rationality testing - Is there a way to prove numbers irrational in general?

I'm familiar with the typical proof that $\sqrt2\not\in\mathbb{Q}$, where we assume it is equivalent to $\frac ab$ for some integers $a,b$, then prove that both $a$ and $b$ are divisible by $2$, repeat infinitely, proof by contradiction, QED. I'm also familiar with the fact that if you repeat this procedure for any radical, you can similarly prove that each $a^n,b^n$ are divisible by the radicand, where $n$ is the root of the radical (though things get tricky if you don't reduce first). In other words, the proof that $\sqrt2\not\in\mathbb{Q}$ generalizes to prove that any number of the form $\sqrt[m]{n}\not\in\mathbb{Q}$.

Further, since adding or multiplying a rational and an irrational yields an irrational, this is a proof for all algebraic irrationals. (Ex. I can prove $\phi=\frac{1+\sqrt5}{2}$ irrational just by demonstrating that the $\sqrt5$ term is irrational.)

Because this relies on the algebra of radicals, this won't help for transcendental numbers. Is there a proof that generalizes to all irrational numbers, or must each transcendental number be proven irrational independently?