Famous fractions: Can any "special" numbers be approximated by simple ratios like $3.14\ldots$ as $22/7$?

I got interested in flexagons as a child and needed to draw strips of equilateral triangles. The fact that $$\frac {\sqrt 3}2 \approx \frac {13}{15}$$ came in very handy. It is better than a part in a thousand.

The fact that $2^{10}=1024 \approx 1000=10^3$, which we use in counting computer things can be expressed as $$\log_2(10) \approx \frac {10}3$$ but we don't really use the fraction much.

I'll describe a way to derive rational approximations of irrational numbers and then use it to recover some well-known examples.

Given a number like $\pi$, we set aside the integer part ($a_0 = 3$), and we seek to approximate the fractional part ($0.14159\!\ldots$) by some Egyptian fraction $\frac{1}{a_1}$, or just as well its reciprocal, $7.06251\!\ldots$ by an integer $a_1$. Rounding down gives the familiar approximation $\pi \approx 3 + \frac{1}{7} = \frac{22}{7}$. Iterating this process gives a series of approximations, $$3, 3 + \frac{1}{7}, 3 + \frac{1}{7 + \frac{1}{15}}, 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1}}}, 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \frac{1}{292}}}}, \ldots .$$ We call the formal limit of these fractions the continued fraction for $\pi$, and for readability sometimes just write out this quantity as the sequence $$[3; 7, 15, 1, 292, \ldots] .$$ Simplifying the above fractions gives the convergents of $\pi$, a sequence of improving approximations: $$3, \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \frac{103993}{33102} \ldots .$$ Evidently passing from a convergent just before a large number (e.g., $292$ in the above expansion) to the next one results in a relatively small adjustment, so, very roughly speaking, convergents given by evaluating a fraction just before the one corresponding to a large number gives a relatively good approximation for the size of the denominator. Indeed, the convergent given by stopping just before $292$ gives the famous approximation $$\pi \approx \frac{355}{113}$$ discovered by Yu Chongzhi in the 5th Century A.D. and sometimes known as the Milü (密率); it is accurate to about one part in $10^7$.

Some more examples:

• The continued fractions of some familiar numbers exhibit obvious patterns. For example, it follows from the fact that the Golden Ratio $\phi$ satisfies $\phi^2 = \phi + 1$ (and is greater than $1$) that its continued fraction is $[1; 1, 1, 1, \ldots]$. Its successive convergents are the successive ratios $F_{n + 1} / F_{n}$ of Fibonacci numbers: $$1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \frac{13}{8}, \ldots .$$

• The convergents of the disintegration constant $\log 2$ are $$0, 1, \frac{2}{3}, \frac{7}{10}, \ldots ,$$ and the occurrence of $\frac{7}{10}$ can be used to derive the Rule of 70.

• As lulu mentioned in the comments, approximations of $\log_2 3$ are important in music theory: One sometimes wants to work with two tones whose frequency ratio is close to $3 : 2$. In an equally tempered scale of $n$ notes to an octave, this means approximating $\log_2 3$ with some rational number $\frac{m}{n}$. The convergents of $\log_2 3$ are $$1, 2, \frac{3}{2}, \frac{8}{5}, \frac{19}{12}, \frac{65}{41}, \ldots ,$$ and the presence of the ratio $\frac{19}{12}$ corresponds to the fact that an interval of a perfect fifth in the familiar chromatic ($12$-note, evenly tempered) scale is a good approximation of a $3 : 2$ ratio of frequencies.

• The continued fraction for $\sqrt{2}$ is $[1; 2, 2, 2, \ldots]$, and its convergents are $$1, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \ldots ,$$ and one can show that the numerator and denominator from every second convergent ($\frac{3}{2}, \frac{17}{12}, \ldots$) are solutions to the classical Pell equation $$x^2 - 2 y^2 = 1 ,$$ and the others are solutions to the Pell equation $x^2 - 2 y^2 = -1$. Similar observations hold for other square roots of integers.

• We can use this technique to produce approximations of any irrational we like. For example, the continued fraction for $e$ turns out to be $[2; 1, 2, 1, 1, 4, 1, 1, 6, 1, \ldots]$, and truncating the continued fraction just before the $6$ gives the approximation $$e \approx \frac{193}{71} ,$$ which is accurate to about one part in $10^5$.