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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?

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2 Answers

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.

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up vote 9 down vote accepted

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.

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Looks well and nice. +1 –  DonAntonio May 31 '12 at 23:17
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