# Is the root of $x^5-x-1$ rational or irrational?

I am wondering whether the unique real root of the polynomial $x^5-x-1$ ($1.1673\ldots$) is rational or irrational. Is it possible to show that it is either rational or irrational? Also, can it be expressed in any other way than "the root of $x^5-x-1$"? For example, by $n^{th}$ roots?

• Do you know the rational roots theorem? – Umberto P. May 16 '17 at 19:27
• Have you heard of the rational root theorem? If there is a rational root of the polynomial, then it must be one of the roots suggested by the theorem. – John Lou May 16 '17 at 19:28
• If it were rational, it would be $1$ or $-1$, by the rational roots theorem. – Bernard May 16 '17 at 19:28

If it is $p/q$, a reduced fraction, then $p^5-pq^4=q^5$, hence $p|q$, a contradiction to being in reduced form unless $p|1$. In that case, $q^5\pm q^4=1$ which is also impossible, since then $q|1$.
If it were rational then according to Rational Roots Theorem, the roots would be $\pm 1$ now when you substitute 1 or -1 in equation the equation doesn't give 0, so this equation has no rational roots.
By the rational roots theorem, it has no rational roots. In "Solving Solvable Quintics" by Dummit it is specifically given a case for polynomials $x^5 + ax + b$ to be solvable in radicals; to be solvable, the resolvent $f_{20}(x)$ must have a rational root, where
$$f_{20}(x) = x^6 + 8ax^5 + 40a^2x^4 + 160a^3x^3 + 400a^4x^2 + (512a^5-3125b^4)x + (256a^6-9375ab^4)$$
If I have not tragically erred in calculation, this polynomial has no rational roots, so $x^5-x-1$ cannot be solved in radicals.