Talk:Algebraic numbers

Solutions of algebraic equations by radicals

If an algebraic equation is solvable by addition, subtraction, multiplication, division, exponentiation and root extraction in terms of the coefficients of the polynomial with integer coefficients, then the solutions are obviously arithmetic. Now, does this hold?

an algebraic equation is NOT solvable by (...) in terms of the coefficients (...)

if and only if

the solutions are NOT arithmetic, i.e. NOT expressible by (...) in terms of the integers

— Daniel Forgues 06:01, 11 January 2013 (UTC)

If an integer solution exists with those operations then the number is arithmetic (to use Stevenson's terminology). I'm not exactly sure how you intend to restrict it when you say "in terms of the coefficients" and so I can't speak to their equivalence. Charles R Greathouse IV 06:28, 11 January 2013 (UTC)

Actually... do you have Stevenson's precise definition? The definition I'm accustomed to ("solvable by radicals") uses merely the basic four plus root extraction, that is, square, cube, ... roots. If he allows unlimited exponentiation then this expands the scope considerably: he can express $2^{2^{1/2}}$ which is transcendental. I assumed the usual definition above and in the article. Charles R Greathouse IV 06:38, 11 January 2013 (UTC)

It allows only exponentiation with integer exponent and radicals with integer index, otherwise we also get transcendental numbers. (He says that all arithmetic numbers are algebraic, so that must be implied. The definition I wrote is his definition verbatim.)

When I say "in terms of the coefficients" I mean a formula expressed "in terms of the coefficients" for equations of a given form, so maybe there is no such formula for equations of some form, although the solutions might be arithmetic. Could this be the case? — Daniel Forgues 07:19, 11 January 2013 (UTC)

Is there a name for transcendental numbers that can be expressed without a power series, like $2^{2^{1/2}}$? — Daniel Forgues 07:24, 11 January 2013 (UTC)

What does it mean to express a transcendental number as a power series? A power series is a different kind of object from a real number...

You may be thinking of "closed-form number" which has been defined variously.

Charles R Greathouse IV 17:40, 11 January 2013 (UTC)

I meant that that every complex number can be expressed as the limit of some power series (probably in a [countably?] infinite number of ways?). So $2^{2^{1/2}}$ has been called a "closed-form number," if I understand... Now, is the set of "closed-form numbers" a countably (or uncountably) infinite proper subset of the complex numbers? Do the "closed-form numbers" have a density of 0 among the complex numbers? — Daniel Forgues 03:40, 12 January 2013 (UTC)

I still don't know what you mean regarding power series. Why not just define them in some standard way, say via Dedekind cuts (and the imaginary unit)?

There are only countably many closed-form numbers (for any of the various definitions), so their density is 0.

Charles R Greathouse IV 01:15, 13 January 2013 (UTC)

Admittedly, a Dedekind cut of the rational numbers uniquely defines a real number. E.g., the Dedekind cut of the rational numbers representing the irrational number $\scriptstyle {\sqrt {2}}\,$ is uniquely given by

$A=\{a\in \mathbb {Q} :a^{2}<2\lor a\leq 0\},\,$

$B=\{b\in \mathbb {Q} :b^{2}>2\land b>0\}.\,$

That was an easy one. How would we express the Dedekind cut for $\scriptstyle \pi \,$? — Daniel Forgues 05:55, 13 January 2013 (UTC)

Powers series do not uniquely define a real number, since e.g. any fixed base $b$ representation of a given real number amounts to a power series and we have a countably infinite number of integer bases $\scriptstyle b\,\geq \,2\,$ to choose from. And there are still an (countable?) infinity of other powers series whose limit is the given real number... so you are right. — Daniel Forgues 05:52, 13 January 2013 (UTC)

Now, how do we know whether or not a given real number is a "closed-form number" of some form? What about $\pi$ or $e$, how do we prove that there does not exist any closed form (I suspect there isn't any closed form for those...) for those transcendental numbers? — Daniel Forgues 05:52, 13 January 2013 (UTC)

Nonarithmetic numbers

I suspect that almost all algebraic numbers are nonarithmetic, is that true? — Daniel Forgues 06:10, 11 January 2013 (UTC)

I'm not sure how to interpret that. Numbers which are solvable in terms of radicals are countably infinite, as are algebraic numbers not solvable in terms of radicals. So what would "almost all" mean?

Surely it's true that almost all polynomials of a given height are not solvable, so in that sense perhaps it's true. But that assumes that picking coefficients from polynomials at a given height is a reasonable way to choose algebraic numbers and I'm not sure if it is for your purpose.

Charles R Greathouse IV 18:04, 11 January 2013 (UTC)

I am tempted to believe that "the density of arithmetic numbers among the algebraic numbers" is 0. (I don't know how one would proceed to evaluate this, and whether or not this is well-defined...) — Daniel Forgues 03:25, 12 January 2013 (UTC)

Not well-defined, as such. You'd need to specify an ordering. Charles R Greathouse IV 01:15, 13 January 2013 (UTC)

One of the examples

The number ${\frac {{\sqrt[{101}]{19}}+127{\sqrt[{29}]{257+{\sqrt[{67}]{13+{\sqrt {-71}}}}}}}{87-64{\sqrt[{997}]{97}}}},$ has real part 6.8191274... and imaginary part 0.000000809389532... Alonso del Arte 18:46, 11 January 2013 (UTC)

"root extraction of natural numbers"

I'm not keen on the phrasing, because this makes it sound like the radicand is restricted to natural numbers rather than the index. But I can't find a good phrasing. Any suggestions?

Charles R Greathouse IV 20:10, 11 January 2013 (UTC)

Root extraction with integer index. — Daniel Forgues 07:36, 13 January 2013 (UTC)