Algebraic and Transcendental Numbers

# Algebraic and Transcendental Numbers

 Definition: A number $\alpha \in \mathbb{R}$ is said to be Algebraic if there exists a nonzero polynomial $p \in \mathbb{Z}[x]$ such that $p(\alpha) =0$. If no such polynomial exists, then $\alpha$ is said to be Transcendental.

It is easy to check that every $a \in \mathbb{Z}$ is algebraic since $p_a(x) = x - a$ is a polynomial satisfying $p_a(a) = 0$. Similarly, it is easy to check that every $\frac{a}{b} \in \mathbb{Q}$ is algebraic since $p_{a,b}(x) = bx - a$ is a polynomial satisfying $p_{a,b} \left ( \frac{a}{b} \right ) =0$. So every rational number is algebraic. The converse is not true in general though. There are irrational numbers which are also algebraic.

For example, let $\alpha = \sqrt{2}$. Then $p(x) = x^2 - 2$ is a polynomial for which $p(\alpha) = 0$ and so $\sqrt{2}$ is an irrational algebraic number.

If $\alpha$ is algebraic then there exists a polynomial $p \in \mathbb{Z}[x]$ for which $p(\alpha) = 0$. Note that if $f \in \mathbb{Z}[x]$ is any other polynomial then:

(1)
\begin{align} \quad (pf)(\alpha) = p(\alpha)f(\alpha) = 0f(\alpha) = 0 \end{align}

So there are infinitely many polynomials with $\alpha$ as a root. Of course this must be a polynomial with minimum degree for which $\alpha$ is a root by the Well-Ordering Principle.

 Definition: If $\alpha$ is an algebraic number then the Degree of $\alpha$ is the minimum degree of all such nonzero polynomials $p \in \mathbb{Z}[x]$ for which $p(\alpha) = 0$.

For example, the degree of $\sqrt{2}$ is $2$. The degree of $\sqrt{3}$ is also $2$.

 Theorem 1: There are countably many algebraic numbers.

We will omit a formal proof of Theorem 1 above.

Intuitively it is clear that there are only countable many polynomials in $\mathbb{Z}[x]$. In particular, for each $n \in \mathbb{Z}^+$ let $\mathbb{Z}_n[x]$ denote the set of polynomials of degree equal to $n$. Then each $\mathbb{Z}_n[x]$ is countable since each $p(x) \in \mathbb{Z}_n$ can be represented by an ordered $(n+1)$-tuple $(a_0, a_1, ..., a_n)$, $a_n \neq 0$ corresponding to the polynomial $p(x) = a_0 + a_1x + ... + a_nx^n$ and so:

(2)
\begin{align} \quad |\mathbb{Z}_n[x]| = |\underbrace{\mathbb{Z} \times \mathbb{Z} \times ... \times \mathbb{Z}}_{n \: \mathrm{times}} \times (\mathbb{Z} \setminus \{ 0 \}) | \end{align}

A countable (in this case finite) union of countable sets is countable and so $\mathbb{Z}_n[x]$ is countable. For each $p(x) \in \mathbb{Z}_n[x]$, there are at most $n$ distinct real roots by the Fundamental theorem of Algebra, and so the set of algebraic numbers who are roots of polynomials of degree $n$ is countable. Taking the union of all algebraic numbers who are roots of polynomials in $\mathbb{Z}[x]$ (including $0$ of course) is thus countable.