Wikipedia:Reference desk/Archives/Mathematics/2018 January 13

= January 13 =

natural irrational numbers
Much ado is made over the natural numbers. It seems like a similar concept should exist for numbers like $$2\sqrt3 + 3\sqrt[3]{2}$$. AFAICT you should have a mathematical group for addition, subtraction, and multiplication for numbers "$$\sqrt[n]{x_1, x_2 ... x_n}$$" = the sum of $$\sqrt[n]{x_i}$$ for positive $$x_i$$ and $$-\sqrt[n]{-x_i}$$ for negative $$x_i$$. Operations using different n should return numbers with the least common multiple of the n's, at most. I didn't immediately find something like this by searching, but do such/similar numbers have a name and any interesting properties? Wnt (talk) 07:55, 13 January 2018 (UTC)


 * I think you are interested in algebraic numbers, and maybe also algebraic integers for their arithmetic properties. In fact, starting from positive integers and constructing new numbers performing the four operations and extractions of n-th  roots do produce all algebraic numbers (this is the Ruffini-Abel theorem). Actually there are several collections of numbers as you are describing, named algebraic fields, and their properties and characterization is the object of the algebraic number theory.--pm a  09:27, 13 January 2018 (UTC)
 * The Ruffini-Abel theorem shows there are algebraic numbers which can't be produced using the four aritthmetic operations and roots on the natural numbers. For instance the solutions of x^5+x+1=0 are algebraic but can't be produced that way. Dmcq (talk) 14:13, 13 January 2018 (UTC)


 * $$2\sqrt3 + 3\sqrt[3]{2}$$ is a closed-form algebraic expression, and hence is an algebraic solution (also called a solution in radicals). Algebraic solutions are a subset of the algebraic numbers, the latter of which, as Dmcq says, also include solutions of algebraic equations that, by the Abel-Ruffini theorem, cannot be in your form. Loraof (talk) 19:22, 13 January 2018 (UTC)


 * See also Radical extension. Loraof (talk) 19:32, 13 January 2018 (UTC)


 * I have to admit that so far I remain confused, probably due to ignorance. Algebraic numbers seem to include oddities like $$\sqrt[3]{\sqrt2 + 1}$$ that aren't in the set I had in mind... well, then again I don't know that, but I wouldn't think so.  Also, I personally have no idea whether any polynomial has to exist that has a solution $$\sqrt2 + \sqrt[3]{3} + \sqrt[5]{5}$$, for example.  And algebraic fields seem to be about linear combinations, whereas the numbers I have in mind here consist only of the presence or absence of each integer value, with no integer and its inverse appearing in the same expression.  For example, the last one I mentioned is equal to $$\sqrt[30]{32768} + \sqrt[30]{59049} + \sqrt[30]{15625}$$ and thus might be written "$$\sqrt[30]{15625, 32768, 59049}$$"  The numbers have to be unique in "reduced form" because if any two match up you just multiply by 2root and replace them with that.  It seems simple enough and pretty enough I feel like it must be old hat somewhere. Wnt (talk) 22:44, 13 January 2018 (UTC)


 * According to Wolfram Alpha, the minimal polynomial of $$\sqrt2 + \sqrt[3]{3} + \sqrt[5]{5}$$ has degree 30. -- ToE 03:58, 14 January 2018 (UTC)


 * More generally, an expression of the form $$\sqrt[n]{x_1} + \sqrt[n]{x_2} + \dots + \sqrt[n]{x_k}$$ will satisfy an equation of degree nk, and so will be algebraic. The notation you have, $$\sqrt[30]{15625, 32768, 59049}$$ to mean $$\sqrt[30]{15625} + \sqrt[30]{32768} + \sqrt[30]{59049}$$, seems interesting, and you seem to asking whether there is a use for the set of numbers to which it applies. I won't claim the answer is a definite no, but historically it seems more useful to study either more specific forms (as in quadratic fields) or more general forms (as in Galois theory). Not that there's anything wrong with trying to find out interesting things about a middle ground, but math is not always cooperative when it comes to actually doing it. To start with, you might look at the Kronecker–Weber theorem. --RDBury (talk) 05:11, 14 January 2018 (UTC)
 * I think the minimal polynomial article implies every one of these numbers does solve a polynomial, because you should just be able to set up a bunch of factors $$(x - \sqrt[30]{15625} - \sqrt[30]{32768} - \sqrt[30]{59049})(x + \sqrt[30]{15625} - \sqrt[30]{32768} - \sqrt[30]{59049})(etc.)$$. For every product where you multiply an x (for example) by the positive square root there's another where you multiply the other x by the negative square root, so the radical terms all ought to cancel out leaving a polynomial with (I think) integer coefficients.  Though I'll admit, I didn't actually check that in this case let alone prove it. Wnt (talk) 02:33, 15 January 2018 (UTC)
 * $$\prod (x \pm \sqrt[30]{15625} \pm \sqrt[30]{32768} \pm \sqrt[30]{59049})$$ won't work. Not only will there be 8th power constant terms you don't want (e.g. 156258/30), but the middle terms don't cancel out as fully as you might wish.  For instance, $$\prod (x \pm a \pm b) =  x^4 - 2 a^2 x^2 - 2 b^2 x^2 - 2 a^2 b^2 + a^4 + b^4$$.  More generally, this construction will lead to those terms containing any odd powers (of  variable or constant) cancelling out, but you will be left with terms containing only even powers.  (Drawing two marbles from a sack containing an equal number of red and blue marbles will more likely yield a mixed pair than a matched pair.)  Minimal polynomial (field theory) states:
 * If α = √$\overline{2}$ + √$\overline{3}$, then the minimal polynomial in Q[x] is a(x) = x4 &minus; 10x2 + 1 = (x &minus; √$\overline{2}$ &minus;  √$\overline{3}$)(x + √$\overline{2}$ &minus; √$\overline{3}$)(x &minus; √$\overline{2}$ + √$\overline{3}$)(x + √$\overline{2}$ + √$\overline{3}$).
 * The minimal polynomial in Q[x] of the sum of the square roots of the first n prime numbers is constructed analogously, and is called a Swinnerton-Dyer polynomial.
 * The key for this method is the square roots which are taken care of by the even powers. -- ToE 14:59, 15 January 2018 (UTC)
 * Abstractly, one can consider the tower of fields obtained by adjoining separately each of $$\sqrt{2}, \sqrt[3]{3}, \sqrt[5]{5}$$. Because 2, 3, and 5 are coprime, and it can be shown that the two ways of constructing the field extension are the same, one obtains the degree of the overall extension generated by adjoining $$\sqrt{2} + \sqrt[3]{3} + \sqrt[5]{5}$$ as the product of the degrees of the individual extensions (2, 3, and 5).--Jasper Deng (talk) 08:30, 15 January 2018 (UTC)

Let $$\omega$$ be a primitive 30rd root of unity, say $$\omega=\cos(2\pi/30)+i\sin(2\pi/30)=1^{30^{-1}}$$. Consider the polynomial $$\prod_{l=0}^{29}\prod_{m=0}^{29}\prod_{n=0}^{29} (x -\omega^l \sqrt[30]{15625}-\omega^m \sqrt[30]{32768}-\omega^n \sqrt[30]{59049})$$. Bo Jacoby (talk) 23:11, 15 January 2018 (UTC).