Inverse of an algebraic function is irrational under a condition

Question

I've recently been looking to Function Theory/Real Analysis and I've encountered a bit of a problem I'm not entirely sure how to approach. This isn't from a textbook, rather something I have noticed that may be the case or may not be. Here's the problem:

Let $f: \mathbb{R} \to \mathbb{R}$ be an algebraic function with an existent inverse over $\mathbb{R}$. If $\alpha \in \mathbb{R}$ is such that $f(\alpha) \in \mathbb{Q}$, is $f^{-1}(\alpha) \not\in \mathbb{Q}$? Is the converse, inverse and contrapositive of this statement true?

EDIT: As pointed out, $f(x) = x$ is a counterexample to this, but since this is obvious, I'll elect to exclude this as a function to be considered.

I've been tempted to generalise everything and brute force from there, although I highly doubt that is the correct approach, let alone a reasonable way to approach this. I feel like I could be missing something really obvious here however I have not been able to even get a proper foothold. Any help or guidance would be greatly appreciated!

EDIT 2: Found a simple counterexample again, so this question isn't really that troubling anymore.

Answer 1

Focus on examples. Take the counterexample of $f(x) = x = f^{-1}(x)$.

• For the forward direction, try $\alpha = 1$, which is rational.
• For the converse, try $\alpha=\pi$, which is irrational.
• For the inverse, try $\alpha=\pi$.
• For the contrapositive, try $\alpha=1$.

Each of these give a counterexample to the relevant statement. Remember that, for the statement $P \implies Q$, we have

• the converse being $Q \implies P$
• the inverse being $\neg P \implies \neg Q$
• the contrapositive being $\neg Q \implies \neg P$