Consider the set of all numbers which are constructable via finite addition, multiplication, and exponentiation on the Rationals. This set includes numbers like -1001, sqrt(2), and (3000[sup]1/10[/sup] + 75/82). This set is clearly a subset of the Algebraic numbers; I used to think that it was the Algebraic numbers, since whenever teachers give you examples of Algebraics, they have this form. Then I discovered sin(10°), which is a root of 8x[sup]3[/sup] - 6x + 1. Convinced that it must have had a form constructable from the Rationals, I sought it, and was met only with failure. I eventually came to the realization that nobody had ever told me that all Algebraics are in the set I mentioned earlier; I had just assumed it. So now I’m curious about this set.
Is it actually a proper subset of the Algebraics? Is there a proof (for sin(10°), or anything) that shows an Algebraic number not to be in the set?
If it is a proper subset, do its elements have any special properties that the Algebraics in general do not?
Is there a simple way to state a means by which any Algebraic number (and no other) may be constructed, such as “finite addition, multiplication, exponentiation, and taking-the-sine-of on the Rationals”?
Oh, sorry. I didn’t mean to imply that there was anything special about the sine function, or any trig function. I just offered sin(10°) as an example of an Algebraic number which I believe may not be in the set of constructables. I understand that transcendental functions will at points take on Algebraic values; in truth, though, I wasn’t really concerned about that. That first paragraph is sort of a tangent - what I really want to know is the answers to my questions at the bottom.
I’ve forgotten a lot of whatever Galois theory I used to know, but yeah, that construction you mentioned gives only a proper subset of the algebraic numbers. Abel proved that there are fifth degree (and above) polynomial equations which cannot be solved by taking radicals. Your example of sin(10[sup]o[/sup]) actually does have an expression in radicals–you can use the cubic equation to solve any cubic polynomial using radicals.
Obviously these constructed numbers do have some special properties over other algebraics, namely, they can be constructed this way. The associated Galois groups have special properties as well, but I’m afraid I don’t remember the details offhand.
Of course, the definition of the algebraic numbers is the set of all roots of polynomials with rational coefficients, but that’s probably not what you mean by “construct”. I don’t know of any collection of operations with which you can construct them, they certainly wouldn’t be “common” operations, but there may be such a collection, I don’t know.
“you can use the cubic equation to solve any cubic polynomial using radicals.”
Really? That’s actually one of the things I was looking for, but I assumed it didn’t exist. The only cubic equations I’ve seen, when there were three real roots, gave them in terms of sines or cosines, so you were back where you started (Cf. Eqs. 60-62 on this page). Can you provide a link or an explanation of the cubic equation which doesn’t involve this?
I don’t think that sin(1°) is the root of any third- or fourth-degree polynomial, but it’s Algebraic, so maybe that’s a better example.
Thanks for the comments on Galois groups. I don’t know what you’re talking about, but I know what to look for now. Do you happen to know of a fifth-degree equation, offhand, one of the roots of which is not constructable? (I think I was pretty clear what I meant by constructable, but if you want me to explain it again, I will.)
x[sup]5[/sup] -6x + 3 = 0 has no roots that are constructible by radicals.
There’s a formula for cubic polynomials and quartic polynomials using radicals. Quintic polynomials are where such a solution fails. I gotta run right now, but if noone else does, I’ll hunt up a link to the cubic and quartic formulas later (they’re pretty common, and shouldn’t be hard to find).
When you apply that cubic formula to Achernar’s simple 8x3 - 6x + 1, you get imaginary numbers in the result (you can skip down the page, since it is already in the depressed cubic form). Of course, they will end up canceling, since the equation does have three real roots, but just try it yourself. Even Mathematica failed me.
The problem with the cubic formula is that when all three roots are real, the solution may involve the cube root of a complex number. Sometimes it isn’t possible to simplify this to get an answer in closed form with no imaginary numbers in it, even though all the roots are real. Since the solution of the quartic involves an auxiliary cubic, this same problem may carry over. So it is possible to solve any cubic or quartic with radicals, but they may involve radicals of complex numbers, even though the roots are all real. This is the reason the solutions are sometimes given in trigonometric form, since that way you can avoid the complex cube roots.
It isn’t immediately obvious that the field generated from the rationals by appending all exponentiations to rational powers is a subfield of the algebraic numbers, but in fact it is. In other words, given a complicated expression containing integers, addition, multiplication, and various radicals, with radicals imbedded in radicals, etc., how do you construct an algebraic equation of which it is the root? In simple cases (for example, where there is only one radical), you set x equal to the expression, get the radical on one side of the equation all by itself, and take both sides to the proper power to eliminate the radical. That is, if it is a cube root, then cube both sides, etc. For more complicated expressions, where there are several radicals, this procedure can be used to eliminate one of them, but at the same time it might create more complicated ones on the other side.
and the solutions are of the form (s - t). I put it in polar notation so that it’d be easy to take the cube roots. Now then, of course each of the numbers there has three cube roots, and each one corresponds to a different root of the polynomial. One of them is as follows:
s = (1 / 2)e[sup]i280°[/sup] = -(1 / 2)e[sup]i100°[/sup]
t = (1 / 2)e[sup]i260°[/sup] = (1 / 2)e[sup]-i100°[/sup]
x = s - t = -(1 / 2)(e[sup]i100°[/sup] + e[sup]-i100°[/sup]) = -cosh(i100°) = -cos(100°) = cos(80°) = sin(10°)
Okay okay. Technically, then, you can write sin(10°) like this:
But even though this expression meets my original condition of being constructable from the Rationals, I don’t really like it for two reasons. First, it’s not really what I had in mind - imaginary numbers cancelling out. I was really looking for something I could punch into a scientific calculator with a broken sin key. Second, it’s triple-valued. I guess there’s not a single-valued expression that fits the bill? Richard seems to say there won’t be, in general, so oh well…
Cabbage, is there any known way to express any of the roots of x[sup]5[/sup] - 6x + 3?
Richard, I was under the impression that construction of a polynomial for which any given Algebraic is a root was tedious, but straightforward. Can you give an example for me to waste my time trying to do?
Just out of curiosity, where did you get the polynomial 8x[sup]3[/sup] - 6x + 1? I ask because I’m looking at my algebra book, and it uses the polynomial 8x[sup]3[/sup] - 6x + 1 to show that a 60[sup]o[/sup] angle can’t be trisected with a compass and straightedge.