# Origin of Irrational Numbers and Their Approximations

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## Abstract

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## 1. Introduction

## 2. Sulbasutras

## 3. Aryabhata’s Method for Extracting Square and Cube Roots

## 4. Great Pyramid at Gizeh and Rhind Mathematical Papyrus

## 5. Babylonians Tablet YBC 7289

## 6. Shatapatha Brahmana

## 7. Pythagoreans (Followers of Pythagoras) Crisis of Incommensurability

## 8. Democritus of Abdera (around 460–362 BC, Greece)

## 9. Theodorus of Cyrene (about 431 BC, Libya, Greece)

## 10. Geometric Proof of Irrationality of $\sqrt{2}$

- The following inquisitive geometric proof of Apostol [23] (also for similar proofs see earlier books by Kiselev [24], and Conway and Guy [25]) is in line with the above proof. A circular arc with center at the uppermost vertex and radius equal to the vertical leg of the triangle intersects the hypotenuse at a point, from which a perpendicular to the hypotenuse is drawn to the horizontal leg (see Figure 9). Each line segment in the diagram has integer length, and the three segments with double tick marks have equal lengths. (Two of them are tangents to the circle from the same point). Therefore the smaller isosceles right triangle with hypotenuse on the horizontal base also has integer sides.

## 11. Eudoxus of Cnidus (around 400–347 BC, Greece)

## 12. Aristotle (around 384–322 BC, Greece)

## 13. Euclid of Alexandria (around 325–265 BC, Greece, Egypt)

- In the above proof we can ignore all geometric arguments, and directly proceed to algebraic Equation (18), where p and q are in its lowest term, and hence are of different parity. Then, showing that $\sqrt{2}$ is irrational is equivalent to proving that (18) is impossible. For this, the Website http://www.cut-the-knot.org/proofs/sq_root.shtmlcontains29proofs (accessed on 3 March 2021),
- Joseph Louis Lagrange (1736–1813, French, Italian) in his Lectures on Elementary Mathematics of 1898 argues that if p and q are in its lowest terms, then ${p}^{2}$ and ${q}^{2}$ are also in its lowest terms. Since fraction ${p}^{2}/{q}^{2}$ is built from the fraction $p/q$ it cannot be a whole number $2.$ A similar reasoning appeared in 1831 in the work of Augustus De Morgan (1806–1871, British-India).
- Whittaker and Watson in their book [26] of 1920, and later Gardner [27], and Laczkovich [28] in their books assume that in $\sqrt{2}=p/q$ the integer q is the smallest possible such number. Their main argument is essentially to use the equality ${(2q-p)}^{2}=2{(p-q)}^{2}$ which is true if and only if (18) holds. Thus, it follows that$$2\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{{(2q-p)}^{2}}{{(p-q)}^{2}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{or}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\sqrt{2}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{2q-p}{p-q},$$It is interesting to note that$$\sqrt{2}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}(\sqrt{2}+1)-1\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{\sqrt{2}-1}-1\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{p/q-1}-1\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{2q-p}{p-1}.$$
- Rademacher and Toeplitz in their book of 1957 ([29], Chapter 4) assert that (18) implies p is even, so q must be odd. However, the square of an even number is divisible by $4,$ which leads to conclude that q must be even. Thus, we have Aristotle type contradiction.

- By the FTOA, p and q can be factored uniquely into their prime factors, so let $p={p}_{1}{p}_{2}\cdots {p}_{r}$ and $q={q}_{1}{q}_{2}\cdots {q}_{s}.$ Putting this back in Equation (18), we get$${({p}_{1}{p}_{2}\cdots {p}_{r})}^{2}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}2{({q}_{1}{q}_{2}\cdots {q}_{s})}^{2},$$$${p}_{1}{p}_{1}{p}_{2}{p}_{2}\cdots {p}_{r}{p}_{r}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}2{q}_{1}{q}_{1}{q}_{2}{q}_{2}\cdots {q}_{s}{q}_{s}.\phantom{\rule{2.em}{0ex}}$$Now among the primes ${p}_{i}$ and ${q}_{i},$ the prime 2 may occur (it will occur if either p or q is even). If it does occur, it must appear an even number of times on the left side of Equation (19) (since each prime there appears twice), and an odd number of times on the right side (because 2 already appears there once). However, then we have a contradiction: since the factorization into primes is unique, the prime 2 cannot appear an even number of times on one side of the equation and an odd number on the other. Thus, Equation (18) is impossible.
- From the uniqueness of the factorization, one can argue directly that ${p}^{2}$ has even number of prime factors, whereas $2{q}^{2}$ has odd number of prime factors, which is absurd.
- Some of the above illustrations can be extended to prove the result: If $N\in I\phantom{\rule{-1.99997pt}{0ex}}N,$ then $\sqrt{N}$ is a rational number if and only if $\sqrt{N}$ is an integer.

- Dedekind in his proof assumed that if N is not a square of an integer, then there exists a positive integer $\lambda $ such that ${\lambda}^{2}<N<{(\lambda +1)}^{2}.$ Again, if N is rational, then there exist $p,q\in I\phantom{\rule{-1.99997pt}{0ex}}N$ such that ${p}^{2}-N{q}^{2}=0,$ where q is the least possible integer possessing the property that its square multiplied by N is the square of $p.$ Since $\lambda q<p<(\lambda +1)q,$ it follows that the integers $s=p-\lambda q$ and $t=Nq-\lambda p$ are positive, and we have ${t}^{2}-N{s}^{2}=({\lambda}^{2}-N)({p}^{2}-N{q}^{2})=0,$ which contradicts the assumption on $q.$
- On the Website https://www.quora.com/If-p-is-a-natural-number-but-not-a-perfect-nth-power-how-does-one-prove-that-the-nth-root-of-p-is-not-rational (accessed on 3 March 2021), Thomas Schürger (2019) has provided a very simple proof of the following general result: The kth, $k\in I\phantom{\rule{-1.99997pt}{0ex}}N,\phantom{\rule{3.33333pt}{0ex}}k\ge 2$ root of a nonnegative integer $N\ge 2$ is rational if and only if N is a perfect kth power. One direction of this statement is clearly true: the kth root of a kth power is rational. Let us prove the other direction via proof by contradiction. Let us assume that N is not a perfect kth power, and $\sqrt[k]{N}$ is rational, i.e.,$$\sqrt[k]{N}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{p}{q}$$$$\frac{N}{1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{{p}^{k}}{{q}^{k}}$$
- Some of the above arguments need slight modification to prove: If r and s are distinct primes, then $\sqrt{rs}$ and ${log}_{r}s$ are irrational. For example, to show ${log}_{r}s$ is irrational, we assume contrary, i.e., ${log}_{r}s=p/q,$ where $p,q\in I\phantom{\rule{-1.99997pt}{0ex}}N.$ We can assume that $\mathrm{gcd}(p,q)=1.$ Then ${r}^{p/q}=s$ and so ${\left({r}^{p/q}\right)}^{q}={s}^{q}.$ Therefore, ${r}^{p}={s}^{q}.$ Since $r|{r}^{p},$ it follows that $r|{s}^{q}$ and so $r|s,$ which is a contradiction.
- We shall follow Dov Jarden (1911-1986, Israel) work of 1953 to show that there exist irrational numbers a and b such that ${a}^{b}$ is rational. Consider the irrational numbers $a=b=\sqrt{2}.$ If the number ${a}^{b}={\sqrt{2}}^{\sqrt{2}}$ is rational, we are done. If ${\sqrt{2}}^{\sqrt{2}}$ is irrational, we consider the numbers $a={\sqrt{2}}^{\sqrt{2}}$ and $b=\sqrt{2}$ so that ${a}^{b}={\left({\sqrt{2}}^{\sqrt{2}}\right)}^{\sqrt{2}}={\sqrt{2}}^{\sqrt{2}\sqrt{2}}={\sqrt{2}}^{2}=2$ is rational. Note that in this proof we could not find irrational numbers a and b such that ${a}^{b}$ is rational.

## 14. Archimedes of Syracuse (287–212 BC, Greece)

- Again on the Website https://mathpages.com/home/kmath038/kmath038.htm (accessed on 3 March 2021) a clever observation is that if a is a bound (upper or lower) of $\sqrt{3},$ then $(5a+9)/(3a+5)$ is a closure bound on the opposite side (lower or upper). This suggest the iterative scheme$${a}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{5{a}_{n}+9}{3{a}_{n}+5},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{a}_{0}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{5}{3},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}n\ge 0.\phantom{\rule{2.em}{0ex}}$$Since$${a}_{n+1}^{2}-3\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}{\left(\frac{5{a}_{n}+9}{3{a}_{n}+5}\right)}^{2}-3\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}-\frac{2({a}_{n}^{2}-3)}{9{a}_{n}^{2}+30{a}_{n}+25}\phantom{\rule{3.33333pt}{0ex}}\simeq \phantom{\rule{3.33333pt}{0ex}}-\frac{{a}_{n}^{2}-3}{51.98\cdots}$$$${a}_{1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{26}{15},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{a}_{2}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{265}{153},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{a}_{3}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1351}{780},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{a}_{4}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{13775}{7953},\cdots .$$Thus, ${a}_{2}$ and ${a}_{3},$ respectively, give the lower and upper Archimedes bounds of $\sqrt{3}.$

- On the same Website and on the Website https://www.mathpages.com/home/kmath190/kmath190.htm (accessed on 3 March 2021), following Babylonians’ the basic ladder rule for generating a sequence of integers to yield the square root of a number N the following recurrence relation has been discussed$${s}_{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\left(2a\right){s}_{n-1}+(N-{a}^{2}){s}_{n-2},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}n\ge 2\phantom{\rule{2.em}{0ex}}$$$${q}^{2}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\left(2a\right)q+(N-{a}^{2}),$$$$\frac{{s}_{n+2}}{{s}_{n+1}}-a\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{\left(2a\right){s}_{n+1}+(N-{a}^{2}){s}_{n}}{{s}_{n+1}}-a\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}a+(N-{a}^{2})\frac{{s}_{n}}{{s}_{n+1}}\phantom{\rule{3.33333pt}{0ex}}<\phantom{\rule{3.33333pt}{0ex}}a+(N-{a}^{2})\frac{1}{\sqrt{N}+a}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sqrt{N}.$$Similarly, for the second sequence it suffices to show that if $a+(N-{a}^{2})({s}_{n}/{s}_{n+1})>\sqrt{N},$ which is the same as $({s}_{n}/{s}_{n+1})>1/(\sqrt{N}+a),$ then $a+(N-{a}^{2})({s}_{n+1}/{s}_{n+2})<\sqrt{N}.$ However, this is the same as proving $({s}_{n+2}/{s}_{n+1})>\sqrt{N}+a.$ Now from (31) it follows that$$\frac{{s}_{n+2}}{{s}_{n+1}}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}2a+(N-{a}^{2})\frac{{s}_{n}}{{s}_{n+1}}\phantom{\rule{3.33333pt}{0ex}}>\phantom{\rule{3.33333pt}{0ex}}2a+(N-{a}^{2})\frac{1}{(\sqrt{N}+a)}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}2a+(\sqrt{N}-a)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}(\sqrt{N}+a).$$

- Davies in his preprint [32] combined a simple proposition:$$If\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\frac{v}{u}\phantom{\rule{3.33333pt}{0ex}}<\phantom{\rule{3.33333pt}{0ex}}\frac{y}{x}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}then\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\frac{v}{u}\phantom{\rule{3.33333pt}{0ex}}<\phantom{\rule{3.33333pt}{0ex}}\frac{v+y}{u+x}\phantom{\rule{3.33333pt}{0ex}}<\phantom{\rule{3.33333pt}{0ex}}\frac{y}{x}$$$$\begin{array}{cccccccc}{\displaystyle \left(\frac{1}{1},\frac{2}{1}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{3}{2},\frac{2}{1}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{5}{3},\frac{2}{1}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{5}{3},\frac{7}{4}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{12}{7},\frac{7}{4}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{19}{11},\frac{7}{4}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{19}{11},\frac{26}{15}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{45}{26},\frac{26}{15}\right)}\hfill \\ \phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{-8.53581pt}{0ex}}\hfill \\ {\displaystyle \left(\frac{71}{41},\frac{26}{15}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{71}{41},\frac{97}{56}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{168}{97},\frac{97}{56}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left({\frac{265}{153}}^{*},\frac{97}{56}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{265}{153},\frac{362}{209}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{627}{362},\frac{362}{209}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{989}{571},\frac{362}{209}\right)\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}}\hfill & {\displaystyle \phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\left(\frac{989}{571},{\frac{1351}{780}}^{*}\right)}\hfill \end{array}$$While the above list of pairs of approximations of $\sqrt{3}$ contain lower and upper bounds of Archimedes, an extended algorithm for the computation of $\sqrt{N}$ for an arbitrary integer N has no merit.
- For the lower bound on the Website https://math.stackexchange.com/questions/894862/archimedes-approximation-of-square-roots (accessed on 3 March 2021), posted in 2015, the secant method has been suggested. Recall from the standard numerical analysis text books, the secant method for finding a simple root ${a}^{*}$ of the equation $f\left(x\right)=0$ is$${a}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{{a}_{n-1}f\left({a}_{n}\right)-{a}_{n}f\left({a}_{n-1}\right)}{f\left({a}_{n}\right)-f\left({a}_{n-1}\right)},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}n\ge 1\phantom{\rule{2.em}{0ex}}$$$${a}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{{a}_{n-1}{a}_{n}+N}{{a}_{n-1}+{a}_{n}},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}n\ge 1.\phantom{\rule{2.em}{0ex}}$$It is interesting to note that if in (35), we take ${a}_{n-1}={a}_{n},$ then it is the same as (11). Applying (35) with $N=3,\phantom{\rule{3.33333pt}{0ex}}{a}_{0}=5/3$ (which is less than $\sqrt{3}$), and ${a}_{1}=26/15$ (which is greater than $\sqrt{3}$), see Table 1, we immediately get ${a}_{2}=265/153,$ which is the lower bound in (22). From (35), we also compute ${a}_{3}=13775/7953\simeq 1.73205079844,$ which is a better lower bound than in (22).
- For the lower bound in (22) on the Website https://hsm.stackexchange.com/questions/771/what-is-so-mysterious-about-archimedes-approximation-of-sqrt-3 (accessed on 3 March 2021), posted in 2015, following Babylonians tables are constructed for ${n}^{2}$ and $3{n}^{2},\phantom{\rule{3.33333pt}{0ex}}n\ge 1$ and it was noticed that $70225={\left(265\right)}^{2}\simeq 3{\left(153\right)}^{2}=20227.$
- Upper bound in the inequality (22) is the same as obtained in Sulbasutras, see (5). Unfortunately, historians never found place to write this fact.
- For two positive numbers $a,b$ three classical Pythagorean means are the arithmetic mean (AM) $=(a+b)/2,$ the geometric mean (GM) $=\sqrt{ab},$ and the harmonic mean (HM) $=2ab/(a+b).$ These means were studied with proportions by Pythagoreans and later generations of Greek mathematicians because of their importance in geometry and music. The following inequalities and equality between these means are straightforward and well-known in the literature$$min\{a,b\}\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}HM\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}GM\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}AM\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}max\{a,b\},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{GM}^{2}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}AM\xb7HM.$$Based on the above inequalities, we have the following three algorithms HMA, GMA, and AMA$${c}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{2{a}_{n}{b}_{n}}{{a}_{n}+{b}_{n}},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{b}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sqrt{{a}_{n}{b}_{n}},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{a}_{n+1}=\frac{{a}_{n}+{b}_{n}}{2},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}n\ge 0$$$${a}_{n+1}-{a}_{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{{a}_{n}+{b}_{n}}{2}-{a}_{n}=\frac{{b}_{n}-{a}_{n}}{2}\phantom{\rule{3.33333pt}{0ex}}\le \phantom{\rule{3.33333pt}{0ex}}0,$$$${b}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sqrt{{a}_{n}{b}_{n}}\phantom{\rule{3.33333pt}{0ex}}\ge \phantom{\rule{3.33333pt}{0ex}}\sqrt{{b}_{n}{b}_{n}}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}{b}_{n},$$$${c}_{n+1}-{c}_{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{2{a}_{n}{b}_{n}}{{a}_{n}+{b}_{n}}-{c}_{n}\ge \frac{2{a}_{n}{b}_{n}}{2{a}_{n}}-{c}_{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}{b}_{n}-{c}_{n}\ge 0,$$$${c}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{2{a}_{n}N}{{a}_{n}^{2}+N},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{b}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sqrt{N},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{a}_{n+1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{2}\left({a}_{n}+\frac{N}{{a}_{n}}\right),\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}n\ge 0$$Here ${a}_{0}$ is some positive rational number. Clearly, AMA is the same as (11). We note that the equation $(a+3/a)/2=1351/780$ gives $a=26/15,$ and $(a+3/a)/2=26/15$ holds for $a=5/3.$ Thus, if we employ AMA for $N=3$ with ${a}_{0}=5/3$ (which is a reasonable choice, see (5)) then ${a}_{2}$ is the same as the upper bound of the inequality (22). We further note that the equation $6a/({a}^{2}+3)=265/153,$ which is the same as $265{a}^{2}-918a+795=0$ has no rational roots, and hence lower bound of (22) cannot be obtained from HMA for $N=3.$
- A proof of (22) based on very simple inequalities is as follows:$$\frac{1351}{780}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{15}\left(26-\frac{1}{52}\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{15}\sqrt{{26}^{2}-1+\frac{1}{{52}^{2}}}\phantom{\rule{3.33333pt}{0ex}}>\phantom{\rule{3.33333pt}{0ex}}\frac{1}{15}\sqrt{{26}^{2}-1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sqrt{3}$$$$\frac{265}{153}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{15}\left(26-\frac{1}{51}\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{15}\sqrt{{26}^{2}-1-\frac{50}{{51}^{2}}}\phantom{\rule{3.33333pt}{0ex}}<\phantom{\rule{3.33333pt}{0ex}}\frac{1}{15}\sqrt{{26}^{2}-1}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sqrt{3}.$$

## 15. Apollonius of Perga (around 262–200 BC, Greece)

## 16. Bakhshali Manuscript (about 200 BC)

## 17. Marcus Vitruvius Pollio (about 75–15 BC, Italy)

## 18. Theon of Smyrna (about 70–135 AD, Turkey-Greece)

- For $n\ge 0,$ explicit solutions of the system (45) are$${a}_{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{2\sqrt{2}}\left[{(1+\sqrt{2})}^{n+1}-{(1-\sqrt{2})}^{n+1}\right]\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\mathrm{and}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{b}_{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{2}\left[{(1+\sqrt{2})}^{n+1}+{(1-\sqrt{2})}^{n+1}\right].$$Now we define ${T}_{0}=0,\phantom{\rule{3.33333pt}{0ex}}{T}_{1}=1,\phantom{\rule{3.33333pt}{0ex}}{T}_{n}={a}_{n-1}{b}_{n-1},\phantom{\rule{3.33333pt}{0ex}}n\ge 2$ (recall ${a}_{n-1},{b}_{n-1},\phantom{\rule{3.33333pt}{0ex}}n\ge 2,$ respectively, are the denominator and numerator of column 2 in Table 3) then from the above expressions it follows that$${T}_{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{4\sqrt{2}}\left[{(1+\sqrt{2})}^{2n}-{(1-\sqrt{2})}^{2n}\right]\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{4\sqrt{2}}\left[{(3+2\sqrt{2})}^{n}-{(3-2\sqrt{2})}^{n}\right],\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}n\ge 1$$$${T}_{n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}6{T}_{n-1}-{T}_{n-2},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{T}_{0}=0,\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{T}_{1}=1.\phantom{\rule{2.em}{0ex}}$$In 1778, Euler showed that$${T}_{n}^{2}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{32}{\left[{(3+2\sqrt{2})}^{n}-{(3-2\sqrt{2})}^{n}\right]}^{2},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}n\ge 1$$$$\begin{array}{ccccccccc}{T}_{1}^{2}& {T}_{2}^{2}& {T}_{3}^{2}& {T}_{4}^{2}& {T}_{5}^{2}& {T}_{6}^{2}& {T}_{7}^{2}& {T}_{8}^{2}& \cdots \\ \phantom{\rule{-11.38109pt}{0ex}}\\ {1}^{2}& {6}^{2}& {35}^{2}& {204}^{2}& {1189}^{2}& {6930}^{2}& {40391}^{2}& {235416}^{2}& \cdots \\ \phantom{\rule{-11.38109pt}{0ex}}\\ {t}_{1}& {t}_{8}& {t}_{49}& {t}_{288}& {t}_{1681}& {t}_{9800}& {t}_{57121}& {t}_{332928}& \cdots \end{array}$$For more details on this work see Website https://en.wikipedia.org/wiki/Square_triangular_number (accessed on 3 March 2021).

## 19. Liu Hui (around 220–280, China)

## 20. Bhaskara II or Bhaskaracharya (Working 486, India)

## 21. Ab$\overline{\mathrm{u}}$ K$\overline{\mathrm{a}}$mil, Shuj${\overline{\mathrm{a}}}^{\prime}$ ibn Aslam ibn Muammad ibn Shuj${\overline{\mathrm{a}}}^{\prime}$ (850–930, Egypt)

## 22. Abu Abd Allah Muhammad ibn Isa Al-Mahani (about 820–880, Iran-Iraq)

## 23. Abu Ja’far al-Khazin (900–971, Iran)

## 24. Al-Hashimi (10th Century, Iraq)

## 25. Abu Abdallah al-Hassan ibn al-Baghdadi (10th Century, Iraq)

## 26. Omar Khayyám (1048–1131, Iran)

## 27. Nilakanthan Somayaji (around 1444–1544, India)

## 28. Nicolas Chuquet (around 1445–1488, France)

## 29. Michael Stifel (1486–1567, Germany)

## 30. Guillaume Gosselin (1536–1600, France)

## 31. Zhu Zaiyu (1536–1611, China)

## 32. Francois Viéte (1540–1603, France)

## 33. Simon Stevin (1548–1620, Elgium)

## 34. John Wallis (1616–1703, Britain)

## 35. Jacob Bernoulli: Introduction of e

## 36. Continued Fractions

## 37. Leonhard Euler: Irrationality of e

## 38. Johann Heinrich Lambert: Irrationality of $\pi $

## 39. Joseph Liouville (1809–1882, France)

## 40. Karl Theodor Wilhelm Weierstrass: Sequential Definition of Irrationality

## 41. Gustav Conrad Bauer (1820–1906, Germany)

## 42. Charles Hermite: Transcendence of e

## 43. Leopold Kronecker (1823–1891, Poland-Germany)

## 44. Julius Wilhelm Richard Dedekind’s Cut

## 45. Paul Gustav Heinrich Bachmann (1837–1920)

## 46. Georg Cantor (1845–1918, Russia-Germany)

## 47. Carl Louis Ferdinand von Lindemann: Transcendence of $\pi $ (1852–1939, Germany)

## 48. Friedrich Engel (1861–1941, Germany)

## 49. David Hilbert (1862–1943, Germany)

## 50. Godfrey Harold Hardy: Statistical Distribution of the Digits

## 51. Aleksander Osipovich Gelfond (1906–1968, Russia)

## 52. Roger Apéry (1916–1994, Greek-French)

## 53. More on Numbers

## 54. Conclusions

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

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n | $\mathit{N}=2$ | $\mathit{N}=2$ | $\mathit{N}=3$ | $\mathit{N}=3$ | $\mathit{N}=3$ |
---|---|---|---|---|---|

0 | $\frac{4}{3}$ | $\frac{3}{2}$ | 2 | $\frac{5}{3}$ | $\frac{3}{2}$ |

1 | $\frac{17}{12}$ | $\frac{17}{12}$ | $\frac{7}{4}$ | $\frac{26}{15}$ | $\frac{7}{4}$ |

2 | $\frac{577}{408}$ | $\frac{577}{408}$ | $\frac{97}{56}$ | $\frac{1351}{780}$ | $\frac{97}{56}$ |

3 | $\frac{665857}{470832}$ | $\frac{665857}{470832}$ | $\frac{18817}{10864}$ | $\frac{3650401}{2107560}$ | $\frac{18817}{10864}$ |

Algorithm (25) | Algorithm (26) | |||
---|---|---|---|---|

$\mathit{n}$ | $\mathit{N}=\mathbf{2}$ | $\mathit{N}=\mathbf{3}$ | $\mathit{N}=\mathbf{2}$ | $\mathit{N}=\mathbf{3}$ |

0 | $\frac{4}{3}$ | $\frac{3}{2}$ | $\frac{5}{3}$ | $2$ |

1 | $\frac{46}{33}$ | $\frac{27}{16}$ | $\frac{28}{21}$ | $\frac{5}{3}$ |

2 | $\frac{5812}{4125}$ | $\frac{1929}{1120}$ | $\frac{1078}{735}$ | $\frac{37}{21}$ |

3 | $\frac{91785094}{64964625}$ | $\frac{9644721}{5575360}$ | $\frac{1450204}{1044435}$ | $\frac{1915}{1113}$ |

N = 2 | N = 3 | |||
---|---|---|---|---|

$\mathit{n}$ | $({\mathit{s}}_{\mathit{n}+\mathbf{1}}/{\mathit{s}}_{\mathit{n}})-\mathbf{1}$ | $\mathbf{1}+({\mathit{s}}_{\mathit{n}}/{\mathit{s}}_{\mathit{n}+\mathbf{1}})$ | $({\mathit{s}}_{\mathit{n}+\mathbf{1}}/{\mathit{s}}_{\mathit{n}})-\mathbf{1}$ | $\mathbf{1}+\mathbf{2}({\mathit{s}}_{\mathit{n}}/{\mathit{s}}_{\mathit{n}+\mathbf{1}})$ |

2 | $\frac{3}{2}$ * | $\frac{7}{5}$ * | $2$ | $\frac{5}{3}$ |

3 | $\frac{7}{5}$ | $\frac{17}{12}$ | $\frac{5}{3}$ | $\frac{7}{4}$ |

4 | $\frac{17}{12}$ * | $\frac{41}{29}$ * | $\frac{7}{4}$ | $\frac{19}{11}$ |

5 | $\frac{41}{29}$ | $\frac{99}{70}$ | $\frac{19}{11}$ | $\frac{26}{15}$ |

6 | $\frac{99}{70}$ * | $\frac{239}{169}$ * | $\frac{26}{15}$ | $\frac{71}{41}$ |

7 | $\frac{239}{169}$ | $\frac{577}{408}$ | $\frac{71}{41}$ | $\frac{97}{56}$ |

8 | $\frac{577}{408}$ * | $\frac{1393}{985}$ * | $\frac{97}{56}$ | $\frac{265}{153}$ |

9 | $\frac{1393}{985}$ | $\frac{3363}{2378}$ | $\frac{265}{153}$ | $\frac{362}{209}$ |

10 | $\frac{3363}{2378}$ * | $\frac{8119}{5741}$ * | $\frac{362}{209}$ | $\frac{989}{571}$ |

11 | $\frac{8119}{5741}$ | $\frac{19601}{13860}$ | $\frac{989}{571}$ | $\frac{1351}{780}$ |

12 | $\frac{19601}{13860}$ * | $\frac{47321}{33461}$ * | $\frac{1351}{780}$ | $\frac{3691}{2131}$ |

13 | $\frac{47321}{33461}$ | $\frac{114243}{80782}$ | $\frac{3691}{2131}$ | $\frac{5042}{2911}$ |

14 | $\frac{114243}{80782}$ * | $\frac{275807}{195025}$ * | $\frac{5042}{2911}$ | $\frac{13775}{7953}$ |

15 | $\frac{275807}{195025}$ | $\frac{665857}{470832}$ | $\frac{13775}{7953}$ | $\frac{18817}{10864}$ |

16 | $\frac{665857}{470832}$ * | $\frac{1607521}{1136689}$ * | $\frac{18817}{10864}$ | $\frac{51409}{29681}$ |

n | $\mathit{N}=2$ | $\mathit{N}=3$ | $\mathit{N}=41$ | $\mathit{N}=105$ | $\mathit{N}=481$ |
---|---|---|---|---|---|

0 | $1$ | $1$ | $6$ | $10$ | $21$ |

1 | $\frac{17}{12}$ | $\frac{7}{4}$ | $\frac{11833}{1848}$ | $\frac{3361}{328}$ | $\frac{1698568}{77448}$ |

2 | $\frac{665857}{470832}$ | $\frac{18817}{10864}$ | ${A}_{1}$ | ${A}_{2}$ | ${A}_{3}$ |

0 | $2$ | $2$ | $7$ | $11$ | $22$ |

1 | $\frac{17}{12}$ | $\frac{97}{56}$ | $\frac{2017}{315}$ | $\frac{12737}{1243}$ | $\frac{1862441}{84920}$ |

2 | $\frac{665857}{470832}$ | $\frac{708158977}{408855776}$ | ${A}_{4}$ | ${A}_{5}$ | ${A}_{6}$ |

$\mathit{N}=2$ | $\mathit{N}=3$ | $\mathit{N}=5$ | $\mathit{N}=7$ | |
---|---|---|---|---|

$\mathit{n}$ | $({\mathit{a}}_{\mathbf{0}},{\mathit{b}}_{\mathbf{0}})=(\mathbf{2},\mathbf{3})$ | $({\mathit{a}}_{\mathbf{0}},{\mathit{b}}_{\mathbf{0}})=(\mathbf{1},\mathbf{2})$ | $({\mathit{a}}_{\mathbf{0}},{\mathit{b}}_{\mathbf{0}})=(\mathbf{4},\mathbf{9})$ | $({\mathit{a}}_{\mathbf{0}},{\mathit{b}}_{\mathbf{0}})=(\mathbf{3},\mathbf{8})$ |

0 | $\frac{3}{2}$ | $\frac{2}{1}$ | $\frac{9}{4}$ | $\frac{8}{3}$ |

1 | $\frac{17}{12}$ | $\frac{7}{4}$ | $\frac{161}{72}$ | $\frac{127}{48}$ |

2 | $\frac{577}{408}$ | $\frac{97}{56}$ | $\frac{51841}{23184}$ | $\frac{32257}{12192}$ |

3 | $\frac{665857}{470832}$ | $\frac{18817}{10864}$ | $\frac{5374978561}{2403763488}$ | $\frac{2081028097}{786554688}$ |

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Agarwal, R.P.; Agarwal, H.
Origin of Irrational Numbers and Their Approximations. *Computation* **2021**, *9*, 29.
https://doi.org/10.3390/computation9030029

**AMA Style**

Agarwal RP, Agarwal H.
Origin of Irrational Numbers and Their Approximations. *Computation*. 2021; 9(3):29.
https://doi.org/10.3390/computation9030029

**Chicago/Turabian Style**

Agarwal, Ravi P., and Hans Agarwal.
2021. "Origin of Irrational Numbers and Their Approximations" *Computation* 9, no. 3: 29.
https://doi.org/10.3390/computation9030029