Exercise 1.5.9

Question

A real number $x \in \mathbf{R}$ is called algebraic if there exist integers $a_{0}, a_{1}, a_{2}, \ldots, a_{n} \in \mathbf{Z}$, not all zero, such that
  $$
  a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}=0
  $$
  Said another way, a real number is algebraic if it is the root of a polynomial with integer coefficients. Real numbers that are not algebraic are called \emph{transcendental} numbers. Reread the last paragraph of Section 1.1. The final question posed here is closely related to the question of whether or not transcendental numbers exist.

\begin{enumerate}[label=(\alph*)] 
  \item Show that $\sqrt{2}, \sqrt[3]{2}$, and $\sqrt{3}+\sqrt{2}$ are algebraic.
  \item Fix $n \in \mathbf{N}$, and let $A_{n}$ be the algebraic numbers obtained as roots of polynomials with integer coefficients that have degree $n$. Using the fact that every polynomial has a finite number of roots, show that $A_{n}$ is countable.
  \item Now, argue that the set of all algebraic numbers is countable. What may we conclude about the set of transcendental numbers?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $x^2 - 2 = 0$, $x^3 - 2 = 0$ are obvious. Now consider $\sqrt 3 + \sqrt 2$. The key is setting $x = \sqrt 3 + \sqrt 2$ then using algebra on $x$ to concoct an integer, and thus find the polynomial with $x$ as a root.

We have $x^2 = 5 + 2\sqrt6$ meaning $x^2 - 5 = 2\sqrt{6}$ and thus $(x^2 - 5)^2 = 24$ so $(x^2 - 5)^2 - 24 = 0$ is a polynomial with $\sqrt 3 + \sqrt 2$ as a root.
  \item Basically $A_n \sim \mathbf{Z}^n \sim \mathbf N^n \sim \mathbf{N}$.
    \begin{enumerate}[label=(oman*)] 
    \item $A_n \sim \mathbf{Z}^n$ since integer polynomials of degree $n$ are identical to an ordered list of $n$ integers.
    \item $\mathbf Z^n \sim \mathbf N^n$ since $f : \mathbf{N}^n 	o \mathbf{Z}^n$ is just the piecewise application of $g : \mathbf{N} 	o \mathbf{Z}$.
    \item $\mathbf N^n \sim \mathbf N$ since it is the intersection of finite sets $\bigcup_{n=2}^\infty \mathbf \{(a,b) : a+b = n\}$.
     \end{enumerate}

In general if $V$ is countable then $V^n = (v_1, \dots, v_n)$ is also countable.
  \item By ef{ex:countable_union} the set of all algebraic numbers $\bigcup_{n=1}^\infty A_n$ is countable.
   \end{enumerate}

Exercise 1.5.9

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Exercise 1.5.9

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