Exercise 2.1.8. Show that the sequence $1, \frac{1}{10}, \frac{1}{10^{2}}, \frac{1}{10^{3}}, \dots$ is not a Cauchy sequence in $(\mathbb{Q}, |\cdot|_{p})$, for every prime number $p$.
Proof. Firstly, notice that the sequence is given by $\{\frac{1}{10^{n}}\}$ for $n = 0, 1, 2, \dots$. Now notice that $\frac{1}{10^{n}} = 2^{-n}5^{-n}$ for all $n$. We will make a distinction of cases. First, suppose that $p = 2$. In this case, the $p$-adic absolute value of any element in the sequence is equal to $|2^{-n}5^{-n}|_{p} = 2^{n}$. For all $m > n$, we see that $|2^{-m}5^{-m}|_{p} = 2^{m}$. Thus, now with the reverse triangle inequality we find that $|x_{m} - x_{n}|_{p} \geq ||x_{m}|_{p} - |x_{n}|_{p}| = |2^{m} - 2^{n}| = 2^{m} - 2^{n} > 0$ for $m > n \geq 0$. Now, if we let $\varepsilon = 2^{m} - 2^{n} > 0$, we find without loss of generality for $m > n$ that $|x_{m} - x_{n}|_{p} \geq 2^{m} - 2^{n} = \varepsilon$. We see that this sequence is not a Cauchy sequence with respect to $|\cdot|_{p}$. For the case of $p = 5$, we find the same result with $\varepsilon = 5^{m} - 5^{n}$.
Now suppose that $p$ is a prime number with $p \notin \{2, 5\}$. In this case, we see that $|x_{n} - x_{m}|_{p} = 1$, since it is not possible to extract a power greater than 1 of a prime number from this difference of fractions. Therefore, if we take $\varepsilon = 1$, we deduce that $|x_{m} - x_{n}|_{p} = \epsilon = 1$ for all $m, n$. In this case, the sequence is not a Cauchy sequence with respect to $|\cdot|_{p}$.
We conclude that for every prime number $p$, the sequence $1, \frac{1}{10}, \frac{1}{10^{2}}, \frac{1}{10^{3}}, \dots$ is not a Cauchy sequence in $(\mathbb{Q}, |\cdot|_{p})$.
Is this correct?