2
$\begingroup$

The third root of 2 is a real algebraic number. It is not constructible.

If we had a set generated from all the constructible numbers and all finite roots (not just square roots) would this set be the same as the real algebraic numbers?

Alternatively, is there any real algebraic number that is not constructible and not a finite root of some constructible number?

$\endgroup$
  • $\begingroup$ No, all constructible numbers are roots, so you would miss out on most algebraic numbers, since almost none of them are solvable by radicals in general, so take, e.g. a root of $x^5-4x+2$ $\endgroup$ – Adam Hughes Jul 17 '18 at 3:50
0
$\begingroup$

Absolutely there are! For example, there is no way to express the solutions of $x^5-x+1$ as radical expressions. More simply, I think that $\sqrt[3]{2} + \sqrt{3}$ satisfies your criteria.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.