# Did Galois show $5^\sqrt{2}$ can't solve a high-order integer polynomial?

Suppose you have an unlimited quantity of the number one, and the operators plus, minus, multiply, divide, and power. Consider the (countable) set $S$ you generate by combining these:

• Using just one and plus, you can construct the natural numbers.

• Using minus, you can construct the integers.

• Using divide, you can construct the rational numbers.

• Using power, you can construct nth roots like $2^{1/2} = \sqrt{2}$

So far, so good. However, you can now go further and construct things like $5^\sqrt{2}$ and far more bizarre things. Questions:

• Did Galois show $S$ is a proper subset of the algebraic numbers? I know he showed arbitrary 5th-degree polynomials don't have "closed- form" solutions, but I believe his definition of "closed form" was more limited.

• If not, let $T$ be the set of numbers Galois considers "closed form". Are there members of $S-T$ that solve high order polynomials?

• $S$ seems like an "obvious" set to me. Does it have a name, and do people study it?

• I chose $5^\sqrt{2}$ as a "random example": it seems obvious that it's non-algebraic, but I can't seem to prove it.

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$5^{\sqrt 2}$ is transcendental, via Gelfond-Schneider. – J. M. Sep 3 '11 at 16:50
Solutions to the quintic do have closed forms, but in general one requires hypergeometric or theta functions to express them (much like trigonometric functions turn up in solving the cubic). – J. M. Sep 3 '11 at 16:53
Galois's work does not deal with $S$. The set $S$ contains transcendental numbers, for instance your $5^{\sqrt{2}}$. Since $S$ is countable, "most" transcendental numbers are not in $S$. I know nothing about the intersection of $S$ and the algebraic numbers. There is a limited number of results that could be thought of as being about $S$, having to do with algebraic independence. – André Nicolas Sep 3 '11 at 17:47
Galois' work is not really about these matters. – Mariano Suárez-Alvarez Sep 3 '11 at 19:50
Isn't this closure you described the Exponential closure of the rationals? – Asaf Karagila Sep 4 '11 at 7:37

It may be that your set $S$ is the set some people call "the elementary numbers." There are some papers about this: Tim Chow, What is a closed-form number?, Amer Math Monthly, 1999; aargh, my internet connection just vanished, so I can't copy out any more, but I typed "elementary number" into Google Scholar and a bunch of likely papers came up.