# Prove linear combinations of logarithms of primes over $\mathbb{Q}$ is independent

Suppose we have a set of primes $p_1,\dots,p_t$. Prove that $\log p_1,\dots,\log p_t$ is linear independent over $\mathbb{Q}$. Now, this implies $\sum_{j=1}^{t}x_j\log(p_j)=0 \iff x_1=\dots=x_t=0$.

I think I have to use that fact that every $q\in\mathbb{Q}$ can be written as $\prod_{\mathcal{P}}$, where $n_p$ is a unique sequence ($n_2$,$n_3$,$\dots$) with domain $\mathbb{Z}$. Here, $\mathcal{P}$ denotes the set of all integers.

Now how can I use this to prove the linear independency?

If $\sum_{j=1}^{t}x_j\log(p_j)=0$ then $\sum_{j=1}^{t}y_j\log(p_j)=0$ where $y_j \in \Bbb Z$ is the product of $x_j$ by the common denominator of the $x_j$'s.
Therefore $\log\left(\prod_{j=1}^t p_j^{y_j}\right) = 0$, which implies $\prod_{j=1}^t p_j^{y_j} = 1$, and this is only possible if $y_j=0$ for all $j$. Indeed, you have $$\prod\limits_{\substack{1 \leq j \leq t\\ y_j \geq 0}} p_j^{y_j} = \prod\limits_{\substack{1 \leq i \leq t\\ y_i < 0}} p_i^{-y_i}$$ and uniqueness of prime powers decomposition implies $y_j=0$ for all $j$.
The converse is easy to see: if $x_j=0$ for all $j$, then $\sum_{j=1}^{t}x_j\log(p_j)=0$.
• This result implies that $\dim_{\Bbb Q}(\Bbb R) \geq \aleph_0$. – Watson Feb 8 '17 at 12:41