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 $p$.



irrationality of square roots of all non perfect square naturals
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 $2$.


