What is the 3rd most irrational number?
The answer will surprise you!


Most of us know about the golden ratio, \(\phi = \frac{1+\sqrt{5}}{2} \) and how it can be considered the most irrational number, but most people never ask the question, “What is the second and third most irrational number?” In this post, I discuss discuss some special irrational numbers, and why they can be considered the 2nd and 3rd most irrational numbers.



Finding good rational approximations to reals.

In number theory, the field of Diophantine approximation deals with the approximation of real numbers by rational numbers. It is named after Diophantus of Alexandria. The problem on how well a real number can be approximated by rational numbers is an extremely well studied problem .

Let us, say that the rational number $\frac{p}{q}$ is a good approximation of a real number $x$ , only if the absolute value of the difference between $a/b$ and $α$ can not be improved with any other rational rational number with a smaller denominator. That is,

$$ | x-\frac{p}{q} | < | x-\frac{p’}{q’} | \quad \text{for every } p’,q’ \; \text{such that } 0< q'<q  \tag{1}$$

Often, a very similar definition which is generally more amenable to study because it does not include fractions, is used

$$ | p-q x | < |p’- q’ x | \quad \text{for every } p’,q’ \; \text{such that } 0< q'<q \tag{2} $$

Continued fractions can be used to compute the best rational approximations of a real number. This is because the successive convergents of the continued fraction $x$ produce the best approximations of $x$ possible, (as per the second definition above).

For example, for the constant $\pi= 3.14159265358…$, the regular continued fraction is:

$$ \pi  = 3 + \cfrac{1}{7+\cfrac{1}{15+\cfrac{1}{1+\cfrac{1}{292+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}}}} $$

Because the numerators for all regular continued fractions are always 1, we can succinctly describe this continued fraction simply by listing the denominators (and the initial whole number)

$$\pi = [3;7,15,1,292,1,1,1,2,1,3,1,\ldots] $$

These produce the following progressively accurate rational approximations of $\pi$.

$$ \{ 3, \frac{22}{7}, \frac{333}{106},\frac{355}{113},\frac{103993}{33102}, \frac{104348}{33215}, \frac{208341}{66317}, \frac{312689}{99532},\frac{833719}{265381}, \ldots \} $$

From this, it is clear why $\frac{22}{7}$ is frequently used as an approximation of $\pi$, and why rationals more accurate than $\frac{355}{113}$ are almost never used.

How accurate are these approximations?

The obvious measure of the accuracy of a rational approximation of a real number $x$ by a rational number $p/q$, is $\epsilon = |x-p/q| $. However, this quantity can always be made arbitrarily small by increasing the absolute values of $p$ and $q$. The previous section showed how we can efficiently find $p,q$ that produce successively more and more accurate rational approximations to $x$.

Therefore, the accuracy of the approximation is usually estimated by comparing this quantity to some the size of the denominator $q$, typically a negative power of it.

The following table shows the same approximations of $\pi$ along with their error $\epsilon$ when multiplied by $q$ and by $q^2$.

p/q   & \epsilon & \epsilon q & \epsilon  q^2\\ \hline
\frac{3}{1} & 1 \times 10^{-1} & 1 \times 10^{-1} & 0.142 \\ \hline
\frac{22}{7} & 1 \times 10^{-3} & 9 \times 10^{-3} & 0.062 \\\hline
\frac{333}{106} & 8\times 10^{-5} & 9 \times 10^{-3} & 0.935 \\\hline
\frac{355}{113} & 3\times 10^{-7} & 3 \times 10^{-5}  & 0.003 \\ \hline
\frac{103993}{33102} & 6\times 10^{-10} & 2 \times 10^{-5} & 0.633 \\ \hline
\frac{104348}{33215} & 3\times 10^{-10} & 1 \times 10^{-5} & 0.366 \\ \hline
\frac{208341}{66317} & 1\times 10^{-10} & 8 \times 10^{-6}  & 0.538 \\ \hline
\frac{312689}{99532} & 3\times 10^{-11} & 3 \times 10^{-6} & 0.289 \\ \hline
\frac{833719}{265381} & 9\times 10^{-12} & 2 \times 10^{-6}  & 0.614 \\ \hline
\ldots  & \ldots  & \ldots   & \ldots \\ \hline

From this one can see that the values in the columns $\epsilon$, as well as $\epsilon q$ become arbitrarily small, but those in the final $\epsilon q^2$ column , don’t.  Said another way, for good approximations, the error term will go down in proportion to square of the denominator.

Thus, it seems that by using the error term $\epsilon q^2$ we can more meaningfully compare how well each of these approximate  $\pi$. That is, of those listed, $355/133$ is by far best, followed by $22/7$, then $3/1$ then $312689/99532$, etc,…

Dirichlet’s Theorem

Noting that for an irrational number, the equivalent continued fractions are infinite, it means that we can produce an infinite number of convergent rational approximations. Looking at the table above, we see that for the first 10 convergents, $\epsilon q^2 < 1$. The question then naturally arises, “How often is $\epsilon q^2$?”

Dirichlet proved that for any irrational $x$, there are an infinite number of rational approximations such that $\epsilon < 1/q^2$. That is, there are an infinite number of fractions $p/q$ such that :

$$ \left| x- \frac{p}{q} \right | < \frac{1}{q^2} \tag{3} $$

Furthermore, if $p/q$ are regular continued fraction convergents, then $\epsilon q^2$ will always be less than 1.

Although this property has  ‘long been known from the theory of continued fractions’, it is named after Dirichlet after he provided a very elegant proof of it in 1838.

Hurwitz’s Theorem

In 1891, Hurwitz showed that Dirichlet’s theorem could be improved. Furthermore, he proved that it was a strict upper bound. That is, it could not be improved any further.  That is, there are an infinite number of fractions $p/q$ such that :

$$ \left| x- \frac{p}{q} \right | < \frac{1}{\sqrt{5} q^2} \tag{4} $$

Hurwitz showed that the reason that this is a strict upper bound and can not be improved is because if we consider the real number $x = (1+\sqrt{5})/2$, commonly called the golden ratio, and usually denoted $\phi$, then for $C> \sqrt{5}$, there are are only a finite number of rational numbers $p,q$ such that:

$$ \left| \phi – \frac{p}{q} \right | < \frac{1}{C q^2} \tag{5} $$

The continued fraction representation of $\phi$ is simply $[1;1,1,1,1,1,1,1,1,\ldots]$ and its convergent fractions are:

$$ \{ \frac{1}{1}, \frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \frac{13}{8}, \frac{21}{13}, \frac{34}{21},\ldots \} $$

The error terms $\sqrt{5}q^2 \epsilon$ for these convergents are,

$$ 1.38,0.854,0.1.056,0.978,1.008,0.996,1.001,0.9995,1.0001,0.999994,1.000002,0.999999,\ldots $$

We see that the error terms tend to 1, which is an indication that $\sqrt{5}q^2$ is the bound.


Equivalence Relations

It is very well known that the golden ratio $\phi$, can be considered the most irrational number. From the commentary above, we can see that this is most notably because it provides the limiting bound for Hurwitz’s theorem.  Technically speaking, the fact that the golden ratio has the continued fraction comprising only of ones, is a consequence rather than a rigorous reason why it is the most irrational number.

One lesser known fact is that there are infinitely many other real numbers that can be considered equally irrational, because they force strictness on the upper bound of the Hurwitz theorem. For example, all of the following numbers are equally irrational and thus could claim the title of ‘the most irrational number’:

$$ \{ \frac{1}{ \phi+3}, \; \frac{ \phi-1}{\phi}, \; \frac{ \phi+1}{2 \phi+1} ,\; \frac{3 \phi+2}{4 \phi+3}, \; \frac{ \phi+2}{ \phi+3} ,\; \frac{3 \phi-2}{2 \phi-1},\; \frac{ \phi-4}{\phi-3},\; \ldots \}  $$

There are an infinite number of them, and they are all considered equivalent, as they are all of the form

$$ \frac{a \phi+b}{c \phi+d}, \quad \textrm{for integers } a,b,c,d \;\;  \textrm{such that } ad-bc=\pm 1. \tag{6}$$

The continued fractions for these values are:

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

$$ \frac{\phi-1}{\phi}  \simeq 0.216542 = [0; 2,1,1,1,1,1,1,1,\ldots]  $$

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

$$\frac{3\phi+2}{4\phi+3}  \simeq 0.216542 = [0; 1,2,1,1,1,1,1,1,\ldots]  $$

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

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

Serrat showed that with the exception of a finite initial sequence,  equivalent fractions have identical continued fraction expansions.

So although all of the numbers in this class are equally irrational, it is reasonable to select from all of those, the number whose continued fraction is the simplest to be the canonical number in this class. Namely, $\phi = [1;1,1,1,1,1,\ldots]$.

The second most irrational number

If we consider all the real numbers excluding those of the form described in Eqn 6, then it turns out that Hurwitz’s theorem may be made even stronger still. That is, if $x$ is not equivalent to $\phi  = (1+\sqrt{5})/2$, then there are an infinite number of fractions $p/q$ such that :

$$ \left| x- \frac{p}{q} \right | < \frac{1}{\sqrt{8} q^2} \tag{4} $$

Again, this inequality is strong, because for any number equivalent to $x = \sqrt{2}$, the value of $C=\sqrt{8}$ can not be improved.

Thus the second most irrational class of numbers are those that are equivalent to $x=\sqrt{2}$.

The continued fraction for $\sqrt{2}$ is $[1;2,2,2,2,2,2,\ldots]$,

therefore using Serrat’s theorem, the number $1+\sqrt{2} = [2;2,2,2,2,….]$ is equivalently irrational.

Thus, we can say that the second most irrational number is $1+\sqrt{2}$.


The third most irrational number

Repeating this line of argument again, if we exclude all the real numbers that are equivalent to $\phi$, or $1+\sqrt{2}$, then Hurwitz’s theorem can be made even stronger.

That is, if $x$ is not equivalent to $\phi  = (1+\sqrt{5})/2$ or $1+\sqrt{2}$, then there are an infinite number of fractions $p/q$ such that :

$$ \left| x- \frac{p}{q} \right | < \frac{1}{C q^2} \quad \textrm{where } C=\sqrt{221}/5. \tag{4} $$

Again, this inequality is strong, because for any number equivalent to $x = (9+\sqrt{221})/10$, the value of $C=\sqrt{221}/5$ can not be improved.

Thus the second most irrational class of numbers are those that are equivalent to $x=\sqrt{2}$.

The continued fraction for $(9+\sqrt{221})/10$ is $[2;2,1,1,2,2,1,1,2,1,1,2,2,1,1,2,\ldots]$,

Thus we can say that the third most irrational number is $(9+\sqrt{221})/10$.

In a similar fashion, we can say:

  • the fourth most irrational number is $(23+\sqrt{1517})/26 = [2; 2,1,1,1,1,2,2,1,1,1,1,2 ,…] $
  • the fifth most irrational number is $(5+\sqrt{7565})/58 = [1; 1,1,2,2,2,2,1,1,2,2,2,2,…]$
  • and so on…


Markov Spectrum

Recalling equation 5, the first class of irrationals are those that enforce the upper bound when $C=\sqrt{5}$, and those relating to the second class correspond to the bound where $C=\sqrt{8}$, etc.

The question arises, ‘what is the pattern?’. The answer is these numbers  form the Lagrange / Markov spectrum, and are all of the form

$$ L_n  = \sqrt{9-\frac{4}{m_n^2}}, $$ where $m_n$ is the n-th term of the Markov sequence.  The first few terms of the Markov sequence are $ \{ 1,2,5,13,29,34,89,169,…\}$.

Thus, $L_n = \{ \sqrt{5}, \sqrt{8}, \sqrt{221}/5, \sqrt{1517}/13, … \} $



Much of this post was based on a wonderful paper “The hyperbolic geometry of Marjov’s theorem on Diophanitne approximation and quadratic forms” by Boris Springborn.



My name is Martin Roberts. I have a PhD  in theoretical physics. I love maths and computing. I am open to new opportunities – consulting, contract or employment – as a data scientist, so let’s have a chat on how we can work together! Contact details can be found here.



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