(related to Chapter: Algebraic and Transcendent Numbers)
Let $\frac{a}b\in\mathbb Q$ be a rational number with $0 < a$ and $0 < b.$ In the proof of the greatest common divisor for the calculation of the greatest common divisor $\gcd(a,b)$, the Euclidean algorithm can be written with a sequence of rational numbers defined as follows:
$$\begin{array}{rclclcl} \frac{a}{b} & = & q_0 & + & \frac{r_1}{b} & \text{with} & q_0\in\mathbb N,\; 0 < r_1 < b\\ \frac{b}{r_1} & = & q_1 & + & \frac{r_2}{r_1} & \text{with} & q_1\in\mathbb N,\; 0 < r_2 < r_1\\ \frac{r_1}{r_2} & = & q_2 & + & \frac{r_3}{r_2}& \text{with} & q_2\in\mathbb N,\; 0 < r_3 < r_2\\ &\vdots&\\ \frac{r_{n-3}}{r_{n-2}} & = & q_{n-2}& + & \frac{r_{n-1}}{r_{n-2}} & \text{with} & q_{n-2}\in\mathbb N,\; 0 < r_{n-1} < r_{n-2}\\ \frac{r_{n-2}}{r_{n-1}} & = & q_{n-1}& + & \frac{r_{n}}{r_{n-1}} & \text{with} & q_{n-1}\in\mathbb N,\; 0 < r_{n} < r_{n-1}\\ \frac{r_{n-1}}{r_{n}} & = & q_{n} &&& \text{with} & q_{n}\in\mathbb N \end{array}$$
Therefore, we can put the ratios in each other and get the following representation of the number $\frac ab$:
$$\frac{a}b=q_0+\cfrac{1}{q_1+\cfrac{1}{q_2+\cfrac{1}{\ddots+\cfrac{1}{q_{n-2}+\cfrac{1}{q_{n-1}+\cfrac{1}{q_n}}}}}}$$
Please note that because $r_{n} < r_{n-1}$, we have $q_n > 1$ and thus $q_n\neq 0.$ Thus, this representation is well-defined. It is possible to represent any positive rational number this way. But how about irrational numbers. In order to answer this question, we will introduce a new concept, the concept of continued fractions.
