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Given a linear function $T: \Bbb R \rightarrow \Bbb R$ and $T(x+y) = T(x) + T(y)$, do we have that $T(x) = s(x)$ for some $s \in \Bbb R$

I'm doing an exam review for my analysis class but the professor said this has to do with rings and fields and I'm not sure how to rigorously show this using analysis.

My idea for the proof is to let $x = 0,1$ but I feel like those are the trivial cases. Any help is appreciated!

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    $\begingroup$ $T:\mathbb{R}\to\mathbb{R}$ can satisfy $T(x+y)=T(x)+T(y)$ without being of the form $Tx=sx$ for some $s\in\mathbb{R}$. To see this, take a basis of $\mathbb{R}$ over $\mathbb{Q}$, say containing $1$ and $\pi$. Define $T$ linearly by mapping $1$ to $\pi$ and vice-versa, and fixing the other elements of the basis. However, if $T$ is continuous then $T$ is of the form you described. $\endgroup$ – Luiz Cordeiro Feb 16 '16 at 22:30
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    $\begingroup$ @LuizCordeiro Yes, this is so, but it is said given a linear map $T$, which I do not know what is meant by. Because you are talking for an aditive map, which is not multiplicative over $\mathbb R$ $\endgroup$ – Svetoslav Feb 16 '16 at 22:32
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    $\begingroup$ @Kundor this is also what I am wandering. $\endgroup$ – Svetoslav Feb 16 '16 at 22:33
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    $\begingroup$ @Svetoslav I see. At first I tought it was just some mispelling. The word 'linear' is really ambiguous here, but I guess it mean of the form $T(x)=ax+b$ (but for me this is bad nomenclature; I prefer to call them "affine") in which case the Svetoslav's first comment applies (and makes the exercise quite trivial. If we use linear-algebraic nomenclature, then the exercise becomes even more trivial. $\endgroup$ – Luiz Cordeiro Feb 16 '16 at 22:39
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    $\begingroup$ @LuizCordeiro I guess, since it was in analysis class, the professor actually meant $ax+b$. But still, as he also mentioned rings and fields, it might be as well the linear-algebraic stuff. $\endgroup$ – Svetoslav Feb 16 '16 at 22:56

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