If an operation applied to a rational number (root taking, limit of an infinite sequence of rational numbers) leads to a number that is not a rational number, i.e. if it cannot be written as a ration of 2 integers, which is representable as a finite or periodic decimal fraction, then this number is considered as an irrational number. Here the term irrational has historical reasons, as a demarcation from the rational(numbers that are ratios) and has no secondary meaning irrational = unreasonable or unthinkable.

If one applies the operations mentioned above to irrational numbers, then this does not yield another type of number.

Rational numbers constitute a countable set - they can be ordered in such a way, that they constitute a countable sequence. The irrational numbers however do not constitute a countable set. In this sense there are more irrational as rational numbers.

The need to introduce numbers that are not rational, the Pythagoreans (Pythagoras, 570 - 510 b.C., mathematician and natural philosopher in the Greek colony metapont in southern Italy ) recognized during their reflections about the calculation of right triangles with the hypotenuse $c$ and the legs $a$ and $b$.

In the domain of integers there are only few solutions, the Pythagorean triples, that are often used in homework problems (3, 4, 5; 6, 8,10; 5, 12, 13; 8, 15, 17; 7, 24, 25; 9, 12, 15; 10, 24, 26; etc)

Theorem of Pythagoras: ${a}^{2}+{b}^{2}={c}^{2}\to c=\sqrt{{a}^{2}+{b}^{2}}$

Example of an integer solution: $c=\sqrt{{3}^{2}+{4}^{2}}=\sqrt{25}=5$

Example of a rational solution:$\sqrt{{\left(\frac{3}{2}\right)}^{2}+{2}^{2}}=\sqrt{\frac{25}{4}}=\frac{5}{2}$

Example of an irrational solution: $c=\sqrt{{1}^{2}+{1}^{2}}=\sqrt{2}$

Numbers, that are in general obtained as solutions of polynomial equations with rational coefficients, i.e. that are their roots, are designated as algebraic numbers. The include both rational as well as irrational numbers. Rational numbers are thus rare special cases of irrational numbers

Irrational numbers, that are not a root of a polynomial with rational coefficients , are called transcendental numbers.

Here transcendental simply means going beyond the rational numbers and does not have any mystical background whatsoever.

The most common transcendental numbers are the circle number $\pi $ and the Euler number $e$ ( written in blocks of in the following(

$\pi =3.14159\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}26535\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}89793\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}23846\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}26433\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}83279\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}50288\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}41971\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}69399\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}37510$

$e=2.71828\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}18284\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}59045\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}23536\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}02874\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}71352\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}66249\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}77572\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}47093\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}69995\dots \dots $ It is a characteristic feature of transcendental numbers, that they are limits of infinitely often repeated operations (additions, multiplications, formation of continued fraction, root taking, etc.) (see below).

Using the example of the number $\pi $ it will be demonstrated, how this transcendental number of high practical importance can be obtained as the limit of a sequence. We follow the famous train of thought due to Archimedes.

Using the theorem of Pythagoras and the formula for the area of a triangles with baseline $a$ and height $h$ i.e. $F=\frac{1}{2}ah$ the mathematicians and surveyors of Egypt and antiquity were able to reduce the area of an arbitrary surface that is bounded by straight lines to that of a square of the same area, whose length is given by a square root, i.e mostly an irrational number; even today the unit of surface area for arbitrarily bounded surfaces is still the “square meter”.

The “Quadrature of the circle”, as paradigm of calculating the area of a surface, that is bounded by curved lines, stayed however unsolved for a long time. Archimedes

The famous inventor and mathematician Archimedes (287 - 221 b.C.), who lived in the greek colony Syracuse in Sicily found a royal road to this end, that was only developed further nearly 2000 years later, and that represents the start of working with convergent, infinite sequences and with limits.

His method, which starts with a polygon that is inscribed or circumscribed to a circle (Figure 3.2), will be demonstrated in short due to its historical significance. He uses the theorem of Pythagoras, the formula for the area of a right triangle and symmetry considerations. From the above it follows, that the baselines of the triangles constituting the polygons with $n$ corners are given as a simple function of $n$ when doubling $n$. The following diagrams visualize the procedure. The first regular polygon, a yellow square, is circumscribed around the circle filled in gray, a second one, without colour is inscribed.

The inscribed polygon has a smaller area then the circumscribed one; The true value for the circle lies between the two. It is immediately evident, that halving the angle of division to obtain an octagon, which is blue filled, will make the differences smaller, and that this goes on with further doubling of $N$ (a polygon with 16 corners also is shown in red). The sketch shows the first steps of the calculation for inscribed polygons with ${2}^{N}$ corners, with $N>2$.

The square, with which the calculation starts, consists of 4 equal right triangles, whose cathetuses for the unit circle under consideration have length $1$. According to the theorem of Pythagoras the hypotenuse of each triangle has the length $\sqrt{2}$. The height ${h}_{4}$ is obtained via the theorem of Pythagoras using ${s}_{4}\u22152$ and the hypotenuse $1$ of the lower triangle. The distance ${z}_{4}$ is the difference to the radius $1$. The transition to the octagon again proceeds with the theorem of Pythagoras via ${s}_{4}\u22152$ and ${z}_{4}$. As the following calculation shows, this algorithm can be repeated in the same way in factors of $2$ towards a subdivision of the surface of the circle into ever smaller triangles. Thus this procedure results in recursion formulas, with which one can obtain the results of the $N$-th step from those of the $N-1$-th step. We give the results for the inscribed polygon with $n$ sides.

$$\begin{array}{c}\text{radius}r\text{}=1;\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\text{index}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}n={2}^{N},\text{with}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}N=2,\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}3,\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}4,\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}5,......\hfill \\ {s}_{4}=\sqrt{1+1}=\sqrt{2;}\phantom{\rule{0em}{0ex}};\phantom{\rule{0em}{0ex}}{h}_{4}=\sqrt{1-{\left(\frac{{s}_{4}}{2}\right)}^{2}}\phantom{\rule{0em}{0ex}};\phantom{\rule{0em}{0ex}}{z}_{4}=1-{h}_{4}=1-\sqrt{1-{\left(\frac{{s}_{4}}{2}\right)}^{2}}\hfill \\ {s}_{8}=\sqrt{{\left(\frac{{s}_{4}}{2}\right)}^{2}+{z}_{4}^{2}}=\sqrt{2}\sqrt{1-\sqrt{1-{\left(\frac{{s}_{4}}{2}\right)}^{2}}};\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{h}_{8}=\sqrt{1-{\left(\frac{{s}_{8}}{2}\right)}^{2}}=\frac{1}{\sqrt{2}}\sqrt{1+\sqrt{1-{\left(\frac{{s}_{4}}{2}\right)}^{2}}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\hfill \\ \hfill \end{array}$$

$$\begin{array}{c}\text{recursiveformula}\hfill \\ {s}_{N}^{i}=\sqrt{2}\sqrt{1-\sqrt{1-{\left(\frac{{s}_{N-1}^{i}}{2}\right)}^{2}}};\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{h}_{N}^{i}=\sqrt{1-{\left(\frac{{s}_{N}^{i}}{2}\right)}^{2}}\hfill \end{array})=\frac{1}{\sqrt{2}}\sqrt{1+\sqrt{1-{\left(\frac{{s}_{N-1}^{i}}{2}\right)}^{2}}}\phantom{\rule{0em}{0ex}}\text{circumference}U,\text{surface}F:\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\hfill \\ \text{}{U}_{N}^{i}=n{s}_{N}={2}^{N}{s}_{N}={2}^{N}\sqrt{2}\sqrt{1-\sqrt{1-{\left(\frac{{s}_{N-1}}{2}\right)}^{2}}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\hfill \\ {F}_{n}^{i}=n\frac{{s}_{N}{h}_{N}}{2}={2}^{N}\frac{{s}_{N}{h}_{N}}{2}=\frac{{2}^{N}}{2}\sqrt{2-{\left(\frac{{s}_{N-1}}{2}\right)}^{2}}\hfill \\ \hfill $$

In the following the equations have been written out starting from the inscribed square $n=4$ to the polygon with $n=64$ corners. On realizes the iterated characters of the repeated root taking of the side length $\sqrt{2}$ of the triangles making up the inscribed square.

$$\begin{array}{cccc}\hfill \begin{array}{c}{s}_{4}=\sqrt{2}\hfill \\ {s}_{8}=\sqrt{2-\sqrt{2}}\hfill \\ {s}_{16}=\sqrt{2-\sqrt{2+\sqrt{2}}}\hfill \\ \hfill \\ {s}_{32}=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}\hfill \\ \hfill \\ {s}_{64}=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}\hfill \\ \hfill \end{array}\hfill & \hfill \begin{array}{c}{h}_{4}=\frac{1}{2}\sqrt{2}\hfill \\ {h}_{8}=\frac{1}{2}\sqrt{2+\sqrt{2}}\hfill \\ {h}_{16}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}}\hfill \\ {h}_{32}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}\hfill \\ {h}_{64}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}\hfill \\ \hfill \end{array}\hfill & \hfill \begin{array}{c}{F}_{4}=\frac{4}{4}2\hfill \\ {F}_{8}=\frac{8}{4}\sqrt{2}\hfill \\ {F}_{16}=\frac{16}{4}\sqrt{2-\sqrt{2}}\hfill \\ \hfill \\ {F}_{32}=\frac{32}{4}\sqrt{2-\sqrt{2+\sqrt{2}}}\hfill \\ {F}_{64}=\frac{64}{4}\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}\hfill \\ \hfill \end{array}\hfill & \hfill \begin{array}{c}=2,0000\hfill \\ \hfill \\ =2,8284\hfill \\ \hfill \\ =3,0614\hfill \\ \hfill \\ =3,1214\hfill \\ \hfill \\ =3,1365\hfill \\ \hfill \end{array}\hfill \\ \hfill \hfill \end{array}$$

These formulae fascinate via their aesthetic symmetry!

With a simple spreadsheet this calculation that was rather tedious for Archimedes can be done quite quickly up to a high number of corners.

Then one sees, how quickly the surface areas of the inscribed and circumscribed polygon approximate the number $\pi \left(3.14159\dots \right)$ and the corresponding circumference approximates $2\pi $. In Figure 3.4 they are shown for the square up to the polygon with $8192$ corners (corresponding to $N=2$ to $N=13$). In addition the respective differences of the surface area from $\pi $ are given (logarithmic right hand scale).

Already for the $10$-th approximation (polygon with $1024$ corners) the difference is only $1{0}^{-5}$. Archimedes himself started with a hexagon and took the calculation up to a polygon with $92$ corners and obtained his value for the circular number of $3.141635$ (the symbol $\pi $ for this number was only introduced in the 18th century); we suggest that you retrace the calculation of Archimedes.