An earlier exercise asks for a proof of the following result:
Corollary 1.9 Let $\alpha$ be a real number and $N$ a positive integer. Then there exists a rational number $p/q$ such that $1\le q\le N$ and $$|\alpha - {\frac{p}{q}}|\le {\frac{1}{2N}}$$.
$\it{Proof}$. Consider the intervals $\{I_k\}$ indexed by $k\in {\bf{Z}}$ where $I_k = [k/N, (k+1)/N)$. These mutually disjoint intervals cover the real line, and so $\alpha$ must lie in a unique such interval. The closest end point of the interval $I_k$ to $\alpha$ furnishes required rational number $p/q$.
In this context, Burger defines the notion of a $\it{sharp}$ inequality as follows: "Given an integer $N$ there exist real numbers $\alpha$ for which we would have $\it{equality}$ in the upper bounds of the two inequalities in Corollary 1.9. We say the inequalities are $\it sharp$ for such $\alpha$." (page 9)
QUESTION: (Exercise 1.12) Given a fixed integer $N$ for what values of $\alpha$ are the inequalities of Corollary 1.9 sharp?
Burger's definition of $\it{sharp}$ : "That is given an integer $N$, there are $\alpha$ for which we would have $\it{equality}$ in the upper bounds of the two inequalities in Corollary 1.9. Thus we say that the inequalities are sharp for those $\alpha$."
These two inequalities are: $1\leq q\leq N$, and $|{\alpha} - (p/q)|\leq {\frac{1}{2N}}$. So equality in the two upper bounds would be $q = N$ and $|{\alpha} - (p/N)|={\frac{1}{2N}}.$ This happens for all real numbers ${\alpha} = {\frac{2k+1}{2N}}$ for $k\in{\bf{Z}}$, since either rational $p/q = k/N$ or $p/q = (k+1)/N$ works.