Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 2G (Topics in Analysis)

For each degree and coefficient bound there are finitely many integer polynomials. Hence $\mathbb{Z}[x]$ is countable. Every nonzero polynomial has finitely many roots, and every algebraic number is a root of one of these polynomials. A countable union of finite sets is countable.

Choose a subsequence $a_{n_j}$ so sparse thatfor every $j$; this is possible because the tails of the convergent positive series tend to zero. Distinct binary sequences $({\epsilon }_j)$ then have distinct sums ${\sum }_j{\epsilon }_j{a}_{n_j}$: at their first disagreement, the corresponding term exceeds the entire remaining tail. Thus these subsums form an uncountable set. The algebraic numbers are countable, so at least one subsum is transcendental. Set ${\theta }_{n_j}={\epsilon }_j$ and all other ${\theta }_n=0$.

Letand suppose $x=p/q$. Choose $n\ge \max (q,2)$ such that some ${\theta }_j=1$ for $j>n$. Then $n!x$ is an integer, while its first $n$ terms also sum to an integer. Their difference is strictly positive and satisfiesThis cannot be the difference of two integers, so $x$ is irrational.