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*.

| | | | created: 2019-06-23 21:00:27 | modified: 2019-06-27 17:22:15 | by: *bookofproofs* | references: [701], [8189]

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[8189] **Kraetzel, E.**: “Studienbücherei Zahlentheorie”, VEB Deutscher Verlag der Wissenschaften, 1981

[701] **Scheid Harald**: “Zahlentheorie”, Spektrum Akademischer Verlag, 2003, 3. Auflage

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