# Abstract Nonsense

## 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.