I came across the following problems about p-adic norms:
Problem. Show that $\prod_{p} |x|_p = \frac{1}{|x|}$ where the product is taken over all primes $p = 2,3,5, \dots$ and $x \in \mathbb{Q}$.
We have the following: $|x|_2 = 2^{-\max \{r: 2^{r}|x \}}$, $|x|_3 = 3^{-\max \{r: 3^{r}|x \}}, \ \dots$ so that $\prod_{p} |x|_p = \frac{1}{2^{\max \{r: 2^{r}|x \}}} \cdot \frac{1}{3^{\max \{r: 3^{r}|x \}}} \cdots$
Then it seems that by the Fundamental Theorem of Arithmetic the result follows. Is this the right idea?
Problem. If $x \in \mathbb{Q}$ and $|x|_p \leq 1$ for every prime $p$, show that $x \in \mathbb{Z}$.
We know that $\prod_{p} |x|_p = \frac{1}{|x|} \leq 1$
Then suppose for contradiction that $x \notin \mathbb{Z}$? Or maybe the product is a null sequence?
Added. Suppose $x \notin \mathbb{Z}, \ x \in \mathbb{Q}$. Then $x = \frac{r}{s}$ where at least one prime $p$ divides $s$. Then $\text{ord}_{p} x = \text{ord}_{p} r- \text{ord}_{p} s$. So $|x|_{p} = p^{-\text{ord} _{p} x} = p^{ \text{ord}_{p} s- \text{ord}_{p} r}$
$= \frac{p^{\text{ord}_{p} s}}{p^{\text{ord}_{p} r}} > 1$
which is a contradiction?