How many fractions of the form $\frac{n}{n+1}$ are less than $7 / 9$, knowing that $n$ is a positive integer?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
The correct option is (c).

Solution 1: The fractions of the form $\frac{n}{n+1}$, with $n$ a positive integer, are

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-037.jpg?height=160&width=616&top_left_y=1799&top_left_x=800)

Observe that we have $\frac{1}{2} \cdot \frac{35}{45} = \frac{7}{9}$. Therefore, $\frac{4}{5}$ is greater than $\frac{7}{9}$, and since the sequence is increasing, from $\frac{4}{5}$ onwards, all fractions in this sequence are greater than $\frac{7}{9}$. Thus, there are only three fractions of the form $\frac{n}{n+1}$ that are less than $\frac{7}{9}$, namely, $\frac{1}{2}, \frac{2}{3}$, and $\frac{3}{4}$.

Solution 2: Converting everything to decimal numbers, we have $7 / 9 = 0.777 \ldots$ and $1 / 2 = 0.5, 2 / 3 = 0.666 \ldots, 3 / 4 = 0.75, 4 / 5 = 0.8, 5 / 6 = 0.8333 \ldots$ Therefore, the sequence is increasing, and only $1 / 2 = 0.5, 2 / 3 = 0.666 \ldots$, and $3 / 4 = 0.75$ are less than $7 / 9 = 0.777 \ldots$