Exercise 7.4

Question

\begin{enumerate}[label=(\alph*)]
\item A real number $x$ is said to be \boldit{algebraic} (over the rationals) if it satisfies some polynomial equation of positive degree
\begin{gather*}
x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = 0
\end{gather*}
with rational coefficients $a_i$.
Assuming that each polynomial equation has only finitely many roots, show that the set of algebraic numbers is countable.
\item A real number is said to be \boldit{transcendental} if it is not algebraic.
Assuming the reals are uncountable, show that the transcendental numbers are uncountable.
(It is a somewhat surprising fact that only two transcendental numbers are familiar to us: $e$ and $\pi$.
Even proving these two numbers transcendental is highly nontrivial.)
\end{enumerate}

Dan Whitman · Accepted Answer

\def\vq{\mathbf{q}}
(a)
\begin{proof}
First consider arbitrary degree $n \in {{\mathbb{Z}}_+}$.
Then for each $\vq = \seqfin{q}{n} \in {\mathbb{Q}}^n$, there is a corresponding polynomial equation in $x$:
\begin{gather*}
x^n + \sum_{i = 1}^n q_i x^{i-1}  = x^n + q_n x^{n-1} + \cdots + q_2 x + q_1 = 0 \,,
\end{gather*}
which is assumed to have a finite number of solutions.
So let $X_\vq$ be the finite set of real numbers that are solutions.
(We note that the polynomial corresponding to the vector $\vq = (0,\ldots,0) \in {\mathbb{Q}}^n$ becomes $0=0$ so that any real number $x$ satisfies it.
Similarly the polynomial corresponding to $\vq = (q_1, 0, \ldots, 0) \in {\mathbb{Q}}^n$ for nonzero $q_1$ corresponds to the equation $q_1 = 0$, which has no solutions.
Of course, $X_\vq = \varnothing$ is still finite in this case.
For the infinite-solution case, we could simply remove the zero vector from ${\mathbb{Q}}^n$ without changing the argument in any substantial way.
This is also taken care of if we really do assume that \emph{any} polynomial has a finite number of solutions as we are evidently doing here.)

Now, we clearly have that ${\mathbb{Q}}^n$ is countable by Theorem~7.6 since it is a finite product of countable sets (since it was shown in \href{./exercise_7-1}{Exercise~7.1} that ${\mathbb{Q}}$ is countable).
Thus the set $A_n = \bigcup_{\vq \in {\mathbb{Q}}^n} X_\vq$ is countable by Theorem~7.5 since it is a countable union of finite (and therefore countable) sets.
Of course, this is the set of all algebraic numbers from polynomials of degree $n$.
Then $A = \bigcup_{n \in {{\mathbb{Z}}_+}} A_n$ is the set of all algebraic numbers, which is also then countable by Theorem~7.5 since each $A_n$ was shown to be countable.
\end{proof}

(b)
\begin{proof}
As in part~(a), let $A \subset {\mathbb{R}}$ be the set of algebraic numbers so that clearly, by definition, $T = {\mathbb{R}} - A$ is the set of transcendental numbers.
Note that clearly ${\mathbb{R}} = A \cup T$ so that, if $T$ were countable, then ${\mathbb{R}}$ would be too since it is a finite union of countable sets.
This of course contradicts the (hitherto unproven) fact that ${\mathbb{R}}$ is uncountable so it must be that $T$ is also uncountable as desired.
\end{proof}

Exercise 7.4

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Exercise 7.4

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