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As always, Quanta Magazine reporting is just garbage. The name for such numbers is ‘algebraic’.
by 9991 3y ago
As always, Quanta Magazine reporting is just garbage. The name for such numbers is ‘algebraic’.
- gjm11 3y agoThe 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.
- JadeNB 3y ago> 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.
- throwaway81523 3y agoNo 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 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.
- bubblyworld 3y agoYeah, 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'.
- throwaway81523 3y agoIt 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.
- bubblyworld 3y agoThat'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).
- throwaway81523 3y agoI'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?
- bubblyworld 3y agoPardon me, you're completely right! Thanks for reminding me to refresh my basics before I write =)
- bubblyworld 3y agoI highly recommend reading the article, you might be pleasantly surprised =)
- 9991 3y agoI wrote them off a long time ago, but after this comment thread, I’ll give them another shot. Thanks!