1
$\begingroup$

Can an irrational number be constructed which is a) not any known transcendental number b) not a surd? If yes, then how can I construct one? A detailed answer regarding the theory behind this and some references will be appreciated. I modified the a) part to what it is now, because I am guessing numbers are either algebraic or not algebraic(I.e. transcendental). Is this correct? I am thinking along the lines of constructing a sequence which converges to the desired number, but then how to construct a sequence to a desired limit?

$\endgroup$
7
  • 1
    $\begingroup$ What is a known transcendental number BTW? $\endgroup$
    – user99914
    Nov 4, 2015 at 6:15
  • $\begingroup$ Err, e or pi or golden ratio? $\endgroup$
    – Non-Being
    Nov 4, 2015 at 6:16
  • 1
    $\begingroup$ What do you mean by a surd? If you mean something you can write down in terms of $n^{th}$ roots then the answer is yes. Some quintic polynomial should work. $\endgroup$ Nov 4, 2015 at 6:16
  • $\begingroup$ @user286490 e and pi are transcendental and the golden ratio is a surd $\endgroup$ Nov 4, 2015 at 6:17
  • $\begingroup$ Oh okay, and yes, by surd I mean what you. $\endgroup$
    – Non-Being
    Nov 4, 2015 at 6:18

1 Answer 1

1
$\begingroup$

The answer is yes. If you know about Galois theory, you need an extension of $\mathbb{Q}$ that has a non-solvable Galois group (like $S_5$). If you don't know about Galois theory, then the roots of the polynomial $x^5-80x+5$ are irrational numbers but they are not surds and not transcendental.

$\endgroup$
8
  • $\begingroup$ Okay, but can I know what the number looks like? $\endgroup$
    – Non-Being
    Nov 4, 2015 at 6:23
  • $\begingroup$ Is it possible to prove that the roots are not surds without knowledge of Galois theory? $\endgroup$
    – cr001
    Nov 4, 2015 at 6:23
  • $\begingroup$ @cr001 I don't know of any way to show that the roots are not surds without using Galois theory. I think it would be quite difficult. $\endgroup$ Nov 4, 2015 at 6:26
  • $\begingroup$ @user286490 what do you mean by what the number looks like? You could use a CAS or wolfram to get approximate complex roots of the equation. Another polynomial that will work is in this link $\endgroup$ Nov 4, 2015 at 6:27
  • $\begingroup$ @Sam Weatherhog Thanks. But using Galois theory there is a way of proving numbers are not surds? $\endgroup$
    – cr001
    Nov 4, 2015 at 6:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .