Proof of rational numbers I am recently starting in the subject of pure mathematics and this problem has eluded me. 
Any rational number $\frac{p}{q}$ can be expressed in a simple series form:
$$\frac{1}{1},\frac{2}{1},\frac{1}{2},\frac{3}{1},\frac{2}{2},\frac{1}{3},\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4},...$$
Show that $\frac{p}{q}$ is the $[\frac{1}{2}(p+q-1)(p+q-2)+q]th$ of the series (where $p$ and $q$ are coprime and $p \neq 0$).
 A: Hint. Divide the numbers in rows:
$$\frac{1}{1}$$
$$\frac{2}{1},\frac{1}{2}$$
$$\frac{3}{1},\frac{2}{2},\frac{1}{3}$$
$$\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4}.$$
In the first $n$ rows there are $1+2+\dots+n=n(n+1)/2$ fractions.
A: First of all I think that the first number in the sequence should be $\frac 11$.
Now consider the sum of the numerator and the denominator and write it as a sequence and we have:
$$2,3,3,4,4,5,5,5,5,6,6,6,6,6...$$
Therefore if $p+q=r$ we have that the fraction $\frac{p}{q}$ might appear after $1+2+3+...+{r-2} = \frac{(r-1)(r-2)}{2} = \frac{(p+q-1)(p+q-2)}{2}$ elements in the sequence, as the sum of denominator and numerator before that element is less than $r$.
Now note that the first denominator of the fractions whose sum is $p+q$ is $1$, then $2$ and so on. So $q$ will appear in the denominator after $q$ elements, starting from the $\frac{(p+q-1)(p+q-2)}{2}$-th element. Therefore we have that the $\frac{(p+q-1)(p+q-2)}{2} + q$ elements is the fraction with denominator $q$ and sum of denominator and numerator $p+q$. In other words it's the rational number $\frac{p}{q}$
