Cauchy's Functional Equation Revisited

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

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

has solutions f(x)=cxf(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 ff be a function f:RRf:\mathbb{R} \to \mathbb{R} such that

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

Consider the graph of y=f(x) 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.

Note by Sharky Kesa
4 years, 11 months ago

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Wikipedia :P

Deeparaj Bhat - 4 years, 11 months ago

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How do you know Wikipedia is correct? Anyone can edit it.

Sharky Kesa - 4 years, 11 months ago

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Anyone can edit wikipedia :v

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

Because you guys are more trusted than wikipedia

Jason Chrysoprase - 4 years, 11 months ago

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

Deeparaj Bhat - 4 years, 11 months ago

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