# The Classic Greek Ladder and Newton's Method

### Introduction

For many students in early mathematics courses, their familiarity with approximations is limited to $$\sqrt{2}\approx{1.414}$$, $$\sqrt{3}\approx{1.732}$$, $$\pi\approx{\frac{22}{7}}$$, and maybe a few more. But a topic of number theory, Diophantine Approximations (honoring Diophantus, a mathematician of Alexandria who lived circa 207 - 291 AD and wrote books called Arithmetica), involves approximating irrational numbers by ordinary reduced fractions. One of the approximation "tools" of ancient mathematicians is a construct called Greek ladders. Maybe Greek ladders will ignite your interest in approximations by ordinary fractions.

The phrase "classic Greek Ladder" is taken here to mean the infinite array that begins $\begin{array}{cc} 1 & 1\\2 & 3\\5 & 7\\12 & 17\\29 & 41\\70 & 99\end{array}$ where each rung $$\langle a \quad b \rangle$$ is followed by $$\langle a+b \quad 2a+b \rangle$$ and the approximations to $$\sqrt{2}$$ are the fractions $$b/a$$ .

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Recall  that the classic Greek Ladder starts with first rung $$\langle 1 \quad 1 \rangle$$, each rung $$\langle a \quad b \rangle$$ is followed by $$\langle a+b \quad 2a+b \rangle$$, and the approximations to $$\sqrt{2}$$ are the fractions $$b/a$$.

I first learned about the classic Greek Ladder in [5]. This Greek Ladder is old, its introduction (invention?) often accredited to the mathematician Theon of Smyrna (circa 70-135 AD) [6], but there is reason to believe that it --- or equivalent constructs --- may well have been known much earlier, for on the first page of [1, Chap. XII], we find, "There appeared in India and Greece as early as 400 B.C. approximations $$a/b$$ to $${\sqrt{2}}$$  . . ." and that "Baudhâyana, the Hindu author of the oldest of the works, Sulva-sutras, gave the approximations $$17/12$$ and $$577/408$$ to $$\sqrt{2}$$ ." The first of these is from rung four of the above ladder, the second from rung eight, as you will see by extending that ladder another three rungs. (The Greek Ladder for $$\sqrt{2}$$ can begin with any two numbers, not both zero, for the first rung, but the classic ladder begins as shown.)

This ladder also arises from a "vertical" recursive definition: \begin{align*} a_{1} & =1\text{, }a_{2}=2\text{, and for }n>2\text{, }a_{n}=2a_{n-1}+a_{n-2}\\b_{1} & =1\text{, }b_{2}=3\text{, and for }n>2\text{, }b_{n}=2b_{n-1}+b_{n-2}.\end{align*} You can check that this definition arises simply by examining an arbitrary rung $$\langle a\quad b\rangle$$ and its two successors, using the first definition: $\begin{array}{cc} a & b\\ a+b & 2a+b\\ 3a+2b & 4a+3b \end{array}$ The second definition will be used later to construct a general term of the classic ladder.

To see that $$\sqrt{2}$$ is in fact the limit of the sequence $\frac{1}{1},\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\frac{99}{70},\dots$ look instead at the sequence of their squares: $\frac{1}{1},\frac{9}{4},\frac{49}{25},\frac{289}{144},\frac{1681}{841},\frac{9801}{4900},\dots$ which can also be written as $2-\frac{1}{1^{2}},2+\frac{1}{2^{2}},2-\frac{1}{5^{2}},2+\frac{1}{12^{2}},2-\frac{1}{29^{2}},2+\frac{1}{70^{2}},\dots$ Now the limit of the sequence $-\frac{1}{1^{2}},\frac{1}{2^{2}},-\frac{1}{5^{2}},\frac{1}{12^{2}},-\frac{1}{29^{2}},\frac{1}{70^{2}},\dots$ is clearly zero since the denominators are --- by the "vertical" definition for the numbers $$a_{n}$$ -- increasing without bound. Thus, the sequence of squares has limit $$2$$, so the original sequence of fractions from the Greek Ladder has limit $$\sqrt{2}$$.

To assess Diophantine approximations of quadratic surds, I devised a "rating" system using the symbol $$(P,n)$$ to mean that the reduced fraction $$\frac{b}{a}$$ satisfies $$b^{2}=Da^{2}+P$$, where $$\frac{b}{a}$$ with an $$n$$-digit denominator approximates $$\sqrt{D}$$. For a given $$D$$, $${\left(\pm1,n\right)}$$ is a "best approximation." (The letter $$P$$ is a nod to John Pell [1, Chap. XII] and [2].)  You can see how "good" the classic $$\sqrt{2}$$ ladder is, in that each of its approximations is a best approximation. In [7], you learn that the Greek Ladder for $${\sqrt{k}}$$ is constructed by following each rung $$\langle a\quad b\rangle$$ by the rung $$\langle a+b\quad ka+b\rangle$$. Using this fact, here are some Greek Ladder approximations and their ratings that you can verify: $\begin{array}{ccccc} \sqrt{3}\approx\frac{19}{11} & & \sqrt{5}\approx\frac{11}{5} & & \sqrt{6}\approx\frac{73}{28}\\ \text{rating }\left(-2,2\right) & & \text{rating }\left(-4,1\right) & & \text{rating }\left(625,2\right) \end{array}$ What is the rating of $$1.414$$ as an approximation for $${\sqrt{2}}$$? What about the rating of $$1.732$$ as an approximation for $$\sqrt{3}$$?

This rating system is easily extended to cubic and higher surds. Also, [7] shows that approximations can frequently be improved by use of the Farey mean $$\frac{a+c}{b+d}$$ of the two common fractions $$\frac{a}{b}$$ and $$\frac{c}{d}$$. For example, the Farey mean of $$\frac{7}{5}$$ and $$\frac{17}{12}$$ is $$\frac{24}{17} \approx1.411765$$.

### Comparison with Newton's Method

For this section, we need, for reference, to repeat and extend the classic ladder, this time with rungs numbered in bold:  $\begin{array}{ccccccccccccccc} \mathbf{1} & 1 & 1 & & \mathbf{5} & 29 & 41 & & \mathbf{9} & 985 & 1393 & & \mathbf{13} & 33461 & 47321\\ \mathbf{2} & 2 & 3 & & \mathbf{6} & 70 & 99 & & \mathbf{10} & 2378 & 3363 & & \mathbf{14} & 80782 & 114243\\ \mathbf{3} & 5 & 7 & & \mathbf{7} & 169 & 239 & & \mathbf{11} & 5741 & 8119 & & \mathbf{15} & 195025 & 275807\\ \mathbf{4} & 12 & 17 & & \mathbf{8} & 408 & 577 & & \mathbf{12} & 13860 & 19601 & & \mathbf{16} & 470832 & 665857 \end{array}$

A standard topic in calculus is Newton's Method for approximating roots, so it is natural to ask how the classic Greek Ladder compares with Newton's Method for approximating $$\sqrt{2}$$. To that end, we let $$s\left( x\right) =x^{2}-2$$, so then $$s^{\prime}\left( x\right) =2x$$, and the Newton scheme is to start with an estimate $$x_{0}$$ and compute for each $$x_{n}$$ the next estimate $$x_{n+1}$$, where $x_{n+1}=x_{n}-\frac{s\left( x_{n}\right) }{s^{\prime}\left( x_{n}\right) }$ The classic Greek Ladder begins with the first estimate of $$\frac{1}{1}$$, so
for comparison, we choose the corresponding Newton estimate as $$x_{0}=\frac {1}{1}$$, and this is what ensues:  $\begin{array}{lll} x_{1}=\frac{1}{1}-\frac{\left( \frac{1}{1}\right) ^{2}-2}{2\times\frac{1}{1}}=\frac{3}{2}\quad\text{from Rung 2} & & x_{3}=\frac{17}{12}-\frac{\left( \frac{17}{12}\right) ^{2}-2}{2\times\frac{17}{12}}=\frac{577}{408}\quad\text{from Rung 8}\\x_{2}=\frac{3}{2}-\frac{\left( \frac{3}{2}\right) ^{2}-2}{2\times\frac{3}{2}}=\frac{17}{12}\quad\text{from Rung 4} & & x_{4}=\frac{577}{408} -\frac{\left( \frac{577}{408}\right) ^{2}-2}{2\times\frac{577}{408}} =\frac{665857}{470832}\quad\text{from Rung 16}\end{array}$ So the $$n$$-th Newton iterate equals the $$2^{n}$$-th Greek Ladder approximation for $$n=1$$ to $$n=4$$. The next Newton iterates comport with the approximations from rung $$32$$ of the ladder, which is $$\langle627013566048\quad 886731088897\rangle$$; the $$64$$-th rung; and on up to rungs too wide for a page here. (These computations were obtained by the software Scientific WorkPlace, and if you ask it to "Evaluate" an expression involving fractions, it gives fractional output. "Evaluate Numerically" gives decimal estimates as output.) In any case, it seems that more than a sufficiency of such calculations have now been made so as to provoke the statement of this

Theorem. The classical $$\sqrt{2}$$ Greek Ladder has the property that its rungs numbered $$2^{n}$$ yield fractions equal to the $$n^{th}$$ Newton's Method iterates $$x_{n}$$ approximating $$\sqrt{2}$$, with $$x_{0}=\frac{1}{1}$$.

Proof. From the recursive definition of the columns of the classical Greek Ladder, which is \begin{align*} a_{1} & =1\text{, }a_{2}=2\text{, and for }n>2\text{, }a_{n}=2a_{n-1} +a_{n-2};\\ b_{1} & =1\text{, }b_{2}=3\text{, and for }n>2\text{, }b_{n}=2b_{n-1}+b_{n-2}\end{align*}
we fix on the sequences $$\left\{ a_{i}\right\}$$ and $$\left\{b_{i}\right\}$$. But first, it needs to be shown that this recursive definition is equivalent to the original. To this end, look at the rungs beginning with $$\langle a_{n-2}\quad b_{n-2}\rangle$$ and use the original system progressing therefrom: $\begin{array}{cc} a_{n-2} & b_{n-2}\\ a_{n-1}=a_{n-2}+b_{n-2} & b_{n-1}=2a_{n-2}+b_{n-2}\\ a_{n}=3a_{n-2}+2b_{n-2} & b_{n}=4a_{n-2}+3b_{n-2} \end{array}$ We can see from this that $a_{n}=3a_{n-2}+2b_{n-2}=2\left( a_{n-2} +b_{n-2}\right) +a_{n-2}=2a_{n-1}+a_{n-2}$ which is the relationship in the recursive definition of $$a_{n}$$. A similar analysis gives $$b_{n}$$ in terms of $$b_{n-1}$$ and $$b_{n-2}$$.

The splendid paper by Hendel [4] gives various methods for deriving general terms of such recursive sequences. The section of that paper titled "Number Theory" seems most direct for the purposes at hand and gives $a_{n}=\frac{\left( 1+\sqrt{2}\right) ^{n}-\left( 1-\sqrt{2}\right) ^{n} }{2\sqrt{2}}\text{ and }b_{n}=\frac{\left( 1+\sqrt{2}\right) ^{n}+\left(1-\sqrt{2}\right) ^{n}}{2}.$ So the $$n$$-th rung of the classical Greek Ladder is ${\left\langle \frac{\left( 1+\sqrt{2}\right) ^{n}-\left( 1-\sqrt{2}\right)^{n}}{2\sqrt{2}}\qquad\frac{\left( 1+\sqrt{2}\right) ^{n}+\left( 1-\sqrt{2}\right) ^{n}}{2}\right\rangle}.$ The corresponding approximation for $$\sqrt{2}$$ deriving from the $$n$$-th rung is thus $\frac{\frac{\left( 1+\sqrt{2}\right) ^{n}+\left( 1-\sqrt{2}\right) ^{n}}{2}}{\frac{\left( 1+\sqrt{2}\right) ^{n}-\left( 1-\sqrt{2}\right) ^{n}}{2\sqrt{2}}}=\frac{\sqrt{2}\left( \sqrt{2}+1\right) ^{n}+\sqrt{2}\left(1-\sqrt{2}\right) ^{n}}{\left( \sqrt{2}+1\right) ^{n}-\left( 1-\sqrt{2}\right) ^{n}}$ and we want to see what happens when this fraction from the $$n$$-th rung is subjected to Newton's Method. Here it is: \begin{align*} & \frac{\sqrt{2}\left( \sqrt{2}+1\right) ^{n}+\sqrt{2}\left( 1-\sqrt {2}\right) ^{n}}{\left( \sqrt{2}+1\right) ^{n}-\left( 1-\sqrt{2}\right)^{n}}-\frac{\left( \frac{\sqrt{2}\left( \sqrt{2}+1\right) ^{n}+\sqrt{2}\left( 1-\sqrt{2}\right) ^{n}}{\left( \sqrt{2}+1\right)^{n}-\left(1-\sqrt{2}\right) ^{n}}\right) ^{2}-2}{2\times\frac{\sqrt{2}\left( \sqrt{2}+1\right) ^{n}+\sqrt{2}\left( 1-\sqrt{2}\right) ^{n}}{\left( \sqrt{2}+1\right) ^{n}-\left( 1-\sqrt{2}\right) ^{n}}}\\ & =\frac{\sqrt{2}\left[ \left( 1+\sqrt{2}\right) ^{2n}+\left( 1-\sqrt{2}\right) ^{2n}\right] }{\left( 1+\sqrt{2}\right) ^{2n}-\left(1-\sqrt{2}\right) ^{2n}}\end{align*} which comes from the $$2n$$-th rung, $\left\langle \frac{\left( 1+\sqrt{2}\right) ^{2n}-\left( 1-\sqrt{2}\right)^{2n}}{2\sqrt{2}}\qquad\frac{\left( 1+\sqrt{2}\right) ^{2n}+\left(1-\sqrt{2}\right) ^{2n}}{2}\right\rangle,$ whence the doubling pattern of the theorem is established.

It is implicit in this proof that the doubling takes place for any rung, so the process can begin by starting not just with $$\langle1\quad1\rangle$$ but with any rung whatsoever.

### Extending the Connection

What about Newton's Method and the $$\sqrt{3}$$ Greek Ladder? From [7], the standard version of this ladder begins with rung $$\langle1\quad1\rangle$$, each rung is followed by $$\langle a+b\quad3a+b\rangle$$, and each rung is "reduced." It begins $\begin{array}{ccc} 1 & 1 & \frac{1}{1}=1.00000\\ 1 & 2 & \frac{2}{1}=2.00000\\ 3 & 5 & \frac{5}{3}\approx1.66667\\ 4 & 7 & \frac{7}{4}=1.75000 \end{array} \qquad \begin{array}{ccc} 11 & 19 & \frac{19}{11}\approx1.72727\\ 15 & 26 & \frac{26}{15}\approx1.73333\\ 41 & 71 & \frac{71}{41}\approx1.73171\\ 56 & 97 & \frac{97}{56}\approx1.73214 \end{array}$ Here is what happens when Newton's Method is applied and iterated. $\begin{array}{c} x_{1}=\frac{1}{1}-\frac{\left( \frac{1}{1}\right) ^{2}-3}{2\times\frac{1}{1}}=\frac{2}{1}\quad\text{from Rung 2}\\ x_{2}=\frac{2}{1}-\frac{\left( \frac{2}{1}\right) ^{2}-3}{2\times2}=\frac{7}{4}\quad\text{from Rung 4}\\ x_{3}=\frac{7}{4}-\frac{\left( \frac{7}{4}\right) ^{2}-3}{2\times\frac{7}{4}}=\frac{97}{56}\quad\text{from Rung 8} \end{array}$ So the doubling pattern comports with the case for $$\sqrt{3}$$ to this point. Can that pattern be proved? Let us see.

The above version of the $$\sqrt{3}$$ ladder is standard, but finding a general term for the sequences $$\{a_{n}\}$$ and $$\{b_{n}\}$$ from the rungs $$\langle a_{n}\quad b_{n}\rangle$$ as presented is not immediate, since the rungs have been reduced when possible. Here is the $$\sqrt{3}$$ Greek Ladder without rung reductions: $\begin{array}{cc} 1 & 1\\ 2 & 4\\ 6 & 10\\ 16 & 28 \end{array} \qquad \begin{array}{cc} 44 & 76\\ 120 & 208\\ 328 & 568\\ 896 & 1552 \end{array}$ And now, a recursive definition of the rungs $$\langle a_{n}\quad b_{n}\rangle$$ of this equivalent version of the $$\sqrt{3}$$ Greek Ladder is suggested and given by \begin{align*} a_{1} & =1\text{, }a_{2}=2\text{, and for }n>2\text{, }a_{n}=2a_{n-1} +2a_{n-2};\\ b_{1} & =1\text{, }b_{2}=4\text{, and for }n>2\text{, }b_{n}=2b_{n-1}+2b_{n-2} \end{align*} (It is left to the reader to show that this is equivalent to the standard definition given earlier for the $$\sqrt{3}$$ ladder.) Using the same method of [4] again, the $$n$$-th rung $$\langle a_{n}\quad b_{n}\rangle$$ is $\left\langle \frac{\left( 1+\sqrt{3}\right) ^{n}-\left( 1-\sqrt{3}\right)^{n}}{2\sqrt{3}}\qquad\frac{\left( 1+\sqrt{3}\right) ^{n}+\left( 1-\sqrt{3}\right) ^{n}}{2}\right\rangle$

So let us see what happens when $$\frac{b_{n}}{a_{n}}$$ is chosen as an estimate in Newton's Method. The corresponding approximation for $$\sqrt{3}$$ deriving from the $$n$$-th rung simplifies to $\frac{b_{n}}{a_{n}} = \frac{\sqrt{3}\left( \sqrt{3}+1\right) ^{n}+\sqrt{3}\left( 1-\sqrt{3}\right) ^{n}}{\left( \sqrt{3}+1\right) ^{n}-\left( 1-\sqrt{3}\right) ^{n}}$ When this fraction is subjected to Newton's Method for $$\sqrt{3}$$, you can check that the result is $\frac{\sqrt{3}\left[ \left( 1+\sqrt{3}\right) ^{2n}+\left( 1-\sqrt{3}\right) ^{2n}\right] }{\left(1+\sqrt{3}\right) ^{2n}-\left(1-\sqrt{3}\right) ^{2n}}$ which is the reduced form of the fraction $$\frac{b_{2n}}{a_{2n}}$$ from the $$2n$$-th rung of the ladder, so the doubling is established for the $$\sqrt{3}$$ ladder.

Since the doubling pattern holds for the ladders of $$\sqrt{2}$$ and $$\sqrt{3}$$, it is natural to ask if it holds for the $$\sqrt{k}$$ Greek Ladder, in which each rung $$\langle a\quad b\rangle$$ is followed by the rung $$\langle a+b\quad ka+b\rangle$$ That ladder begins $\begin{array}{ccc} 1 & 1 & \frac{1}{1}\\ 2 & k+1 & \frac{k+1}{2}\\ k+3 & 3k+1 & \frac{3k+1}{k+3}\\ 4k+4 & k^{2}+6k+1 & \frac{k^{2}+6k+1}{4k+4}\\ k^{2}+10k+5 & 5k^{2}+10k+1 & \frac{5k^{2}+10k+1}{k^{2}+10k+5}\end{array}$ giving rise to the recursive definition \begin{align*} a_{1} & =1\text{, }a_{2}=2\text{, and for }n>2\text{, }a_{n}=2a_{n-1}+\left( k-1\right) a_{n-2};\\ b_{1} & =1\text{, }b_{2}=k+1\text{, and for }n>2\text{, }b_{n} =2b_{n-1}+\left( k-1\right) b_{n-2} \end{align*} It is again left to the reader to show equivalence in definitions. Working from [4] again leads to $a_{n}=\frac{\left( 1+\sqrt{k}\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}{2\sqrt{k}}\text{ and }b_{n}=\frac{\left( 1+\sqrt{k}\right)^{n}+\left( 1-\sqrt{k}\right)^{n}}{2}$ So the corresponding Greek Ladder approximation to $$\sqrt{k}$$ is $\frac{b_{n}}{a_{n}}=\frac{\sqrt{k}\left( 1-\sqrt{k}\right)^{n}+\sqrt{k}\left( \sqrt{k}+1\right)^{n}}{\left( \sqrt{k}+1\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}$ and the Newton formula yields \begin{align*} & \frac{\sqrt{k}\left( 1-\sqrt{k}\right) ^{n}+\sqrt{k}\left( \sqrt{k}+1\right)^{n}}{\left( \sqrt{k}+1\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}-\frac{\left( \frac{\sqrt{k}\left( 1-\sqrt{k}\right)^{n}+\sqrt {k}\left( \sqrt{k}+1\right)^{n}}{\left( \sqrt{k}+1\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}\right)^{2}-k}{2\times\frac{\sqrt{k}\left(1-\sqrt{k}\right)^{n}+\sqrt{k}\left( \sqrt{k}+1\right)^{n}}{\left(\sqrt{k}+1\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}}\\ & =\frac{\sqrt{k}\left( \sqrt{k}+1\right)^{2n}+\sqrt{k}\left( 1-\sqrt{k}\right)^{2n}}{\left( \sqrt{k}+1\right)^{2n}-\left( 1-\sqrt{k}\right)^{2n}} \end{align*} But is this the fraction from the $$2n$$-th rung of the ladder? That rung is $\left\langle \frac{\left( 1+\sqrt{k}\right)^{2n}-\left( 1-\sqrt{k}\right)^{2n}}{2\sqrt{k}}\qquad\frac{\left( 1+\sqrt{k}\right)^{2n}+\left(1-\sqrt{k}\right)^{2n}}{2}\right\rangle$ which gives the approximation $\frac{\frac{\left( 1+\sqrt{k}\right)^{2n}+\left( 1-\sqrt{k}\right)^{2n}}{2}}{\frac{\left( 1+\sqrt{k}\right)^{2n}-\left( 1-\sqrt{k}\right)^{2n}}{2\sqrt{k}}}=\frac{\sqrt{k}\left( \sqrt{k}+1\right) ^{2n}+\sqrt{k}\left( 1-\sqrt{k}\right)^{2n}}{\left( \sqrt{k}+1\right)^{2n}-\left( 1-\sqrt{k}\right)^{2n}}$ as it should.

### Suggestions for Further Exploration

Greek ladders seem to beg for further exploration and generalization. For example, can a  Greek Ladder be defined that matches a sequence of Newton's  Method estimates exactly? Can a Greek Ladder be defined and associated with Newton's Method for cube roots, fourth roots, or fifth roots? (A Greek Ladder  that estimates fourth roots is defined in the next section.) Perhaps the reader can find some hidden treasures! Greek ladders are fun and accessible to students at many different levels.

### Conclusion: A Little Story

It seems that Isaac Newton and his assistant Roger Cotes, who was also Newton's student (and, according to the Mathematics Genealogy Project, Newton only had two graduate students), were seeking in one of their research projects a simple rational approximation for $$\sqrt[4]{2}$$. Newton came up with a couple of estimates that were not very good. But very quickly, Cotes offered the simple estimate of $$\frac{44}{37}$$, which is surprisingly good for a two-digit denominator, in that $\left( \frac{44}{37}\right)^{4}\approx1.999879$ Cotes did not reveal how he got this, and many have tried to explain it, but no one seems to have thought of Greek Ladders. Here are just the first three steps of the classical  $$\sqrt[4]{2}$$ Greek ladder, whose initial rung is $$\langle1\quad1\quad1\quad1\rangle$$ and with each rung $$\langle a\quad b\quad c\quad d\rangle$$ being followed by $$\langle a+b+c+d\quad2a+b+c+d\quad 2a+2b+c+d\quad2a+2b+2c+d\rangle$$, with the approximations being $$\frac{d}{c}$$. The first three rungs can be quickly written and are $\begin{array}{cccc} 1 & 1 & 1 & 1\\ 4 & 5 & 6 & 7\\ 22 & 26 & 31 & 37 \end{array}$ So $$\frac{7}{6}$$ and $$\frac{37}{31}$$ are two estimates, and their Farey mean is $\frac{7+37}{6+31}=\frac{44}{37}$ With such a simple computation, invoking the principle of Occam's Razor makes for a good guess that Cotes used this procedure. Cotes died young, and of him, Newton is reported to have said, "If Cotes had lived, we might have known something."

Since this paper appeared, I have discovered how Cotes apparently arrived at his approximation of $$\frac{44}{37}$$ for $$\sqrt[4]{2}.$$ Cotes was seemingly quite attracted to continued fractions as a method for approximating algebraic irrationals. Indeed, according to David Fowler [8, p. 312], Cotes obtained his estimate for the fourth root of 2 by means of continued fractions, thus making my conjecture -- that he may have used a Greek ladder -- utterly false. My Greek ladder conjecture seemed plausible enough, but according to Fowler, it is simply wrong, and the error is regretful.

As a humorous finish, though, we point out that Cotes could have arrived at the famous fraction $$\frac{44}{37}$$ even more simply by noting that Newton had offered as approximations to $$\sqrt[4]{2}$$ the fractions $$\frac{6}{5}$$, $$\frac{13}{11}$$, and $$\frac{25}{21}$$ [9], then computing their Farey mean; to wit, $$\frac{6+13+25%}{5+11+21}=\frac{44}{37}$$.

#### Acknowledgements

I am indebted to Professor Joseph Zund for introducing me to Greek ladders several years ago, and I thank Professors David Pengelley and Fred Richman for their very useful comments.  The material in this article was presented at the MAA Southwestern Annual Conference, Western New Mexico University, Silver City, NM, April 3-4, 2009.

Robert J. Wisner is Professor Emeritus in the Department of Mathematical  Sciences at New Mexico State University. He was founding editor of SIAM Review, the only one of the 14 journals of the Society for Industrial and Applied Mathematics (SIAM) designed to appeal to all of its members.  Wisner also has authored or co-authored numerous K-12 textbooks for Scott Foresman and Company; was Consulting Editor for Mathematics for Brooks/Cole for over 25 years; and recently co-authored a series of interactive pre-calculus textbooks, available on CD from Hardy Calculus.

#### References

[1] Leonard Eugene Dickson, History of the Theory of Numbers, vol. II, Chelsea Publishing Company, New York, 1919, p. 341.

[2] David Fowler, The Mathematics of Plato's Academy, A New Reconstruction (2nd ed.), Oxford University Press, 2003, pp. 57-58.

[3] Shaun Gilberson and Thomas J. Osler, "Extending Theon's Ladder to Any Square Root," College Math. J. 35 (2004) pp. 222-226.

[4] Russell Jay Hendel, "Approaches to the Formula for the $$n$$-th Fibonacci Number," College Math. J. 25 (1994) pp. 139-142.

[5] H. W. Turnbull, The Great Mathematicians, Simon and Shuster, New York, 1962, pp. 28-30.

[6] G. C. Vedova, "Notes on Theon of Smyrna," Amer. Math. Monthly 58 (1951) pp. 675-683.

[7] Robert J. Wisner, "Square Roots: Greek Ladders, Farey Fractions, and Pascal's Triangle," The AMATYC Rev. 23 (2002) no. 2, pp. 53-60.