Why do we consider that $p$ & $q$ are co-primes when proving square root of a prime number is irrational? When we prove that square root of any prime number is irrational, we assume that there exists some rational number $r=\frac{p}{q};\space p,q\in\mathbb{Z}, q\neq0$ s.t. $\frac{p^2}{q^2}=p_1$ where $p_1$ is prime and then prove by contradiction.
Why do we consider that $p$ & $q$ are co-primes?
 A: That's what we end up contradicting at the end of the proof. You get
$$p^2 = p_1q^2$$
and by Euclid's lemma you find that $p$ is a multiple of $p_1$. Plugging in you similarly find that $q$ is a multiple of $p_1$, and this contradicts the fact that they are coprime as they share the factor $p_1$. This is essentially saying that such a fraction could not be "put in simplest form".
A: First recall how do we proceed in the question
We use $p^2=p_1q$
i.e. $p^2|p_1q\Rightarrow p|p_1q\Rightarrow p|p_1$
You can move onto the last step which says $p|p_1$ only when $p$ is prime.
Why? Because Euclids lemma states $p|ab\Rightarrow p|a$ or $p|b$ 
Can you see the parallel used here?
You can also see that $p$ and $q$ in your question have no common factors. AND $p|p_1q$, so all the factors of $p$(if any) should be factors of $p_1$ too.
A: There is no need to assume $p$ and $q$ are coprime:
In $p^2 = p_1q^2$, consider the exponent of $p_1$ in the prime factorization of each side: on the LHS, you get an even exponent; on the RHS, you get an odd exponent.
The same argument proves that the square root of $m$ is rational iff $m$ is a square because the exponent of each prime in the factorization of $m$ must be even.
A: This image is from pg-2 of Abbott's Understanding Analysis that gives the simple yet elegant answer to the question:

A: Every fraction can be rewritten in the form $\frac pq$ where $p$ and $q$ are coprime. (Example: $\frac{12}{16}$ can be rewritten as $\frac34$.)
Proof: Let $r$ be the rational in question. Now, let "$\frac pq$" be the way of writing $r$, where the denominator $q$ is as small as possible. Claim: $p$ and $q$ are coprime. Reason: If not, then we could just "cancel out" the common factor, and end up with another way to write the fraction with a smaller denominator; this contradicts the idea that $q$ was as small as possible.
