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Find all continuous functions $f: \mathbb{R}\to \mathbb{R}$ such that $$f(x)-f(y)\in \mathbb{Q}\quad \text{ for all }\quad x-y\in\mathbb{Q}$$

I found a solution here which is this:

By translation we may assume $f(0)=0$. Fix $a\in {\bf Q}$ and let $g(x):=f(x+a)-f(x)$ then $g$ is continuous and $g:{\bf R}\rightarrow {\bf Q}$. Hence $g$ is constant, its value is $f(a)$ hence $f(x+a)=f(x)+f(a)$ when $a\in {\bf Q}$. Then also $f(a+b)=f(a)+f(b)$ when $a,b\in {\bf Q}$. Then $f(x)=kx$ on ${\bf Q}$, hence $f(x)=kx$ on ${\bf R}$. The only extra condition is $k\in {\bf Q}$. The orginal problem has solutions $f(x)=kx+l$ where $k\in{\bf Q},l\in {\bf R}$.

  1. How did they assume $f(0)=0$?
  2. How did they conclude that $g(x)$ is a constant function?
  3. Not a doubt, but what could have been the thought process/motivation behind taking such $g(x)$ in such a way?
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zaemon_23
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  • For 1) if f(0)=k define g(x)=f(x)-k then g satisfies same, and g(0)=0. Can reverse this to get any possible f. – coffeemath Sep 07 '23 at 20:11
  • The problem is poorly worded. Hard to know for sure, but, I suspect that something was lost and that the original problem had something like $$f(x)-f(y)\in \mathbb{Q}\quad \text{ for all }\quad x,y \in \mathbb R \quad\text{ such that } \quad x-y\in\mathbb{Q}$$ – Lee Mosher Sep 07 '23 at 20:22

1 Answers1

1

Point 1

The map $h(x)= f(x) - f(0)$ is obviously continuous, satisfies the given hypothesis, and verifies $h(0)=0$. Hence we can assume without loss of generality that $f(0)=0$.

Point 2

The image of an interval under a continuous map is an interval (intermediate value theorem). The only intervals included in $\mathbb Q$ are the ones having only one point, therefore $g$ is constant (the image of $\mathbb R$ which is an interval consists in a single point).

Point 3

It seems natural to start fixing one of the two $x,y$ in the given hypothesis.

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  • Ok, regarding the point 1, say in general, we can construct a function according to our need, and if the function has the same properties as the original function, we can say whatever applies to our function would then also apply to the original function, like here right? – zaemon_23 Sep 07 '23 at 20:21
  • @zaemon_23 You got it. – mathcounterexamples.net Sep 08 '23 at 16:10