Going beyond the Golden Ratio.


I show that for the same reason that the golden ratio, $\phi=1.6180334..$, can be considered the most irrational number, that $1+\sqrt{2}$ can be considered the 2nd most irrational number, and indeed why $(9+\sqrt{221})/10$ can be considered the 3rd most irrational number.

 

This post is #2 out 20 of my “20 posts in 20 days” project!
  

Let us imagine a game between two kids, Emily and Sam – both strong and determined in their own way who spend their entire lives trying to outwit each other, instead of doing their homework. (A real life Generative Adversial Network…)

Emily, proudly reminds us that she simultaneously bears the same first name as Emily Davison, the most famous of British suffragettes; Emily Balch, Nobel Peace Prize laureate; Emilie du Chatelet, who wrote the first French translation and commentary of Isaac Newton’s “Principia“; Emily Roebling, Chief Engineer of New York’s iconic Brooklyn Bridge; Emily Bronte author of Wuthering Heights; Emily Wilson, the first female editor of ‘New Scientist‘ publication; and also Emmy Noether, who revolutionized the field of theoretical physics.

On the other side we have Sam (and all his minion friends, who are aptly called Sam-002, Sam-003, Sam-004 ) who is part human / part robot and plays Minecraft or watches Youtube, 24/7.

They agree to play a game where Emily thinks of a number, and then Sam (with the possible help of his minions) has 60 seconds to find any fractions that are equal to Emily’s number.

And so the game begins…

Emily says “8.5″.

Sam & friends quickly reply with “850/10″,… “34/4″,…  “17/2″,… “425/5″,…

They soon realize that all these answers are equally valid because they are all equivalent fractions. Being competitive they want to pick a single winner, so they all agree that the best answer is the one with the lowest denominator. And so, 17/2 is deemed the best answer.

This time, Emily tries to make it harder by picking ‘0.123456‘. After only a slight pause, Sam  slyly says “123456/1000000“.

Emily’s annoyed with this answer. She knows that although the best answer would be the irreducible fraction 1929/15625, Sam’s answer is still valid answer, and furthermore he will always be able to instantly answer like this if she picks any number with a terminating decimal expansion.

So this time, Emily picks “$\pmb{\pi}$”.

This time, it’s Sam turn to get annoyed and claims that this is cheating. They both know that there isn’t an answer because $\pi$ is an irrational number and so there isn’t any fraction that is exactly equal it.

Therefore they eventually agree, that the best answer will be whichever fraction is closest to her number.

With this new rule Sam’s minions suggest the following: ‘3‘,… ‘31/10‘,… ‘22/7‘,…  ‘16/5‘,… ‘3927/1250‘.

Sam looks at all of them condescendingly and immediately calls out “3141592/1000000

Again, Emily gets annoyed. She realizes that Sam has shown that for any irrational number, you can (easily) pick a fraction that is arbitrarily close to it. That is, the rationals are dense everywhere.

They both know that fractions with small denominators are more in the spirit of the game, but Sam says that because his answer was technically valid, it’s her responsibility (and not his) to think of a new rule to improve things.

Emily consider ways of giving each answer a score. Initially, she thought that for each fraction, the score could be the (absolute) difference between her number and the proposed fraction, and then multiplied by the denominator. (The lower the better).

However, after talking to some of her tech friends, she decided to make it even stricter to very severely penalize fractions with large denominators. Such as scoring system would definitely force Sam to have to work hard to find good fractions. And so finally she proposed that for each fraction, the score should be the absolute difference multiplied by the denominator squared. Lowest score wins.

So for the previous value of $\pmb{\pi}$, the Minions calculate their scores:

$$ 3 \rightarrow 0.141, \quad 31/10 \rightarrow 4.1, \quad 22/7 \rightarrow 0.06, \quad 16/5 \rightarrow 1.5, \quad 3927/1250 \rightarrow 36$$

Whilst, Sam comes last with a truly pathetic and embarrassing score of $3141592/1000000 \rightarrow 592$.

Sam now very annoyed, becomes very determined to find a simple method that will get a low score.

Later that week, whilst sitting at his computer, he had a flash of inspiration.

He first noted that $\pi = 3.14159…$.

So ignoring the fractional part of this implied that $\pi \simeq 3$.

For a better approximation he noted that the reciprocal of the previous fractional part equals  $1/0.14159265 = 7.0625133$.

So, ignoring the fractional part of this last number, implies that $\pi \simeq 3 + \cfrac{1}{7}  = \frac{22}{7}$.

For an even better approximation, he noted that the reciprocal of the previous fractional part is $1/ 0.0625133= 15.996594…$

Again, ignoring the fractional part of this implies that

$$ \pi  \simeq 3 + \cfrac{1}{7+\cfrac{1}{15}}=\frac{333}{106} $$

For an even better approximation he noted that the reciprocal of the fractional part of this last number is $1/ 0.996594 = 1.00341723$.

Again, ignoring the fractional part of this implies that

$$ \pi  \simeq 3 + \cfrac{1}{7+\cfrac{1}{15+\cfrac{1}{1}}}=\frac{355}{113} $$

Further, he realised that since the numerators of each of the fractions is 1, and there is always an addition between the various number the only important numbers are the ones going down the diagonal. That is, 3, 7,15 and 1. So he devises a short-hand way of writing these continued fractions.  

$$ \pi  \simeq [3; 7,15,1] =\frac{355}{113} $$

He calls each of these successive rational approximations, ‘convergents‘ because he knows that in the limit they converge to $\pi$.  The score for each of these convergents is:

$$ \frac{3}{1} \rightarrow 0.141, \quad \frac{22}{7} \rightarrow 0.062, \quad \frac{333}{106} \rightarrow 0.935, \quad \frac{355}{113}  \rightarrow 0.003, \quad \frac{103993}{33102} \rightarrow 0.633$$

Sam is super happy again, that he has found a systematic way of consistently low scores. He then writes a computer program to quickly calculate the continued fraction for any given number. For $\pi$ it was:

$$ \pi \simeq [3; 7,15,1,292,1,1,1,2,1,3,1,14,2,1,1,2,2,2,2,1,84,2,1,1,15,3,13,1,4,2,6,6,99,1,,…] $$

Although he didn’t see any ongoing pattern, he was satisfied that (in principle) he could calculate this to as many digits as he desired.

His Minions can then calculate the convergents, each one based on this continued fraction to one level deeper than the one before. When they look at the scores for the first 100 convergents (see figure 1) they notice that all the scores are less than 1. This is great news, and it turns out be a consequence of Dirchlet’s theorem that states that for any irrational number, there are an infinite number of continued fractions that score less than 1. 

They also noticed that the scores varied widely between 0 and 1 with no discernible pattern.

Figure 1. The score for successive convergents of Pi.

Over the next few days, knowing that Sam was using  a computer to calculate his answers, Emily always tried to think of really complicated numbers. These included: 
$$ \log \pi, \; 2^{\pi}, \; e^2, \; \pi^e, \; \sqrt[3]{2}, \; \sin(\pi/7), ….$$

What they both noticed is that for all these numbers, the score distribution for the convergents were all really similar to the above chart. That is, the scores for the convergents always varied fairly evenly (but in an unpredictable manner) between 0 and 1.

For Emily, the most annoying consequence of this pattern, was that lots of the convergents which the Minions calculated, had a score very close to zero. The Minions boasted that this meant that if they continued their process, then for all the numbers that had Emily picked, they could (in theory) find an infinite number of convergents whose scores were arbitrarily close to zero. This made her very annoyed.

She really wanted to find a number that was really difficult to approximate with rationals.

One day out of sheer exasperation, Emily couldn’t be bothered thinking of a really complicated number so she just called out: $\pmb{\sqrt{7}}$.

Sam then calculates the continued fraction as: 

$$ \sqrt{7} = [2; 1,1,1,4,1,1,1,4,1,1,1,4,1,1,1,4,…] = [2; \overline{1,1,1,4}]$$

where the overline indicates that the block pattern repeats indefinitely.

By complete accident, Sam and Emily had just rediscovered a property that Lagrange (1767) and Euler (1763) had first discovered: namely, that any irrational quadratic surd will have a periodic continued fraction, and similarly all periodic continued fraction are quadratic irrationals.

But the biggest shock came when the Minions calculated the scores for the convergents (see figure 2).  The scores followed an incredibly clear and regular pattern.  Many of the scores converged in an alternating manner (just above, then just below, then just above, then just below,….) around a critical threshold equal to 0.188982, which they soon realized was equal to $1/\sqrt{28}$.

They both knew that this would have some profound implications.  

Sam realised that his Minions could find an infinite number of convergent rationals that were all (slightly) less than $1/\sqrt{28}$. But for any other smaller score $S$, (even infinitesimally smaller than this one), Sam and his Minions would only be able to find a finite number of fractions with a score less than $S$.

That is, the special value of $1/\sqrt{28}$ represented a critical threshold between an infinite number of solutions, and only a finite number of solutions. 

No longer could be Minions boast that they could find an endless number of fractions with scores arbitrarily close to zero! Emily had found a number where the the lower limit of scores was not zero. She had finally found a
 badly approximable number (a concept closely related to low discrepancy sequences) and she was ecstatic. 

I

Figure 2. The score for successive convergents of $\sqrt{7}$.

Emily had stumbled on a very counter-intuitive pattern first discovered by Markoff  (in this very specific field of maths his name is traditionally spelled ‘Markoff’ but in all other areas, it is usually spelled ‘Markov’). He had found that of all possible numbers, those that are the irrational roots of a quadratic equation are harder to approximate than all other irrationals.

Furthermore as she explored she uncovered one of the deepest theorems in the field of Number Theory, namely the Thue-Roth-Siegel theorem, from which follows the corollary that numbers that are the roots of a polynomial (algebraic numbers) are harder to approximate than transcendental numbers like $e$ or $\pi$.

That is, despite her intuition,  numbers of the form $a+b\sqrt{c}$ for rationals $a,b,c$ are actually much much harder to approximate numbers than numbers like $e$ or $\pi$!

Emily continued her exploration of quadratic irrationals and soon learnt than whenever she picked a number that was an irrational root of a quadratic equation $ax^2+bx+c=0$ for integers $a,b,c$, the critical lower threshold would involve (if not be identical) to the expression $ 1/\sqrt{b^2-4ac}$. 

This helped explain why in the original example, which corresponded to the quadratic equation $x^2-7=0$, the critical value was equal to $1/\sqrt{28}$. She soon realized that she could leverage this pattern, to find a quadratic equation where the value of discriminant ,  $b^2-4ac$, was as small as possible. This would force Sam in getting a really high (and therefore bad) score. The discriminant could not be 1 or 4 as this would not result in an irrational root. Then using parity logic, she realized that it was not possible for a quadratic with integer coefficients to have a discriminant of 2 or 3. Thus, she needed to find one, where the discriminant was equal to 5.

She soon found the most awesome quadratic equation : $x^2-x-1=0$.

The positive root of this quadratic equation was  $x=(1+\sqrt{5})/2$, which she recognized as the Golden Ratio, $\phi$.

Based on the above pattern, she predicted this value of $x$ would result in a critical threshold of $1\sqrt{5} \simeq 0.4472$, which was Hurwitz’s theorem

Sam found that the continued fraction for this value was very elegant, and his Minions confirmed (figure 3) that indeed the critical score was equal to $1/\sqrt{5}$.

$$\phi =(1+\sqrt{5})/2 =  [1;1,1,1,1,1,1,1,1,…] = [1; \overline{1}] \simeq 1.618033 $$

Figure 3. The score for successive convergents of the the golden ratio.

Emily was absolutely ecstatic because she had now found the ultimate number that forced Sam to have the highest (worst) critical threshold score.

And although so many knew of the golden ratio as being the most aesthetic pleasing aspect ratio for rectangles, where the Greek Parthenon is poster child for this claim (see figure 4), she now had a deeper understanding of its significance.

Figure 4. The Golden Rectangle. Progressively smaller versions of it with the same aspect ratio, can be obtained by partitioning each rectangle into a square and another golden rectangle. 

She had also learnt something that many people had not realised. Namely, the golden ratio is the most irrational number not (directly) because its continued fraction consisted solely of ones; but rather, because its critical score — which separates the world of infinite rationals with merely a finite number of rationals —  was the largest possible critical score.

Not content to rest on her laurels, in the coming days she discovered a whole lot of quadratic equations who discriminant equals 5, and thus also had a critical threshold score of $1/\sqrt{5}$. Here were some of them, along with their continued fractions.

\begin{align}
x^2-x-1=0 &\rightarrow x=(1+\sqrt{5})/2 &= [1;1,1,1,1,1,1,1,1,…] &= [1; \overline{1}] &= 1.618033 \\ \nonumber
x^2+x-1=0 &\rightarrow x=(-1+\sqrt{5})/2 &= [0;1,1,1,1,1,1,1,1,…] &= [1; \overline{1}] &= 0.618033 \\  \nonumber
x^2-3x+1=0 &\rightarrow x=(3+\sqrt{5})/2 &= [2;1,1,1,1,1,1,1,1,…] &= [2; \overline{1}] &= 2.618033 \\ \nonumber
x^2-5x+5=0 &\rightarrow  x=(5+\sqrt{5})/2 &= [3;1,1,1,1,1,1,1,1,…] &= [2; \overline{1}] &= 2.618033 \\ \nonumber

\end{align}

She learnt that if $x$ was an optimal number (such as $\phi$), then for any integers $a,b,c,d$ such that $ad-bc=\pm 1$ (the integer Moebius transformation), then $\frac{ax+b}{cx+d}$ would also be an equally optimal number. So here were some more:

$$ \frac{1}{\phi+3}  \simeq 0.216542 = [0; 4,1,1,1,1,1,1,1,\ldots]  $$

$$\frac{\phi+1}{2\phi+1}  \simeq 0.619034 = [0; 1,1,1,1,1,1,1,1,\ldots] $$

$$ \frac{\phi+2}{\phi+3}  \simeq 0.783458 = [0; 1,3,1,1,1,1,1,1,\ldots] $$

$$ \frac{3\phi-2}{2\phi-1}  \simeq 1.276390 = [1; 3,1,1,1,1,1,1,1,\ldots] $$

That is, she had found a method to construct an infinite number of values of $x$ that were equivalent to $\phi$ and therefore equally irrational as $\phi$. And for each of these numbers, their continued fraction differed at the start, but eventually became identical to the the repeating block of $\phi$. (Serrat’s theorem).

Quickly Sam got tired of playing this game knowing that the critical score would always be $1/\sqrt{5}$, so he forced Emily to modify the rules of the game. He thus invoked the rule, the she was no longer able to say any number that was equivalent to $\phi$, the golden ratio.

Emily was very disappointed because she had grown very fond of $\phi$ and all its amazing properties, but she eventually relented and agreed to the rule change.

Emily then began the search of more special values of $a,b,c$ such that the corresponding quadratic irrational would be really hard to approximate. She soon identified the quadratic equation

$$x^2-2x-1=0 \rightarrow x=1+\sqrt{2} = [2;2,2,2,2,…]=[2;\overline{2}] $$

She predicted that this would have a critical score of $1/\sqrt{8} \simeq 0.353553$, however, she wasn’t certain that this was the best she could do. So she went to her local library, and discovered that Markoff, at the age of 24 had done his Master’s thesis on this very topic. In his thesis, he famously established an explicit one-to-one correspondence between each equivalence class of irrational numbers and sorted Markov triplets. Sorted Markov triplets were those positive integers $0 < m_1 \leq m_2 \leq m$ such that:

$$ m^2+m_1^2+m_2^2=3m m_1 m_2$$

Furthermore, he showed that all the Markov triplets can be constructed through a Markoff Tree (see also Markoff numbers). This tree has many elegant geometric properties (see also here.)

The sequence of integers $m$ that satisfy this equation is:

$$1,2,5,13,29,34,9,169,194,223,433,610,…$$

Then for each term in this sequence, he defined a new sequence (the Markov-Lagrange spectrum). These new values corresponded to the score for each of the successive classes of quadratic irrational numbers.

$$ \mu_m = \frac{\sqrt{9m^2-4}}{m} = \{ \sqrt{5}, \sqrt{8}, \frac{\sqrt{221}}{5}, \frac{\sqrt{1517}}{13},… \}$$

As expected, the first term corresponds to the quadratic equation $x^2-x-1=0$ and the quadratic irrational $x=(1+\sqrt{5})/2$.

And much to Emily’s satisfaction, the second term in the sequence corresponded to the quadratic equation $x^2+2x-1$ and the quadratic irrational $x=1+\sqrt{2}$ – which is the same one she had guessed. This was the evidence she had been looking for.

Sam calculated that the continued fraction for this value.

$$ 1+\sqrt{2} = [2;2,2,2,2,2,2,…]=[2; \overline{2}] = 2.414213562…$$

And his Minions verified that the critical score for this number was indeed $1/\sqrt{8}$.

Sam and Emily, also confirmed that $\sqrt{2}$ was equivalent to $1=\sqrt{2}$, and therefore also had a critical score of $1/\sqrt{8}$.

$$ \sqrt{2} = [1;2,2,2,2,2,…] = [1; \overline{2}] \simeq 1.41421356 $$

Given new found special nature of $\sqrt{2}$, Emily wondered if there was a rectangle with similarly elegant properties as the golden rectangle. It turned out that there was one. Not only that, but it has been applied to one of the most common of household items (seee figure5! The ISO paper A series are all defined in terms of the aspect ratio of $\sqrt{2}$. Each paper size is exactly half the area of the previous one, with $A0$ having an area of exactly one square metre.

For Sam calculated that $\frac{297}{210} = \frac{594}{420} =[1;2,2,2,2,2]$ and that $\frac{1189}{841}= [1;2,2,2,2]$.

Figure 5a. The ISO paper A series sizes are all defined in terms of the aspect ratio of $\sqrt{2}$. Each paper size is exactly half the area of the previous one, with the area of A0 being exactly one square meter.
Figure 5b. A comparison of a recursive rectangle based on the $\sqrt{2}$ (left), with the Fibonacci rectangle, which is based on the Golden Ratio (right). 

 

This time it didn’t take long for Sam to invoke another new rule: Emily could not say any numbers that were equivalent to $\phi = (1+\sqrt{5})/2$ or $1+\sqrt{2}$.

So after agreeing, Emily, took a deep breath, and tried the next number in the Markoff series: $\frac{\sqrt{221}}{8}$. This looked a lot scarier, then the previous 2 terms, but she had faith….

She found a corresponding quadratic function, $5x^2-9x-7$, and its quadaratic irrational $x=(9+\sqrt{221})/10$.

She wasn’t sure what the continued fraction would be. Sam calculated the continued fraction to be:

$$ x=(9+\sqrt{221})/10 = [2;2,1,1,2,2,1,1,2,2,1,1,2,…]= [2;\overline{2,1,1,2}] \simeq 2.38660687$$

And his Minions did indeed calculate that the critical score was $\frac{5}{\sqrt{221}}$, which was the reciprocal of the 3rd Markov term.

Although I am not aware of any equivalent rectangle that exploits the elegance of $(9+\sqrt{221})/10$ (or any of its equivalents), for those visually inclined figure 6 shows the logarithmic spiral, defined according the the polar equation;

$$ (r,\theta) = (\frac{\sqrt{k}}{n}, \frac{2\pi k}{z}); \quad \textrm{for } 1 \leq k \leq n=1000.$$

Different values of $z$ produce different spirals. The top three are based on the three special irrational numbers: $z=(1+\sqrt{5})/2; z= 1+\sqrt{2}$ and $(9+\sqrt{221})/10$, whilst the other six are random values shown for comparison. The key insight is that the closer the value of $z$ is to a rational number, the more visible will be the aliased spiral arms.

Figure 6. A comparison of 3 special logarithmic spirals (top row) compared to six other random seeded logarithmic spirals. Top row: z= (1+\sqrt{5})/2; z= 1+\sqrt{2}; z=(9+\sqrt{221})/10; and six random values: (Middle row) z=2.16325, z=2.18737, z=2.24535; and (bottom row) z=2.31872, z=2.34796, z=2.555538. Note that the closer the value $z$ is to a rational number, the more visible the aliased spiral arms will be.


***

By the end of this game, Emily and Sam had both learnt that the golden ratio, $\phi=(1+\sqrt{5})/2$ was indeed the most irrational number. That is, it was the hardest number to approximate with a fraction. It was a nice corollary that the continued fraction for this $\phi$ was simply $[1;1,1,1,1,\ldots]$, and that the points in the golden spiral were very evenly distributed.

They had also learnt that there were (infinitely) many other numbers that were equally irrational. She could find them either using the integer Moebius transformation, or by finding equivalent continued fractions. 

The real prize in this game, however, was they could finally answer the question “So if the golden ratio is the most irrational number, what is the second and third most irrational number?

The second most irrational number was $1+\sqrt{2}$, whose continued fraction was $[2;2,2,2,2,\ldots]$. Although they probably wouldn’t have been able to guess this fact, the elegance of this number meant they weren’t too surprised at this result.

However, without a doubt, the most surprising discovery from the game was that the third most irrational number was the most unassuming of numbers: $(9+\sqrt{221})/10$, whose continued fraction $[2;1,1,2,2,1,1,2,2,1,1,\ldots]$, was a simple but not completely obvious.

Somehow, although gaining a podium placing, this special number has sunk into the midst of oblivion, now unbeknownst to all but a very small minority. Furthermore, Markov’s later works on Markov chains have shadowed his seminal work in Diophantine approximation theory.

So at your next dinner party, when you are tired of talking about the weather, why not tell them about Andrei Markov and how he proved that then number $(9+\sqrt{221})/10$ was the 3rd most irrational number?

 

Not all characters and events depicted in this article are entirely fictitious. Any similarity to actual events or persons, living or dead, is probably true.”

 

I hope you liked this quick post. This post is #2 out 20 of my “20 posts in 20 days” project.


My name is Martin Roberts. I have a PhD  in theoretical physics. I love maths and computing. I’m open to new opportunities – consulting, contract or full-time – so let’s have a chat on how we can work together!

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My other contact details can be found here.

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2 Comments

    • Martin Roberts

      Sorry about the delay (been on holidays!). The $n$-th most irrational number is equal to the $n$-th Lagrange number, $L_n$, where $L_n = \sqrt{9-\frac{4}{m_n^2}}$, where $m_n$ is the $n$-th Markov number. The first 1000 Markov numbers are listed in OEIS A002559.
      The parent page of OEIS A002559 also gives Mathematica code for calculating arbitrary terms of the Markov-Lagrange spectrum.
      Hope this helps!

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