Show that this function is 'well defined' Let $ord_p(n)$ denote the multiplicity with which the prime $p$ appears in the prime factorisation of $n$.
For a prime $p$ and $x\in\mathbb Q\backslash \{0\}$ define $ord_p(x)=ord_p(n)-ord_p(d)$ where $x=\frac{n}{d}$ with $n,d \in\mathbb Z$.
I'm asked to show that $ord_p(x)$ is well defined, i.e independent of the choice of fraction for $x$.
Do I just need to show that $ord_p(\frac{n}{d})=ord_p(\frac{kn}{kd})$ with $k\in\mathbb Z$?
Do I need to justify that these are the only fraction representations for $x$?
 A: Well, if your original fraction $\frac{n}{d}$ wasn't irreducible, then that expression wouldn't be all the fraction representations. But you can do it without assuming you're working with the irreducible expression, and I recommend you this structure because it's the same for each "prove that this is well-defined": start with the same $x$ with two different fraction representations, and prove that each one gets the same value.
Assume $x = n/d = m/e$, where $n,d,m,e\in \mathbb{Z}\backslash 0$, then you'd have $ne = md$, and if you take the function $ord_p$ here you'd notice that 
$$\text{ord}_p(ne) = \text{ord}_p(md)\Rightarrow \text{ord}_p(n) + \text{ord}_p(e) = \text{ord}_p(m) + \text{ord}_p(d) $$
Where we have used that if you multiply two numbers, then the multiplicity of $p$ would be the sum of the multiplicity of $p$ in each number factorization. If you didn't know this beforehand, you can easily prove it by writing the factorization of each number. Note: this doesn't work when one of the numbers is $0$, but $x$ can't be $0$ so we don't have problems here.
If you rearrange at the end, you obtain $\text{ord}_p(\frac{n}{d})=\text{ord}_p(\frac{m}{e})$, and it's well defined.
