$(X,Y) \sim U [0,1]^2$. Therefore $X,Y \sim U[0,1]$ each and $X$ and $Y$ are independent. What are the distributions of $X-Y$ and $X+Y$?
My approach: It is my intuitive understanding that $X-Y$ should follow a symmetric triangular distribution over $[-1,1]$, i.e. the triangular distribution with $a=-1,b=1,c=0$. I have tested this using discrete uniform distributions, e.g. the discrete uniform distribution described by: $\mathbb{P}(x,y)= \frac{1}{16}$ for $x,y \in \{\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}\}$, $\mathbb{P}(x,y) = 0$, otherwise. We get $X-Y$ follows the distribution given by:
\begin{align*} \mathbb{P}(x) &=\frac{4}{16} \;\text{if}\; x=0,\\ &=\frac{3}{16} \;\text{if}\; |x|=\frac{1}{4},\\ &=\frac{2}{16} \;\text{if}\; |x|=\frac{2}{4},\\ &=\frac{1}{16} \;\text{if}\; |x|=\frac{3}{4},\\ &=0, \;\text{otherwise.}\\ \end{align*}
Similarly, $X+Y$ should follow the symmetric triangular distribution with $a=0,b=2,c=1$.
However I'm not able to prove these. Thank you in advance for your help.
Edit: The above discrete uniform example is just an example I played with to intuitively understand what the solution should be for the given problem. This example is not meant as a solution to the problem. I thought that was clear but apparently it was not - as somebody pointed out in a (now deleted) comment.
Let's consider first $U=X+Y$
Let's set
$$ \begin{cases} u=x+y, \\ v=x, \end{cases}\rightarrow \begin{cases} x=v, \\ y=u-v, \end{cases}$$
The jacobian is 1 thus
$$f_{UV}(u,v)=\mathbb{1}_{[0;1]}(v) \cdot \mathbb{1}_{[v;v+1]}(u)= \mathbb{1}_{[0;1)}(u)\mathbb{1}_{[0;u]}(v) +\mathbb{1}_{[1;2]}(u)\mathbb{1}_{[u-1;1]}(v) $$
That is defined overthe following parallelogram.
Integrate
$$f_U(u)=\int_{\mathcal{V}} f_{UV}(u,v)dv$$
And obtain your triangular density.
Same reasoning for $X-Y$ (here the parallelogram is shifted)