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Homework Statement
A real number [tex]x \in R[/tex] is called algebraic if there exist integers [tex]a_{0},a_{1},a_{2}...,a_{n}[/tex], not all zero, such that
[tex]a_{n}x^{n} + a_{n}_{1}x^{n-1} + ... + a_{1}x + a_{0} = 0 [/tex]
Said another way, a real number is algebraic if it is the root of a polynomial with integer coefficients...
Fix [tex]n \in N[/tex], and let [tex]A_{n}[/tex] be the algebraic numbers obtained as roots of polynomials with integer coefficients that have degree [tex]n[/tex]. Using the fact that every polynomial has a finite number of roots, show that [tex]A_{n}[/tex] is countable. (For each [tex]m \in M[/tex], consider polynomials [tex]a_{n}x^{n} + a_{n}_{1}x^{n-1} + ... + a_{1}x + a_{0} [/tex] that satisfy [tex]\left|a_{n}\right| + \left|a_{n-1}\right| + ... + \left|a_{1}\right| + \left|a_{0}\right| \leq m [/tex].)
Homework Equations
The Attempt at a Solution
I'm not sure how to explain this coherently... here is what I have. I feel like there are some holes.
-Every polynomial has a finite # of roots (therefore it is countable)
-[tex]m \in N[/tex] is the sum of all integer coefficients for the roots of polynomials.
-Let [tex]C_{m}[/tex] be a set containing all possible polynomials whose integer coefficients add up to [tex]m[/tex] for a fixed [tex]n[/tex]. Since there are finite ways to express [tex]m[/tex] as a sum of integers, each [tex]C_{m}[/tex] is countable.
-Every [tex]A_{n}[/tex] is made up of [tex]C_{m}[/tex], so [tex]A_{n}[/tex] is countable (since union of a countable # of countable sets is countable).