irrationality of square roots of all non perfect square naturals I was reading various proofs of the irrationality of $\sqrt{2}$ including a geometric proof by Richard Guy involving similar right triangles. Since then, after thinking about it, I wonder why we are ever taught about the irrationality of square roots of any one particular non-square natural at all, as the proof that all non-square naturals have irrational square roots seems obvious. Am I missing something?
First, the proof in words: squares of non-integral rationals are non-integral rationals. Therefore, the square-root of a non-square natural can not be rational.
Suppose $x \in \mathbb{Q} \setminus \mathbb{N}$. Therefore, there exist $ p, q \in \mathbb{N}$ relatively prime with $q >1$ such that $x = p/q$. But $p,q$ relatively prime with $q > 1$ implies that $p^2$ and $q^2$ are relatively prime with $q^2 > 1$. Therefore, $x^2=(p/q)^2 \in \mathbb{Q}\setminus \mathbb{N}$.  Therefore, $(\mathbb{Q} \setminus \mathbb{N})^2 \cap (\mathbb{N}\setminus \mathbb{N}^2) = \emptyset$ and consequently, $(\mathbb{Q} \setminus \mathbb{N}) \cap \sqrt{(\mathbb{N}\setminus \mathbb{N}^2)} = \emptyset$. Here, by $\mathbb{A}^p$ we mean the set formed by raising elements of the set $\mathbb{A}$ to the power $p$.
 A: The point is that you use this fact from number theory:
$p$ and $q$ are relatively prime $\Longrightarrow$ $p^2$ and $q^2$ are relatively prime.
While this is not hard, this is not trivial either. You either need unique prime factorization or the Euclidean algorithm.
For $n=2$, you just have to prove that if the square of a number is even, then so is the number itself. This is trivial casebash modulo $2$.
A: A definition of $a,b$ relatively prime is no common factors but 1 and -1. A useful equivalent is $as+bt=1$ for some integers $s,t$. Using only this and a little algebra one can prove  that if $\gcd(a,b)=1$ and $\gcd(a,c)=1$ then also $\gcd(a,bc)=1$. In particular $gcd(a,b^2)=1$ and $\gcd(a^2,b^2)=1$ SO if a/b is a fraction in lowest terms which is not an integer, so is $a^2/b^2$. Hence no proper fraction can be a square root of an integer. (this is a a tool in  developing the theory of factorization in integers)
The irrationality of non integer square roots of integers has been discussed in quite some detail on MO recently. 
A: The history of this goes back to Greek mathematics and there is a good historical discussion in Hardy and Wright "An Introduction to the Theory of Numbers" chapter 4 (I have 5th Edition to hand).
I guess pedagogically you might begin with $\sqrt{2}$ as an easy example in order to show, for example, that there is an irrational number before exploring their nature and existence more generally.
And there is also the question of taking care over the properties of numbers on which the proof is based as darij grinberg has posted in an answer which came up as I was typing this one. Hardy and Wright deal with this too, and more generally also with irrational roots of polynomials with integer coefficients.
