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Note the following:

If you take the set of integers $\mathbb Z$: and the operations of $+$ and $-$

Then all equations of the form ($x + a_1 + a_2 + a_3+\cdots+ a_N = b$) where $a$'s and $b$ are contained in $\mathbb Z$ has a solution in $\mathbb Z$.

If you take the set of rational Numbers ($\mathbb Q$?): and the operations of $+, -, \times$ and $/$

Then all equations of the form $$(a_{11}\times a_{12}\times a_{13}\cdots a_{1N} \times x + a_{21}\times a_{22}\times a_{23}\cdots a_{2N}\times x + \cdots +a_{N1}\times a_{N2}\times a_{N3}\cdots a_{NN}\times x= b)$$ where $a$'s and $b$ are in $\mathbb Q$ has a solution in $\mathbb Q$.

Question:

If you consider the set of Algebraic Numbers, we'll call them $A$. and the set of all algebraic operations (meaning $+, -, \times, /, \text{exponentiation}, \text{roots},$ and $\text{inverses of unsolvable polynomials}$)

Is every Algebraic expression solvable in the algebraic numbers?

I believe the answer is no, though I do not know for sure:

Here is what I have dug out so far:

Starting with Integers, and using only Integral operations
If $x + 2 = 3: x = 1$
If $2 + x + 1 = 0: x = -3$, is still a member of integers

Starting with Rationals, and using only Rational operations
If $2x = 3: x = 2/3$
If $2/3\times x + x + 1 = 0: x = -3/5$, is still a member of rational group

Starting with algebraic, and using only algebraic operations
If $x^2 = 2: x = \sqrt3$
If $x^{\sqrt3} + x + 1 = 0$: is x a member of algebraic group?

I have reservations that $x^{\sqrt3} + x + 1 = 0$ is included since the algebraic numbers are only defined as solutions to polynomial equations with positive integral powers. Note: any expression involving rational powers (ex: $x^{10} + x^{13/17} + 2 = 0$) is solvable using algebraic methods (simply make the substitution $x^{1/17} = u$ to create: $u^{170} + u^{13} + 2 = 0$ and solving using algebraic methods). The method however fails to operate once $u$ has irrational expressions for powers as there doesn't appear to be a way to reduce such an equation to a polynomial equivalent.

awllower
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Sidharth Ghoshal
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1 Answers1

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You may be interested in the Gelfand-Schneider theorem which says, in particular, that there cannot be any algebraic roots for the equation $x^\sqrt{3} + x + 1$ because if $x$ is an algebraic number then $x^\sqrt{3}$ is a transcendental number.

Thus you are correct, if you allow non-integer exponents then the algebraic numbers are not closed under the "algebraic operations" as you have defined them. This is why people who talk about algebraic numbers generally only allow integer exponents, because then you do get closure under the "algebraic operations".

Jim
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    But rather than calling the solutions to the equation transcendental... lets say we call them second order algebraic, that is the first order (original) algebraic numbers are enclosed in the set of second order algebraic numbers (solutions to all combinations of first order numbers), and then we create equations from these second order ones, does closure yet occur? or must third order numbers be made etc... It would like a measure of Transcendence, I wonder where pi and e would end up... maybe at the infinity point of the scale? – Sidharth Ghoshal Mar 14 '13 at 20:18