $a+ib$ can only be expressed in radicals if $\cos(\frac{\arg(a,b)}{3})$ is a solution to a polynomial belonging to a solvable group.

Michael Ejercito
  • 196