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Every problem about algebraic-ness (rational-ness) of numbers that I have seen is in one of the below types:

  • The number is algebraic(rational) and proving that it is algebraic(rational) is trivial, like proving that $\sqrt{7+\sqrt{2}}$ is algebraic.
  • The number is not algebraic, and $\cdots$.

My Question:

  • Is there any example of an algebraic(rational) number which is not trivial to prove that the number is algebraic(rational)?
Mohammad Ghiasi
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  • What is "simple"? I don't understand the question. – Vladimir Dotsenko Apr 05 '15 at 17:06
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    I think you'll be interested in http://mathoverflow.net/questions/32967/have-any-long-suspected-irrational-numbers-turned-out-to-be-rational. In particular, I think Legendre's constant (http://en.wikipedia.org/wiki/Legendre%27s_constant, mentioned in the first answer) is a pretty good example. – Noah Schweber Apr 05 '15 at 17:08
  • for example proving that $\sqrt{7+\sqrt{2}}$ is algebraic is simple. @VladimirDotsenko – Mohammad Ghiasi Apr 05 '15 at 17:09
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    Conway's Constant might be an example: It is algebraic, the root of a polynomial of degee $71$. – Joseph O'Rourke Apr 05 '15 at 17:13
  • The fact that you can just give this one example is precisely why I think that this question is not too well formulated, and vote to close it in its present form. I am sure there must be ways to ask a related question that would have mathematical content, but this way it is too vague. – Vladimir Dotsenko Apr 05 '15 at 20:00
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    There certainly are examples that are hard if the student knows just the basic definition of algebraic and no actual related theory behind that property. For example, show $\sqrt[3]{2} + \sqrt[4]{2} + \sqrt[5]{2}$ is algebraic, or if $\alpha$ is a root of $x^3 + x + 1$ and $\beta$ is a root of $x^8 + 6x^3 + 4x - 10$ then prove $\alpha + \beta$ is algebraic. Do you think those are trivial? – KConrad Apr 06 '15 at 15:01
  • @VladimirDotsenko, I think it's pretty clear what the question is about. In elementary situations, like with nested square roots, it's easy to figure out by hand a (nonzero) polynomial with rational coefficients having the number as a root. I don't think the question is seeking examples where the algebraicity is surprising, as with the unexpected appearance of Conway's constant, but rather is seeking examples where algebraicity is not easily checked by someone knowing nothing besides the definition. – KConrad Apr 06 '15 at 15:05
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    How about $${1\over\pi^2}\sum_1^{\infty}{1\over n^2}$$ – Gerry Myerson Apr 07 '15 at 00:59

4 Answers4

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There are plenty of examples. See, for example this page from google books, which talks about non-trivial proofs of algebraicity of functions. In particular, if a function is algebraic, so are its special values at rational arguments (one example: function is algebraic if its power series has integer coefficients and is meromorphic in a domain of conformal radius $> 1$ - this result is due to Borel and Polya).

Igor Rivin
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Let $E/\mathbb{Q}$ be an elliptic curve, let $\Omega$ be its real period (which is the value of an elliptic integral that's easy to write down using the equation of $E$), and let $L(E,s)$ be the $L$-series of $E$, which we know by Wiles' theorem has an analytic continuation to all of $\mathbb{C}$. Then $L(E,1)/\Omega$ is a rational number, but I'd say that that's not so easy to prove. Even harder, suppose that $E(\mathbb{Q})$ has rank 1, let $P$ be a non-torsion point, and let $\hat h(P)$ be the canonical height of $P$. Then $$ \frac{L'(E,1)}{\Omega\cdot\hat h(P)} $$ is a rational number. In this latter problem, we don't actually know that $\hat h(P)$ is not itself rational (this is unknown for even a single example), but the consensus seems to be that $\hat h(P)$ should be transcendental if it's not zero.

Joe Silverman
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Special values of modular functions are algebraic, but this requires the theory of complex multiplication to prove.

anon
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Algebraic numbers like a^(1/m) + b^(1/n) with a, b simple digits and m, n coprime non-negative integers greater than three, are numbers whose algebraicity is very difficult, impossible indeed, to be directly proved. Fortunately each term of the sum is obviously algebraic and we know that the sum of two algebraic numbers is algebraic itself.

Piquito
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    The sum of two algebraic numbers a root of the resultant of their minimal polynomials. For more, see http://math.stackexchange.com/questions/155122/how-to-prove-that-the-sum-and-product-of-two-algebraic-numbers-is-algebraic – Zack Wolske Apr 07 '15 at 00:52