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I am trying to solve the following question but without any result so far:

Let $p_1, p_2, \ldots, p_t$ be different primes. Prove that $\log (p_1), \log (p_2), \ldots, \log (p_t)$ are linearly independent over $\mathbb{Q}$ that is, if $x_1,x_2,\ldots, x_t$ are rational numbers with $$ x_1 \log (p_1) + x_2 \log (p_2) + \cdots + x_t \log (p_t) = 0 $$ then $x_1 = x_2 = \cdots = x_t = 0$.

We are supposed to use the following result:

Let $x \in \mathbb{Q}$ with $x > 0$. Then there is a unique sequence of integers $(n_2,n_3,n_5, \ldots)$, almost all equal to $0$, such that $$ x = \prod_{p \text{ prime}} p^{n_p}. $$

Any tips are welcome for solving it, thanks.

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marked as duplicate by Watson, Vladhagen, Zain Patel, Juniven, C. Falcon Feb 14 '17 at 0:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You need more than that you need that the integer sequence is unique (this is the fundamental theorem of arithmetic). Then join this with the logarithm's functional equation and monotonicity. $\endgroup$ – Adam Hughes Feb 13 '17 at 4:48
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I assume $t > 1$. Assume we have a linear combination $\sum_{i=1}^t x_i\log(p_i) = 0$. By multiplying through by the least common multiple of the denominators of the $x_i$, we get a linear combination with integer coefficients, so without loss of generality, assume $x_i$ are integers.

Then $$x_t\log(p_t) = \log(p_t^{x_t}) = \sum_{i=1}^{t-1}x_i\log(p_i)$$

Since $\log$ is a bijection from positive reals to reals $$\log(x) = \log(\prod_{p\ \text{prime}}p^{n_p}) = \sum_{p\ \text{prime}}n_p\log(p)$$ if and only if $x = \prod_{p\ \text{prime}}p^{n_p}$.

Obviously, in the case of $p_t^{x_t}$, it's the sequence with $n_{p_t} = x_t$ and all other $n_p = 0$. Since this sequence is unique, there is no linear combination in terms of other $\log(p_i)$.

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  • $\begingroup$ Thanks you! That is very helpful. $\endgroup$ – user386617 Feb 13 '17 at 15:47