If $x \geq y>x/2$ then is it true that $x \pmod y < x/2$? I'm not sure how to prove this statement, which I believe is true: 
given $x,y \in Z$ such that $x \geq y > \frac {x}{2}$ then $x$ (mod $y$) < $\frac {x}{2}.$ 
Edit:
Would this be an acceptable sketch of the proof?
Suppose that $x \geq y \geq \frac{x}{2};$ consider the bounds where $y = x$ or $y = \frac{x}{2}$. 
In the case in which $x=y,$ then $x$ (mod $y$) = $x$ (mod $x) = 0$.
On the other hand, if $y = \frac{x}{2},$ then $x$ (mod $y$) = $x$ (mod $\frac {x}{2}) = \frac {x}{2}$.
Therefore, $0 \leq x$ (mod $y$) < $\frac {x}{2}$.

This is true.  Suppose that $0\leq \frac{x}{2} < y \leq x$.  Long divide $x$ by $y$ by the Division Algorithm to write $x=yq+r$ where $0 \leq r <y$.  Here, $r$ is your $x \bmod y$.  We claim that $r < \frac{x}{2}$.
Suppose not, so that $r \geq \frac{x}{2}$.  Then $x=yq+r \geq yq+\frac{x}{2}$.  Subtracting $\frac{x}{2}$ from each side gives $\frac{x}{2} \geq yq$.  But $y >\frac{x}{2}$ so this gives $\frac{x}{2} > \frac{x}{2}q$ so $q < 1$.  Hence $q=0$ and $x=r < y \leq x$, a contradiction.  
