Cauchy's functional equation is given as


Wikipedia states that the solution to this functional equation with $x\in\mathbb Q$ is $f(x)=cx$, where $c$ is an "arbitrary rational" number. Without adding more constraints, one cannot solve for $x\notin\mathbb Q$.

I don't understand why we would have $c\in\mathbb Q$, as from my proof:

For $x\in\mathbb N$:

$$f(x)=f\left(\sum_{k=1}^x1\right)=\sum_{k=1}^xf(1)=xf(1)\tag{By Induction}$$

One can further extend this to $x\in\mathbb Z$ and lastly $\mathbb Q$.

So the problem becomes down to $f(x)=xf(1)$, and since we can have $f(1)$ to be any value we want, $f(x)=cx$ for arbitrary $c$ and rational $x$. But why is it that $f(1)$ be rational? Why not have $c\in\mathbb C$?

And I also want to be sure that we can't solve for $x\notin\mathbb Q$ because the induction step doesn't work over $\mathbb R$. For example, one may say $f(\pi)=100$, and so $f(\pi+x)=100+cx$ for $x\in\mathbb{Q}$, right? One may be able to argue $f(0)=f(\pi)+f(-\pi)$, so I'll say "Yes, $f(-\pi)=-100$, which appears to contradict $f(\pi+x)$, but it doesn't because $x$ must be rational?"


In the context that the Wikipedia article claims the solutions have this form, they are considering the case where $f$ is a function from $\mathbb{Q}$ to $\mathbb{Q}$ (though this assumption is not written down explicitly). So by assumption, $f(1)$ must be in $\mathbb{Q}$. If instead you consider (say) functions $f:\mathbb{Q}\to\mathbb{C}$, then you are correct that $f(x)=cx$ is a valid solution for any $c\in\mathbb{C}$.

When the domain is not restricted to $\mathbb{Q}$, I'm not sure I follow what you're asking--what "induction step" in particular are you talking about? The point is that you have given a proof which shows that $f(x)=xf(1)$ for all $x\in\mathbb{Q}$, and this proof makes essential use of the fact that $x=\pm m/n$ for some $m,n\in\mathbb{N}$ (and indeed at various points uses induction on $m$). So it is obvious that the proof does not apply to irrational numbers, at least not without some major modification.

  • $\begingroup$ Oh, that makes much more sense. Thank you! $\endgroup$ – Simply Beautiful Art May 23 '16 at 23:44

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