Is the set $\{\frac{1}{a\,-\,\pi}\mid a\in\mathbb{Q}\}$ linearly independent over $\mathbb{Q}$? The following problem is from Golan's linear algebra book. I have posted a proposed solution in the answers.
Problem: Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Is the subset $$\left\{\frac{1}{a-\pi}\;\middle\vert\; a\in\mathbb{Q}\right\}$$ linearly independent?
 A: It is. 
For the sake of contradiction, suppose there is a linear combination of elements in the set over $\mathbb{Q}$ that equal 0. We have a sum of the form
$$\sum_1^n \frac{q_j}{a_j-\pi}=0$$
where $q_j, a_j \in \mathbb{Q}$, the $a_j$ are distinct, and $q_j\neq 0$ for all $j$. Consider the following expression:
$$\sum_1^n \frac{q_j}{a_j-x}=0.$$
Our problem is equivalent to showing that $\pi$ is not a root of this.
Clearing denominators yields
$$\sum_{j=1}^n {q_j}\prod_{i\neq j}(a_i-x)=0.$$
This is clearly a polynomial with rational coefficients. The only way $\pi$ can be a root of this is if the polynomial is the zero polynomial, as $\pi$ is a transcendental number. But if this this the zero polynomial, then setting $x=a_1$ yields
$$q_1\prod_{i\neq 1} (a_i-a_1)=0$$
which is a contradiction as all terms are nonzero.
A: Well, if you are going to assume known that $\pi$ is transcendental over $\mathbb Q$ ... you may as well use any other transcendent instead.  For example, prove that
$$
\frac{1}{a-z},\qquad a \in \mathbb Q
$$
in the vector space of meromorphic functions is linearly independent over $\mathbb Q$.  In fact, more is true: linearly independent over $\mathbb C$.  We can see none of these is a linear combination of some others by comparing their poles.
