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Monday 13 May 2013

Algebraic numbers are countable

Q2. Basic Topology (Rudin)

A complex number z is said to be algebraic if there are integers \(a_0 , a_1 , \ldots , a_n \) , not all zero, such that

 \(a_0 z^n + a_1 z^{n-1} + \ldots + a_{n-1} z + a_n = 0 \) .

 Prove that the set of all algebraic numbers are countable. Hint: For every positive integer N there are only finitely many equations with \(  n +  | a_0 | + | a_1 | + \ldots +  | a_n | = N \)

Solution: 

Set A is countable if we can establish one-one correspondence between A and set of positive integers \(\mathbb{N}\)

By Fundamental theorem of Algebra  \(a_0 z^n + a_1 z^{n-1} + \ldots + a_{n-1} z + a_n = 0 \)  has exactly n complex roots.

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