I am given the set $$\{x \in \mathbb{Q} : \text{ord}_p(x) \geq 0\}$$ This is described as the "set of all rationals whose denominator is not divisible by $p$".
I am confused about this description of the set. How does $\text{ord}_p(x) \geq 0$ imply that $x$ is a rational number whose denominator is not divisible by $p$?
I also have to show that the set is closed under multiplication and contains $\mathbb{Z}$, which I am stuck on.
Any help is appreciated.
Given a rational number $x$ written as a fraction $\frac{a}{b}$ such that $a=a'p^n$, $b=b'p^m$, with $a'$ and $b'$ not divisible by $p$, with $n,m \ge 0$, then $\mathrm{ord}_p(x)$ is defined as $n-m$. Note that this can always be done and $\mathrm{ord}_p(x)$ is independent of the fraction representation of $x$. In particular, you can always write $x=\frac{a'}{b'}p^{\mathrm{ord}_p(x)}$ with $a'$ and $b'$ as above. Then it should be clear that your description of the set makes sense (if you assume that numerator and denominator of the fraction representing $x$ should have no common factors).
For the other parts: any number $a \in \mathbb{Z}$ can be written as $\frac{a}{1} \in \mathbb{Q}$. Can you see why $\mathrm{ord}_p(\frac{a}{1}) \ge 0$ for all $p$? As for multiplication, you can use again the above representation $x=\frac{a'}{b'}p^{\mathrm{ord}_p(x)}.$