$x^2 + y^2 - 3 = 0 \implies 3^{1/2}$ is irrational.

Question

The actual problem is from the Book of Proof by Richard Hammack. It relates to a series of problem dealing with the curve $x^2 + y^2 - 3 = 0$.

The problem prior to the one I posted asked to proof that $x^2 + y^2 - 3 = 0$ has no rational solutions. (Show that the curve $x^2+y^2-3=0$ has no rational points) I believe the intended solution by the author shares the same reasoning.

I can manage to prove that $3^{1/2}$ is irrational by contradiction. (Suppose that $3^{1/2}$ is rational, then $3^{1/2} = p/q$, with $p$,$q$ belonging to $\mathbb{Q}$ and $q \ne 0$. Suppose that $p/q$ is a fully reduced fraction, therefore $p$ and $q$ are mutually prime. Note that $3p^2 = q^2$, therefore $3$ is a prime factor of $p^2$ since every prime in $q^2$ must be an even number of times a factor of $q^2$. Therefore, $p$ and $q$ are not coprime, a contradiction. Hopefully this is correct?) However, I'm struggling to see that $x^2 + y^2 - 3 = 0 \implies 3^{1/2}$ is irrational. I will appreciate if anyone can help me understand this fact.

Answer 1

If $x^2+y^2-3=0$ has no rational solutions, it means that if you plug in $y=0$ than the solution of $x^2-3=0$ is irrational, so $x=\sqrt{3}$ is irrational