Continuous functions $f$ with $f(A)$ linearly independent when $A$ is independent Is there any characterization of continuous functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ such that for any linearly independent set $A$ (over the rationals) $f(A)$ is also linearly independent ? 
 A: Trevor Wilson has reduced the possibilities to $\alpha x^\beta$ with $\alpha\ne 0$ and $\beta\ge 0$.  The following argument should reduce the possibilities to $\beta=0$ or $\beta=1$, modulo a lemma that I can't quite seem to prove. Perhaps some other MO reader can plug the gap.
The lemma I need is:

For any $q_1, q_2, q_3\in\mathbb{Q}$, not all zero, and any positive $\beta\ne1$, the equation $$q_1+q_2x^\beta+q_3(1+x^{\beta^2})^{1/\beta} = 0$$ has at most countably many real solutions.

This lemma implies that for any positive $\beta\ne1$, there exists a constant $c>0$ such that the three numbers $x_1:=1$, $x_2:=c$, and $x_3:=(1+c^{\beta})^{1/\beta}$ are linearly independent over $\mathbb{Q}$.  On the other hand, $\alpha x_1^\beta + \alpha x_2^\beta = \alpha x_3^\beta$, so their images under the function are linearly dependent.
A: This may not be a complete answer to the question, but it is too long for a comment.
Given such a function $f$, the restriction $f \restriction [0,\infty)$ is a power function $\alpha x^\beta$ where $\alpha \ne 0$ and $\beta \ge 0$, and similarly the restriction $f \restriction (-\infty,0]$ is a power function of $-x$.  We will prove this for the restriction $f \restriction [0,\infty)$ because the other case is similar.
Because $\sqrt{2}$ is irrational, for any $x > 0$ the set $\{x,\sqrt{2} x\}$ is linearly independent, so the set $\{f(x),f(\sqrt{2} x)\}$ is linearly independent, meaning that its elements are nonzero and either


*

*Their quotient $f(\sqrt{2} x)/f(x)$ is irrational, or

*They are the same element: $f(\sqrt{2} x) = f(x)$, so the quotient $f(\sqrt{2} x)/f(x)$ is 1.
In either case the quotient $f(\sqrt{2} x)/f(x)$ must take a constant value on this interval because it is a continuous function of $x$ whose range is contained in the totally disconnected set $(\mathbb{R} \setminus \mathbb{Q}) \cup \{1\}$.
Therefore the function $x \mapsto f(2 x)/f(x)$ must also take a constant value $K$ on the interval $(0,\infty)$, namely the square of the previous constant.
Observe that for any positive integer $n$ we have $f(2^{1/n}) = K^{1/n}f(1)$, and in fact for any positive rational number $q$ we have $f(2^q) = K^q f(1)$.
This means that $f$ is a multiple of a power function on the set $\{2^q : q \in \mathbb{Q}\}$.  By continuity this holds on the closure of this set, namely the interval $[0,\infty)$.
Note that the constant factor $\alpha$ cannot be zero, and that the exponent $\beta$ must be non-negative for $f$ to be continuous at zero.
Discussion of special cases:
Constant case: If the exponent on the interval $[0,\infty)$ is zero, then
the function is a nonzero constant $\alpha$ on this interval.
Therefore $f(0) \ne 0$, which means that on the interval $(-\infty,0]$ the exponent must also be zero and the function must have the same constant value $\alpha$ by continuity.  So in this case $f$ is a nonzero constant function.
Another case is that the exponent $\beta$ is equal to $1$ on both the intervals $(-\infty,0]$ and $[0,\infty)$, so it is piecewise linear.
In this case the only restriction on the slopes is that they must be nonzero rational multiples of each other.  I don't know if this is the only other case.
