Let $V$ be a vector field over $\mathbb Q$, the rationals. Prove that if $T:V\rightarrow V$ is additive (that is, $T(x+y)=T(x)+T(y)$ for all $x,y \in V$), then $T$ is linear.

The two properties for $T$ to be linear is additivity, which is given, and for $T(cx)=c*T(x)$ for all $c \in \mathbb Q$ and $x \in V$, but I'm not sure how to get that from the givens. I think it has to do with the vector field being over the rationals, but I'm not sure.


First show $T(0)=0$. Then by induction, show $T(nv)=nT(v)$ for $n\in\Bbb N$. Then show $T(-v)=-T(v)$ and conclude $T(nv)=nT(v)$ for $n\in \Bbb Z$. Use this to conclude that $T(\frac nm v)=\frac nm T(v)$.

  • $\begingroup$ How do I prove that $T(0)=0$. I thought I would prove it with $T(x-x)=T(0)=T(x)+T(-x)$, but then I would need the identity that $T(-v)=-T(v)$ which I don't have yet. $\endgroup$ – AndroidFish Sep 21 '16 at 21:18
  • $\begingroup$ @AndroidFish: Any monoid homomorphism between two groups is a group homomorphism. Does this help? (Of course, you still have to prove the above-mentioned fact, but I hope it is sufficiently distilled that there should be only one way to go now.) $\endgroup$ – darij grinberg Sep 21 '16 at 21:20
  • $\begingroup$ @AndroidFish $T(0)=T(0+0)=T(0)+T(0)$. $\endgroup$ – egreg Sep 21 '16 at 21:20


First prove by induction that $T(nx)=nT(x)$ for all natural numbers $n$.

Then extend this property to integers. Finally to prove it for any rational number $c=a/b$, observe $y=cx\iff by=ax$.


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