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Comment by vinnyvichy | Hacker News Reader

The qm article says "A number is totally real if it satisfies a polynomial equation with integer coefficients" etc but I love your spirit and it's totally on quantamags for not stating their inputs up front

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Right, but that's not what they're actually plotting - the quanta bagel is algebraic numbers with colour given by the entropy of their corresponding dynamics! Whereas the one I posted is roots of integer polynomials by density.

The fact that they look similar is striking. It means in some sense roots of integer polynomials generate more complex iterative systems, which is surprising (to me).

As always, Quanta Magazine reporting is just garbage. The name for such numbers is ‘algebraic’.

The quotation from Quanta was incomplete, and the Quanta article doesn't in fact claim that "totally real" means what "algebraic" actually means. Here's what the article actually says (added emphasis mine):

> A number is totally real if it satisfies a polynomial equation with integer coefficients that only has real roots.

Nothing wrong with that.

> As always, Quanta Magazine reporting is just garbage. The name for such numbers is ‘algebraic’.

Any popularization is inevitably going to run into some inaccuracies—or else it's just re-publishing the technical papers—but the opinions of most mathematicians I know are that, far from being garbage, Quanta's reporting is distinctly better than most.

No it's narrower than algebraic. If I have this right, the number is totally real if it's a root of a polynomial with integer coefficients, all of whose roots are real. So for example, the (real) cube root of 2 is not totally real, because x^3-2 has one real root and two complex roots. The idea of totally real is that not only is the root real, but all of its images under the polynomial's Galois group are also real.

See: https://en.wikipedia.org/wiki/Totally_real_number_field

Added: in fact the Quanta article gets this right. It says "A number is totally real if it satisfies a polynomial equation with integer coefficients that only has real roots." Also, Jordana Cepelowicz is a knowledgeable math writer and I believe she has a PhD in math.

Yeah, you've got it right, the Quanta article covered this accurately.

For instance, 'i' is algebraic but not totally real (it isn't a real number!), since it satisfies 'x^2 + 1'.

It does seem like a bit of not totally accurate shorthand: totally real isn't a property of a number, but rather of the extension field where it lives. So sqrt(2) is a root of x^2-2 whose roots are +/- sqrt(2), and the splitting field of x^2-2 is Q(sqrt(2)) which is totally real. But sqrt(2) is also a root of x^4-4 whose roots are +/- sqrt(2) and +/- i sqrt(2), so the splitting field doesn't live inside the reals. The wiki article explains it better. I was unfamiliar with this idea before.

That's true but I think it's implicit here we're working with the algebraic extension generated by the number itself (which always contains the Galois conjugates).

I'm not sure if I'm using the jargon properly, but if r is the real cube root of 2, then I thought the extension generated by r is Q(r), which contains only reals. So (x^3 - 2) factors into (x-r) and the irreducible quadratic (x^2 + r*x + r^2). You have to do some extra thing to reach the splitting field, amirite?

Pardon me, you're completely right! Thanks for reminding me to refresh my basics before I write =)

I highly recommend reading the article, you might be pleasantly surprised =)

I wrote them off a long time ago, but after this comment thread, I’ll give them another shot. Thanks!

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