Uniform random variable is greater by a constant from another uniform random variable I am trying to formulate the following question. 
X and Y are IID , uniform r.v. with ~U(0,1)
What is the probability of
P( X-Y-0.5 > 0) = ? 
0.5 is a constant here and can be different.
I do respect the geometrical solutions but what i would like to see and understand is the generic approach since X and Y can be other distributions. 
 A: This kind of calculation is in general handled with using the joint distribution of the two variables $X$ and $Y$.
You state that $X$ and $Y$ are i.i.d. $U(0, 1)$'s, and this uniquely specifies their joint distribution. Since they are independent, the joint density function factors:
$$f_{X, Y}(x, y) = f_X(x) f_Y(y) = 1 \times 1 = 1$$
Given this, you can calculate any probability involving $X$ and $Y$ by integrating this joint distribution over the region specified in the probability:
$$P(X - Y > 0.5) = \int_{x - y > 0.5} f(x, y) \ dx dy = \int_{x - y > 0.5} 1 \ dx dy = \frac{1}{4}$$ 
In this case it's relatively easy to evaluate this integral by inspection, for more complicated cases evaluating the integral may take some work and creativity.
A: The generic approach is to find the distribution of the rv $X-Y$, denoted $Z$, from the joint distribution of $(X,Y)$, which is a convolution exercise. Since the change of variables from $(X,Y)$ to $(Z=X-Y,Y)$ has Jacobian one
$$\left|\dfrac{\text d(x,y)}{\text d(z,y)}\right|=\left|\dfrac{\text d(z+y,y)}{\text d(z,y)}\right|=\left|\begin{matrix}1 &1\\0 &1\\\end{matrix}\right|=1$$
the joint distribution of $(Z,Y)$ has density$$f_{Z,Y}(z,y)=f_{X,Y}(z+y,y)$$and the marginal distribution of $Z$ has density$$f_Z(z)=\int_{\mathcal Y}f_{X,Y}(z+y,y)\text dy$$leading to$$\mathbb{P}(Z>z_0)=\int_{z_0}^\infty \int_{\mathcal Y}f_{X,Y}(z+y,y)\text dy\, \text dz$$
as for instance
$$\mathbb{P}(Z>0.5)=\int_{0.5}^1\int_{0}^1 \mathbb I_{(0,1)}(y+z)\text dy\, \text dz$$
for two iid $U(0,1)$ rvs. Leading to
$$\mathbb{P}(Z>0.5)=\int_{0.5}^1\int_{0}^{1-z} \mathbb I_{(0,1)}(y+z)\text dy\, \text dz=\int_{0.5}^1 (1-z)\, \text dz=\frac{1}{4}$$
