# Interpreting the ; in a series

This question is linked to this question.

So, suppose I set $n=5$. Given the following formula:

$$\frac{1}{n}, \dots , \frac{n-1}{n}$$

Am I suppose to get:

$$\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \hspace{8.2cm}(1)$$

Or

$$\frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{2}{5}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5} \hspace{5cm} (2)$$ ?

In other words, what purpose to the ";" in:

$$\frac{1}{2}; \frac{1}{3}, \frac{2}{3}; \frac{1}{4}, \frac{3}{4}; \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \dots ; \frac{1}{n}, \dots , \frac{n-1}{n}.$$

serve?

Also, the formula does not mention anything about skipping numbers that are not in lowest common terms. Is skipping this assumed given the definition of $f$? For example, in (2), there is no $\frac{2}{4}$ because it is equal to $\frac{1}{2}$ which is already listed earlier.

Thank you in advance for any help provided.

• you are supposed to understand this as in (1) – W_D Jul 23 '13 at 13:11

The commas separate values for a single given value of $n$. The semi-colons separate for different values of $n$:
$$\underbrace{\frac12}_{n=1};\underbrace{\frac{1}{3}, \frac{2}{3}}_{n=3};\underbrace{\frac14,\frac{2}{4},\frac{3}{4}}_{n=4};\ldots$$
The intention was that this form a single set of unique rational numbers; the $\frac{2}{4}$ can be omitted. The semi-colons distinguish by values of $n$ only for ease of visualization.