It is commonly known that the solution for Cauchy's Functional Equation

\[ f:\mathbb{Q} \rightarrow \mathbb{Q}, \quad f(x)+f(y)=f(x+y)\]

has solutions \(f(x)=cx\). (If you don't know how to prove this, I ask for you to attempt to prove this)

But now, if we instead change this function from being in the rationals to being the reals, we get a strange property.

Let \(f\) be a function \(f:\mathbb{R} \to \mathbb{R}\) such that

\[\begin{align} f(x) + f(y) &= f(x+y) \forall x, y\\ f(1) &= 1\\ \text{ The function is not }f(x) & = x \end{align}\]

Consider the graph of \( y = f(x) \) on the plane. Prove that for any disc, no matter how small, there always exists a point that lies on the graph.

For this proof, you must accept the Axiom of Choice as true.

This question came from a corollary I used in APMO 2016 Q5.

## Comments

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TopNewestWikipedia :P – Deeparaj Bhat · 1 year, 1 month ago

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– Sharky Kesa · 1 year, 1 month ago

How do you know Wikipedia is correct? Anyone can edit it.Log in to reply

For math homework, i usually go here if i'm confused

Because you guys are more trusted than wikipedia – Jason Chrysoprase · 1 year, 1 month ago

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– Deeparaj Bhat · 1 year, 1 month ago

I've seen their proof. It was right in my opinion.Log in to reply

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– Sharky Kesa · 1 year, 1 month ago

I'm saying on the graph of \((x, f(x)\).Log in to reply

– Calvin Lin Staff · 1 year, 1 month ago

K, edited. Makes sense now.Log in to reply