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Does there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y \in\mathbb{R}$; $f(1)=1$; and $f(\sqrt 2)=3$?

Intuitively, I believe the answer is no. However, I'm having a hard time proving this. My initial thought was a proof by contradiction. Somehow using $f(1)=1$ and $f(\sqrt 2)=3$ to show that $f(x+y)\neq(x)+f(y)$ for some choice in $x$ and $y$. This didn't prove fruitful. My next thought was to use properties of functions satisfying the addition property of homomorphisms to prove that $f(\sqrt 2)\neq3$. For example, one such property is $f(nx)=nf(x)$ for all $n \in \mathbb{Z}, x \in \mathbb{R}$. With this, one could theoretically construct a proof by induction that $f(x)=x$ for all $x \in \mathbb{R}$. Once again, this didn't prove fruitful. After these failed attempts at showing such a function doesn't exist, I considered how to show that one does exist. For reference, this question was asked in the context of a Field Theory course. As such, the thought for how to construct a proof is to satisfy the addition property with some linear transformation and show that a correct selection of basis spans all of $\mathbb{R}$. This idea seems feasible, however I am not sure where to begin. Any advice on how to approach this problem is greatly appreciated.

Chappers
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Atsina
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1 Answers1

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Yes, it does exist.

Since $\sqrt{2}$ is irrational, there is a Hamel basis $B$ of $\mathbb R$ over $\mathbb Q$ that contains $1$ and $\sqrt{2}$. Any function $g: B \to \mathbb R$ extends uniquely to a $\mathbb Q$-linear (and thus additive) function $f: \mathbb R \to \mathbb R$. So just take $g(1) = 1$, $g(\sqrt{2}) = 3$, and whatever you want on the rest of $B$.

Robert Israel
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