Abstract Nonsense

Crushing one theorem at a time

Algebraic Numbers Are Countable


Point of post: In this post I will give a proof that the algebraic numbers are countable, and thus we will prove that there are non-algebraic (transcendental) numbers.

Remark: Personally I find my proof much simpler and more intuitive than the usual method by considering the “height” of numbers (viz. Rudin, pg. 43)

Call a number z\in\mathbb{C} algebraic if their exists a_0,\cdots,a_n\in\mathbb{Z} such that a_0+a_1z+\cdots+a_nz^n=0. In other words, the algebraic numbers are those which are the roots of polynomials with integer coefficients. So, formally

\mathbb{A}=\left\{z\in\mathbb{C}:\exists a_0,\cdots,a_n\in\mathbb{Z}\text{ not all zero, such that }a_0+\cdots+a_nz^n=0\right\}

Theorem: \mathbb{A} is countably infinite.

Proof: For each n\in\mathbb{N}-\{1\} define

\mathbb{Z}_*^n=\mathbb{Z}^n-\{\underbrace{0,\cdots,0}_{n\text{ times}})

Then, for each (a_1,\cdots,a_n)\in\mathbb{Z}_*^n define

\Omega_n\left((a_1,\cdots,a_n)\right)=\left\{z\in\mathbb{Z}:a_1+\cdots+a_nz^n=0\right\}

Then, by the Fundamental Theorem of Algebra we have that \text{card }\Omega_n\left((a_1,\cdots,a_n)\right)\leqslant n. We claim that

\displaystyle \mathbb{A}=\overbrace{\bigcup_{n=2}^{\infty}\bigcup_{(a_1,\cdots,a_n)\in\mathbb{Z}_*^n}\Omega_n\left((a_1,\cdots,a_n)\right)}^{U}

To see this let z_0\in\mathbb{A}, then there exists a non-zero polynomial p(z)=a_1+\cdots+a_nz^n such that p(z_0)=0, but this tells us that z_0\in\Omega_n\left((a_1,\cdots,a_n)\right) and so evidently z_0\in U. Conversely, if z_0\in U then there exists some n\in\mathbb{N}-\{1\} (we disregard the case when n=1 since these are just constants and are zero, only when the constant itself is zero) and some (a_1,\cdots,a_n)\in\mathbb{Z}_*^n such that z_0\in\Omega_n\left((a_1,\cdots,a_n)\right), and so by definition a_1+\cdots+a_nz_0^n=0 and since a_1,\cdots,a_n are not all zero we see that z_0\in\mathbb{A}. Thus, U=\mathbb{A}.

Note though that since \mathbb{Z}\simeq\mathbb{Z}^n\simeq\mathbb{Z}_*^{n} for all n\in\mathbb{N} that  for each fixed n_0\in\mathbb{N}-\{1\} we have that

\displaystyle \bigcup_{(a_1,\cdots,a_{n_0})\in\mathbb{Z}_*^{n_0}}\Omega_{n_0}\left((a_1,\cdots,a_{n_0})\right)

is the countable union of finite (non-empty) sets and is thus itself countable. Thus, \mathbb{A} being the countable union of such countable sets is itself countable. \blacksquare

From this we have the following very nice corollary:

Corollary: There exists uncountably real numbers which are non-algebraic (transcendental).

Proof: Suppose to the contrary that \mathbb{R}-\mathbb{A} was countable, then since the finite union of countable sets is countable and

\mathbb{R}=\left(\mathbb{R}-\mathbb{A}\right)\cup\left(\mathbb{R}\cap\mathbb{A}\right)

we see that \mathbb{R} is countable. But, this is a contradiction and so the conclusion follows. \blacksquare

References:

1.  Rudin, Walter. Principles of Mathematical Analysis. New York: McGraw-Hill, 1976. Print.

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October 21, 2010 - Posted by | Set Theory | , , ,

2 Comments »

  1. […] (1): By a previous post we know that there exists transcendental numbers. So, let be such a transcendental number. […]

    Pingback by The Dimension of R over Q « Abstract Nonsense | October 22, 2010 | Reply

  2. […] previously discussed the set of all algebraic numbers at least to the extent to show that the set of all of them is […]

    Pingback by Some Facts About The Ring of Algebraic Integers « Abstract Nonsense | March 3, 2011 | Reply


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