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A psychologist says, that just like any other factor, the workspace designs and interiors can also affect the performance of the employees by such a great extent, you can’t imagine. The workspace basically, when designed effectively, the color combinations, the interior decors, etc., actually leave a lot of positive vibes, and make the people working around feel inspired to do what they are doing. The level of logic behind this is very deep, but actually true, and proven scientifically.
Now, you must be wondering what will it be about the design and stuff. Well, if you have the money to spare, you need not worry about it, just hire a professional interior decorator, he will do the job for you, otherwise, there are some really cool online and offline resources available to study from, and design your own new thing, but if you feel like you don’t have the time to do so, the simple easy solution is to check out these cool Workspace designs for Inspiration, which will give you a brief idea. If you like these workspace Designs for Inspiration then you can easily create the one like anyone of these according to your taste. Let’s check out some of the best Workspace designs for inspiration.Q: 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...
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