I want to intuitively understand why $a+b\sqrt {2}\neq \sqrt {3} $ for $a, b \in \mathbb Q $
I really have no intuition regarding this matter, and have to deal with similar concepts regularly while studying field extensions.
Please do not give algebraic proofs like the squaring of both sides and arriving upon a contradiction. I'm looking for something that involves logic/geometry/anything else that would hone my intuition in such matters.
On
$\sqrt2$ is just an arbitrary number that, when squared, produces $2$. Similarly for $\sqrt 3$. That is, although we define them as the positive roots, there is no algebraic distingtion between the positive and the negative root. Thus for any rational numbers $a,b$ for which $a+b\sqrt 2$ is "a" root of $3$, we should expect that also $a-b\sqrt 2$ (i.e., playing with "the other" root of $2$) gives "a" root ot $3$. It could be the same root of $3$, but then we conclude $a=\sqrt 3$ by adding the equations. Or it could be the "other" root, but then we conclude $b\sqrt 2=\sqrt 3$ by subtracting the equations (and then $2b=\sqrt 6$). Now dig out your favorite proof that $\sqrt 3$ and $\sqrt 6$ are irrational and you are done.
Alternatively, with "hidden" squaring: Assume $a+b\sqrt 2=\sqrt 3$. Then $$ a-b\sqrt 2=\frac{a^2-2b^2}{a+b\sqrt 2}=\frac{a^2-2b^2}{3}\sqrt 3$$ Hence $$2a=\left(\frac{a^2-2b^2}3+1\right)\sqrt 3 $$ and by irrationality of $\sqrt 3$, this requires $a=0$ and then $2b^2=3$, which fails because $\sqrt 6$ is irrational.
I am not sure what a non-algebraic proof would even involve since $\sqrt 2$ and $\sqrt 3$ are only defined algebraically as the roots of some equation(s). However the easiest proof I know of is using the trace. (I am not sure if you have covered it yet...):
$$tr(a+b\sqrt 2) = tr(c\sqrt 3)$$ $$\Rightarrow2a = 0$$
Then it is easy to see that $b\sqrt 2$ cannot equal $c\sqrt 3$ intuitively. The trace is a natural operation to use here since it is the analogue of $z + \bar z = 2Re(z)$ in complex analysis.