# Inverse of an algebraic function is irrational under a condition

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.

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 $$\neg Q \implies \neg P$$
• Oh, I completely forgot about the identity function. That's a good point. I'll edit the question to exclude that function . – LogicAndTruth Mar 26 at 8:35
• In what sense is the negation $\neg P \implies \neg Q$ (which is logically equivalent to the converse)? Surely the negation of $P \implies Q$ is simply $P \land \neg Q$. – 雨が好きな人 Mar 28 at 10:09
• I had some wires crossed, that's my bad. OP meant the inverse, and that's what $\neg P \implies \neg Q$ is. – Eevee Trainer Mar 28 at 18:36