- A real number \(x\) is said to be algebraic (over the rationals) if it satisfies some polynomial equation of positive degree
\[ x ^ {n} + a _ {n - 1} x ^ {n - 1} + \dots + a _ {1} x + a _ {0} = 0 \]
with rational coefficients \(a_{i}\) . Assuming that each polynomial equation has only finitely many roots, show that the set of algebraic numbers is countable.
- A real number is said to be 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.)
Solution
- For each \(n\ge1\), let \(P_n\) be the set of monic degree-\(n\) polynomials with rational coefficients. The coefficient map gives a bijection \[ P_n\cong\mathbb Q^n. \] Since finite products of countable sets are countable, each \(P_n\) is countable. Hence \[ P=\bigcup_{n\ge1}P_n \] is countable.
Each polynomial in \(P\) has only finitely many real roots. The set \(\mathcal A\) of algebraic real numbers is the union, over \(p\in P\), of the finite root set of \(p\). This is a countable union of finite sets, hence countable.
- Let \(T\) be the set of transcendental reals. Then \[ \mathbb R=\mathcal A\cup T. \] If \(T\) were countable, the right side would be a union of two countable sets and hence countable, contradicting the assumed uncountability of \(\mathbb R\). Therefore \(T\) is uncountable.