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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?

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  • $\begingroup$ Do you know the rational roots theorem? $\endgroup$ – Umberto P. May 16 '17 at 19:27
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    $\begingroup$ 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. $\endgroup$ – John Lou May 16 '17 at 19:28
  • $\begingroup$ If it were rational, it would be $1$ or $-1$, by the rational roots theorem. $\endgroup$ – Bernard May 16 '17 at 19:28
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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 $q|1$.

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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.

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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.

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