To celebrate the creation of my new website and blog, and also to test the functionality of its typesetting, we will look at a simple yet elegant proof from my first-year linear algebra class.

This question was given on one of our assignments. I was quite pleased with the following solution which I came up with, although I am certainly not the first to do it this way.

There is a quick argument to be made regarding the countability of $\mathbb{R}$ and $\mathbb{Q}$, but we instead focus on transcendental numbers.

$\textit{Claim:}$ $\dim_{\mathbb{Q}} \mathbb{R}$ is infinite (the vector space of the real numbers over the rational numbers is infinite-dimensional).

$\textit{Proof:}$ Let $V$ be the vector space $\mathbb{R}$ over $\mathbb{Q}$. Consider the set $S = \{1, \pi, \pi^2, ... \}$ which is a subset of $V$.

We first show that $S$ is linearly independent.

Let $n \in \mathbb{N}$ and define $S_n = \{1, \pi, \pi^2, ..., \pi^{n-1}\}$ to be the first $n$ elements of $S$. Suppose by contradiction that $S_n$ is linearly dependent. This means that

for some $a_1, ..., a_n \in \mathbb{Q}$ having at least one non-zero value.

Now consider

Note that $P(x) \in \mathbb{Q}[x]$ and by our assumption, $P(\pi) = 0$. We have that $\pi$ is the root of a polynomial with rational (equivalently integer) coefficients, which is a contradiction since $\pi$ is transcendental. Thus every finite subset $S_n$ is linearly independent, and as a result so is $S$.

Intuitively, this must mean that $V$ has infinite dimension. To show this rigorously, we suppose that $V$ has finite dimension $m$. We know that since $S$ is linearly independent, $|S| \leq m$. But since $S$ has infinite cardinality, $|S| > m$ which is a contradiction. $\ \square$