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? 

You just stated a proof by contradiction.  You assumed that there exists $p$ and $q$ that are relatively prime such that $\frac{p^2}{q^2} = 5$.  You came to the conclusion that $p$ and $q$ cannot be relatively prime, contradicting your assumption.  Therefore, there are no such $p$ and $q$, and therefore no rational number, whose square is 5.
