4 ms·
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
by bubblyworld 3y ago
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'.
- 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 =)