Proof that $\sqrt{5}$ is irrational In my textbook the following proof is given for the fact that $\sqrt{5}$ is irrational:
$ x = \frac{p}{q}$ and $x^2 = 5$. We choose $p$ and $q$ so that the have no common factors, so we know that $p$ and $q$ aren't both divisible by $5$.
$$\left(\dfrac{p}{q}\right)^2 = 5\\ \text{ so } p^2=5q^2$$
This means that $p^2$ is divisble by 5. But this also means that $p$ is divisible by 5. 
$p=5k$, so $p^2=25k^2$ and so $q^2=5k^2$. This means that both $q$ and $p$ are divisible by 5, and since that can't be the case, we've proven that $\sqrt{5}$ is irrational.
What bothers me with this proof is the beginning, in which we choose a $p$ and $q$ so that they haven't got a common factor. How can we know for sure that there exists a $p$ and $q$ with no common factors such that $x=\dfrac{p}{q} = \sqrt{5}$? Because it seems that step could be used for every number
Edit:
I found out what started my confusion: I thought that any fraction with numerator 1 had a common factor, since every integer can be divided by 1. This has given me another question: Are confusions like this the reason why 1 is not considered a prime number? 

Exactly what MSEoris said, you can always reduce a fraction to such a point that they have no common factors, if $\frac{p}{q}$ had a common factor n, then $nk_0 = p$ $nk_1 = q $ then
$\frac{p}{q} = \frac{nk_0}{nk_1} = \frac{k_0}{k_1}$, now if $k_0, k_1$ have a common factor do the same and you will eventually get a fraction with no common factor 
