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:

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


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

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

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


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.

  • $\begingroup$ you are supposed to understand this as in (1) $\endgroup$ – 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.


You are supposed to get (1), the complete enumeration makes it clear. The semicolon is only put for clarity in separating terms with different denumerator. Skipping is implicit because the author is enumerating the values of a set, hence repeating them has no effect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.