Is there an example of $\mathbb{R} \to \mathbb{R}$ function that produces only rationally independent numbers?

Question

Is there a function $f(x): \mathbb{R} \to \mathbb{R}$ such that $$nf(x) \neq mf(y) \quad \forall\ x,y \in \mathbb{R},\ n,m \in \mathbb{Z},\ x\neq y ?$$

Answer 1

Ehh I hardly think something transfinite recursion counts as an "example". How about something like the following:

First note that $\mathbb{R}$ and $(0,1)$ have the same cardinality so it suffices to construct a function $f: (0,1) \to \mathbb{R}$ with the same property.

Take the decimal expansion of a number $a = .a_1a_2a_3a_4\dots$ (ending in say all 0's if it has a finite expansion) and construct a new number $f(a) \le 1$ as follows:

After the decimal point $a_1+1$ zeroes then a 1. Then place $2(a_2+1)$ zeroes followed by a 1. Then place $6 (a_3+1)$ zeroes followed by a 1. Continuing this placing $n! (a_n+1)$ zeroes followed by a 1.

Clearly this is an injective function as we can read off the digits of $a$ from $f(a)$ by looking at the lengths of the strings of zeroes.

Now think about the map $\tilde{f}: \mathbb{N} \times (0,1) \to \mathbb{R}$ given by $\tilde{f}(n,a) = nf(a)$. I claim this is also injective. If we look at the decimal expansion of $\tilde{f}(n,a)$ after a while (depending on how big n is) it eventually looks like a bunch of zeroes followed by $n$ followed by a bunch of zeroes followed by $n$ and so on. Therefore we can figure out what $n$ is and therefore what $a$ is by the previous consideration.

But this function $\tilde{f}$ being injective is exactly the same as the image of $f$ being not being rational multiples as you want.

Answer 2

@AndrésE.Caicedo's point is the following. Suppose $\iota$ bijects some uncountable ordinal $\alpha$ with $\Bbb R$; the axiom of choice guarantees such an $\alpha$ exists, but doesn't identify it. We can choose $\alpha$ to be minimal, i.e. an aleph, so any $\beta,\,\gamma\in\alpha$ satisfy $|\beta||\gamma|=\max\{|\beta|,\,|\gamma|\}<|\alpha|$. Without loss of generality, $\iota(\emptyset)\ne0$. For $\beta\in\alpha$, define $f(\iota(\beta)):=\iota(\gamma)$ with $\gamma\in\alpha$ minimal so that $\iota(\gamma)$ is neither $0$ nor a rational multiple of any $f(\iota(\delta))$ with $\delta\in\beta$. In particular, $f(\iota(\emptyset))=\iota(\emptyset)$. This construction is always well-defined, because the disqualified values are always less numerous than the reals (since $\Bbb Q=\aleph_0$).