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Is there any theory (analogous to Galois theory) for solving equations with irrational exponents like:

$ x^{\sqrt{2}}+x^{\sqrt{3}}=1$


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Well, for one thing, this is now a transcendental equation instead of an algebraic one, and the theory of seeing whether a transcendental equation has (unique) solutions is a lot less developed than in the algebraic case (i.e., one usually proceeds on a case-to-case basis for determining if a transcendental equation has solutions). – J. M. Aug 30 '10 at 0:54
In my physics classes, we would just graph these things and read off the answer. I don't know if there is a better way, but I think this is an interesting question. There must be some transcendental equations with an analytic solution, but I suspect the situation could be like Diophantine equations, where there is just no hope for a general method. – Matt Calhoun Aug 30 '10 at 13:25
I am by no means an expert on Galois theory, far from it, but I didn't think it gave a method to solve polynomials. My understanding is that this theory provides an explanation as to why 5th degree and higher polynomials are not generally solvable by radicals. As for analytic methods for transcendental equations which have nothing to do with Galois theory, check out Newtons method (this approximation will under certain conditions actually converge to the correct answer) – Matt Calhoun Aug 30 '10 at 13:48
The key point is "some"; the difficulty is in finding general analytic solutions for transcendental functions. – J. M. Aug 30 '10 at 13:48
Matt: the thing to understand with Galois theory is that one has to keep adding "tools" to your repertoire to (analytically) solve polynomial equations of increasing degree. For linears, you only need division; for quadratics to quartics, you need radicals; for quintics you need to use hypergeometric or theta functions, and even higher degree polynomial equation need even more complicated functions. Newton is an approximation method; it does not give analytic solutions. – J. M. Aug 30 '10 at 13:52
up vote 11 down vote accepted

The study of such equations is not "abstract algebra" as it is usually understood. The reason is that to even define the function $x^{\sqrt{2}}$, for example, requires analysis; one has to prove certain properties of $\mathbb{R}$ to ensure that such a function exists. This is in marked contrast to the case of integer or rational powers, where one has a purely algebraic definition and the background theory is equational. To define the function $x^{\sqrt{2}}$ one has to either define $e^x$ and the logarithm or consider a limit of functions $x^{p_n}$ where $p_n$ form a sequence of rational approximations to $\sqrt{2}$, and this is irreducibly non-algebraic stuff.

In particular, while polynomials can be studied in an absurdly general setting, transcendental equations like those you describe are more or less restricted to $\mathbb{R}$ (or $\mathbb{C}$ if you really want to pick a branch of the logarithm). The LHS is an increasing function of $x$, so there is at most one root, which probably one can really only compute numerically if it exists. (Its nonexistence can be ruled out by computing local minima in $(0, 1)$.)

This is another question which touches on a theme which has come up several times on math.SE, which is that exponentiation should really not be thought of as one operation. Instead, it is a collection of related operations with various degrees of generality and applicability which happen to share the same algebraic properties, and one should not infer too much about how similar these operations are.

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One should be careful about making strong broad statements such as so-and-so "cannot be called abstract algebra". A large part of elementary calculus can be algebraicized. That's how computer algebra systems work. One has differential algebra and Galois theory for computing integrals and solving ODEs, Hardy-Rosenlicht fields and transseries for computing limits and asymptotics, etc. – Bill Dubuque Aug 30 '10 at 1:30
Fair point; I'll modify the wording. – Qiaochu Yuan Aug 30 '10 at 1:49

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