2) If $a$ and $b$ are two integers with $1 \leq b \leq 3$ then the number of fractions (in lowest terms) of the form $\frac{a}{b}$ such that $1 \leq \frac{a}{b} \leq 2$ is
(A) 4
(B) 5
$\begin{array}{ll}\text { (C) } 6 & \text { (D) } 8\end{array}$
(E) infinite.
2) The answer is (B). The list of fractions $\frac{a}{b}$ with $a, b$ integers satisfying $1 \leq b \leq 3$ and $b \leq a \leq 2 b$ is indeed $\frac{1}{1}, \frac{2}{1}, \frac{2}{2}, \frac{3}{2}, \frac{4}{2}, \frac{3}{3}, \frac{4}{3}, \frac{5}{3}, \frac{6}{3}$ and, of these, 5 are in their simplest form.