Probability that $\frac{x}{y}$ is close to an even number, $0 
Two real numbers $x,y$ are choosen uniformely from $(0,1)$. What is
  the probability that the closest integer to $\frac{x}{y}$ is even? Can
  it be expressed in the form $r+s\pi$ for $r,s \in \mathbb{Q}$.
So, the number $\frac{x}{y}$ could potentially be very small or very large. If $\frac{x}{y}$ is close to an even number, then $2k-0.5<\frac{x}{y}<2k+0.5$ for some $k\in \mathbb{N}$. I used online plot to get an idea of what $\frac{x}{y}$ look like.

 A: 
The lines of constant $x/y$ are just straight lines through the origin. So if $k>=1$, the region $B$ given by $2k-\frac12<x/y<2k+\frac12$ is a triangle, with height $1$ (along the $x-$axis) and base $\dfrac{1}{2k-\frac12}-\dfrac{1}{2k+\frac12}$ (along the $y$-axis), with area $\frac12\left(\dfrac{1}{2k-\frac12}-\dfrac{1}{2k+\frac12}\right)=\dfrac{1}{4k-1}-\dfrac{1}{4k+1}$.
And the region $A$ given by $x/y<\frac12$ is a triangle of area $\frac14$. So the total probability is
$$\frac14+\sum_{k=1}^\infty\left(\dfrac{1}{4k-1}-\dfrac{1}{4k+1}\right)$$
And Leibniz's formula for $\pi$ says that
$$1-\frac13+\frac15-\frac17+\cdots=\frac{\pi}{4}$$
So
$$\sum_{k=1}^\infty\left(\dfrac{1}{4k-1}-\dfrac{1}{4k+1}\right)=\frac13-\frac15+\frac17-\frac19+\cdots$$
is equal to $1-\pi/4$, and we get a total probability of
$$\frac14 + 1 -\frac{\pi}{4}=\frac{5-\pi}{4}$$
A: At $k\in \mathbb{N}$: $0<y<1$, $0<x<1$, $(2k-\frac12)y < x < (2k+\frac12)y$ $\Rightarrow$ $\left(0 < y \leq \frac{1}{2k+\frac12} \land (2k-\frac12)y < x < (2k+\frac12)y\right) \lor$ $\left(\frac{1}{2k+\frac12} < y < \frac{1}{2k-\frac12} \land (2k-\frac12)y < x < 1\right)$
$$P_k=\int_0^{\frac{1}{2k+\frac12}} \int_{(2k-\frac12)y}^{(2k+\frac12)y} dx dy+\int_{\frac{1}{2k+\frac12}}^{\frac{1}{2k-\frac12}} \int_{(2k-\frac12)y }^1 dx dy=\frac{2}{(4k-1)(4k+1)}$$
At $k=0$:  $0<y<1$, $0<x<1$, $(2k-\frac12)y < x < (2k+\frac12)y$ $\Rightarrow$ $\left(0 < y < 1 \land 0 < x < \frac12 y\right)$
$$P_0=\int_0^1 \int_0^{\frac12 y} dx dy=\frac14$$
$$P=\frac14+\sum_{k=1}^\infty \frac{2}{(4k-1)(4k+1)}=\frac{1}{4}+1-\frac{\pi}{4}=\frac{5-\pi}{4}$$
