Is there a function between real vector spaces that satisfies exactly one of the requirements to be a linear transformation? As almost everyone reading this probably knows, a function $T$ from one real vector space to another is defined to be a linear transformation if for all vectors $\mathbf{u}, \mathbf{v}$ in the domain and all real $a$, (i) $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$, and (ii) $T(a\mathbf{u}) = aT(\mathbf{u})$.  Does anyone know of an example of two real vector spaces and a function $T$ from one to the other that satisfies one of the requirements but not both?  I'm assuming there is one, or else it wouldn't be defined that way.
I can show that if $T$ satisfies (i), then (ii) holds for all rational $a$.
Is there such an example where the vector spaces are both finite-dimensional?
I am not interested in complex vector spaces or additional requirements such as norms. 
I browsed the Similar Questions and couldn't find an answer.  I apologize if this is a duplicate question.
Stefan (STack Exchange FAN)    
 A: An example that satisfies $(i)$ but not $(ii)$:
We can construct a function from $\mathbb{R}$ to itself; construct a basis for $\mathbb{R}$ over $\mathbb{Q}$ by taking the set $\{1, \sqrt{k}\}$ for some squarefree $k$, and extend this to a basis using the axiom of choice. Then define $f(1)=1$ and $f$ to be $0$ for all other basis elements. If we also define $f$ to be linear from this vector space of $\mathbb{R}$ over $\mathbb{Q}$, then $f$ is only $\mathbb{Q}$-linear and additive in $\mathbb{R}$, but it can be seen that it's nonhomogeneous, as $f(\sqrt{k}\cdot 1) = 0$ while $\sqrt{k}f(1)=\sqrt{k}$.
A: I'll give one example of a function satisfying (ii) but not (i) that I think is simple. Consider $T:\mathbb R^2 \to \mathbb R$ mapping $(r\cos\theta, r\sin\theta)$ to $r\theta$ for $\theta \in [0, \pi)$ and to $-r(\theta - \pi)$ for $\theta \in [\pi, 2\pi)$.
A: For ii) but not i), I think $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ given by $f(x,y) = (x^3 + y^3 )^{1/3}$ is the simplest example. For i), examples actually exist for when the domain of $f$ is simply $\mathbb{R}$ - see here.
A: (ii) but not (i):
$f:\mathbb{R}^2\to\mathbb{R}^2$,
$f(r\cos\theta,r\sin\theta) = \theta(r\cos\theta,r\sin\theta)$
where $\theta\in[0,\pi), r\in\mathbb{R}$
This is well defined because every point in $\mathbb{R}^2$ can be written in terms of polar coordinates, $f(0,0)=(0,0)$ independently of $\theta$.
$f(ax,ay)=f(ar\cos\theta,ar\sin\theta)=a\theta(r\cos\theta,r\sin\theta)=af(x,y)$.
$f((0,1)+(1,0)) = f(1,1) = \frac{\pi}{4}(1,1)$
$f(1,0) + f(0,1) = 0 + \frac{\pi}{2}(0,1) \ne \frac{\pi}{4}(1,1)$
A: In one dimension, the i) but not ii) half of this reduces to Cauchy's functional equation
$$
f(x+y)=f(x)+f(y)
$$
which has nonlinear solutions, but not nice ones. 
Essentially, i) guarantees that your function is a linear transformation between rational vector spaces. But not every $\Bbb{Q}$-linear transformation is also $\Bbb{R}$-linear, even if its domain and range are $\Bbb{R}$-vector spaces...
