8  Continued Fractions: Fractions of Fractions

Decimals describe a number by recording how far each digit departs from a power of ten. That works fine for everyday measurement, but it can hide arithmetic structure. The decimal 3.14159… tells you almost nothing about why 22/7 is such a good stand-in for pi, and 1.41421… gives no hint that \(\sqrt{2}\) has the most regular pattern of all non-rational numbers. A completely different way of writing any number – the continued fraction – lays that hidden structure bare.

The computer scientist Bill Gosper once described the experience of reading a number through its continued fraction: “It’s completely astounding … it looks like you are cheating God somehow.” (Bailey et al. 2007, Ch. IX)

Every real number has a unique continued fraction, and the pattern of its coefficients encodes how well the number can be approximated by fractions – one of the deepest questions in number theory.

8.1 Beyond Decimal Expansions

Combo Class
Capturing Infinity With Fractions
youtu.be/YCJ99liapCQ

An enthusiastic tour of how infinite continued fractions trap irrational numbers exactly — using nothing but whole numbers and repeated division.

A simple continued fraction is an expression of the form

\[x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cdots}}}\]

where \(a_0\) is an integer (the integer part of \(x\)) and each \(a_1, a_2, a_3, \ldots\) is a positive integer. These numbers are called the partial quotients. The compact notation is \([a_0;\, a_1, a_2, a_3, \ldots]\).

Four examples span the range from simple to surprising:

Number	Decimal	Continued fraction
22/7	3.142857…	[3; 7]
355/113	3.141592…	[3; 7, 16]
\(\sqrt{2}\)	1.414213…	[1; 2, 2, 2, …]
\(\pi\)	3.141592…	[3; 7, 15, 1, 292, …]

Research Example: Why Does 355/113 Approximate \(\pi\) So Freakishly Well?

The table above lists 355/113 as matching π to six decimal places — yet its denominator is only 113. How can such a small fraction get so close? The continued fraction of π hides the answer in a single towering number: the partial quotient 292.

import matplotlib.pyplot as plt

GREEN = '#2ca02c'
BLUE  = '#1f77b4'

pi_cf = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 1, 1]
colors = [GREEN if v == 292 else BLUE for v in pi_cf]
fig, ax = plt.subplots(figsize=(9, 3.5))
ax.bar(range(len(pi_cf)), pi_cf, color=colors, edgecolor='white', linewidth=0.8)
ax.set_yscale('log')
ax.set_xticks(range(len(pi_cf)))
ax.set_xticklabels([f'a{i}={v}' for i, v in enumerate(pi_cf)], fontsize=7, rotation=30, ha='right')
ax.set_ylabel("Partial quotient aₙ (log scale)")
ax.set_title("π = [3; 7, 15, 1, 292, 1, 1, ...]  — The 292 Spike Makes 355/113 Magical", fontsize=11)
ax.annotate('a₄ = 292\n→ 355/113\nmatches π to\n6 decimal places', xy=(4, 292), xytext=(7.5, 30),
            arrowprops=dict(arrowstyle='->', color=GREEN, lw=1.5), color=GREEN, fontsize=8, ha='left')
plt.tight_layout(); plt.show()

Figure 8.1: Partial quotients of π on a log scale. The towering spike at position 4 — value 292 — is why 355/113 is so freakishly accurate. Every other coefficient is tiny by comparison.

The 292 spike is unmistakable — every other partial quotient of π is tiny by comparison. Because 292 is so large, cutting the continued fraction just before it produces 355/113, which matches π to six decimal places with a denominator of only 113. In about 15 lines of Python you have confirmed a 2,000-year-old marvel of rational approximation.

The large coefficient 292 in pi’s expansion is no accident. Cutting a continued fraction off just before a large coefficient always gives an exceptionally accurate approximation. Because 292 is so big, the fraction [3; 7, 15, 1] = 355/113 is extraordinarily close to pi – off by only 0.000000267. We will make this precise in Section 8.3.

Finite continued fractions always represent rational numbers. Infinite ones always represent irrational numbers. The continued fraction therefore acts as a perfect test: it terminates if and only if the number is rational.

(There is one tiny ambiguity: every rational has two finite representations, e.g. [3; 7] = [3; 6, 1]. The convention is to always choose the one that does not end in 1.)

8.2 The Algorithm

Numberphile
Euclid’s Algorithm
youtu.be/6Y3jHHE_hbA

Euclid’s 2,300-year-old algorithm for computing the greatest common divisor — the identical steps that generate every continued fraction expansion.

Finding the continued fraction of a rational number \(p/q\) is the Euclidean algorithm in disguise. The steps are identical to those used to compute \(\gcd(p, q)\) in Section 6.4:

1. Let \(a_0 = \lfloor p/q \rfloor\) (the integer part, i.e. p // q).
2. Replace \((p,\, q)\) by \((q,\; p - a_0 \cdot q)\) (the new remainder).
3. Repeat until the remainder is zero.

Each step produces one coefficient. Because the remainders strictly decrease, rational inputs always terminate in finitely many steps.

import math

def cf_rational(p, q):
    """Continued fraction of rational p/q."""
    coeffs = []
    while q > 0:
        coeffs.append(p // q)
        p, q = q, p % q
    return coeffs

# uses: cf_rational()
import math

print(cf_rational(22, 7))
print(cf_rational(355, 113))
print(cf_rational(1071, 462))

[3, 7]
[3, 7, 16]
[2, 3, 7]

Research Example: Can You See the Euclidean Algorithm as Packing Squares?

The Euclidean algorithm feels like abstract arithmetic — dividing, taking remainders, repeating. But every step is secretly a tiling: you pack the largest possible squares into a rectangle, and the remainder is the leftover strip. Let us make that geometry visible for 22/7.

# uses: cf_rational()
#| label: fig-cfrac-euclidean
#| fig-cap: "The Euclidean algorithm as rectangle tiling: 22/7 = [3; 7]. Three 7×7 squares fill most of the 22×7 rectangle (a₀ = 3); seven 1×1 squares fill the leftover 1×7 strip (a₁ = 7)."
import matplotlib.pyplot as plt
import matplotlib.patches as patches

BLUE   = '#1f77b4'
PURPLE = '#9467bd'


fig, ax = plt.subplots(figsize=(9, 3.2))
for i in range(3):
    r = patches.FancyBboxPatch((i*7, 0), 6.95, 6.95, boxstyle="square,pad=0",
                               linewidth=1.2, edgecolor='white', facecolor=BLUE, alpha=0.85)
    ax.add_patch(r)
    ax.text(i*7+3.5, 3.5, '7', ha='center', va='center', color='white', fontsize=16, fontweight='bold')
for j in range(7):
    r = patches.FancyBboxPatch((21, j), 0.95, 0.95, boxstyle="square,pad=0",
                               linewidth=0.5, edgecolor='white', facecolor=PURPLE, alpha=0.9)
    ax.add_patch(r)
ax.annotate('', xy=(21, -0.5), xytext=(0, -0.5),
            arrowprops=dict(arrowstyle='<->', color=BLUE, lw=1.5))
ax.text(10.5, -1.1, 'a₀ = 3  (three squares of side 7)', ha='center', fontsize=9)
ax.annotate('', xy=(22, 7.5), xytext=(21, 7.5),
            arrowprops=dict(arrowstyle='<->', color=PURPLE, lw=1.5))
ax.text(21.5, 8.1, 'a₁ = 7', ha='center', fontsize=9)
ax.set_xlim(-1, 23.5); ax.set_ylim(-2.0, 9.0); ax.set_aspect('equal'); ax.axis('off')
ax.set_title('22/7 = [3; 7]  —  Each Step of the Euclidean Algorithm Is a Layer of Squares',
             fontsize=10, pad=6)
plt.tight_layout(); plt.show()

Three blue squares, seven purple squares — the whole algorithm drawn as geometry. Every coefficient is just “how many squares fit?” and you already know how to count squares.

Let us verify the first by hand: \(22 = 3 \cdot 7 + 1\), so \(a_0 = 3\). Next, \(7 = 7 \cdot 1 + 0\), so \(a_1 = 7\) and we stop. Result: [3; 7].

For \(\sqrt{D}\) (where \(D\) is a positive integer that is not a perfect square), an exact all-integer method avoids any floating-point rounding. It tracks three integers \(m_n, d_n, a_n\) and updates them by

\[m_{n+1} = d_n\, a_n - m_n, \qquad d_{n+1} = \frac{D - m_{n+1}^2}{d_n}, \qquad a_{n+1} = \left\lfloor\frac{a_0 + m_{n+1}}{d_{n+1}}\right\rfloor,\]

starting from \(a_0 = \lfloor\sqrt{D}\rfloor\), \(m_0 = 0\), \(d_0 = 1\). It can be shown by induction that \(d_n\) always divides \(D - m_{n+1}^2\) exactly, so every quantity stays an integer with no rounding at all.

import math

def cf_sqrt(D, num_terms):
    """Exact integer CF coefficients for sqrt(D)."""
    a0 = math.isqrt(D)      # new: exact integer sqrt
    if a0 * a0 == D:
        return [a0]          # perfect square
    coeffs = [a0]
    m, d, a = 0, 1, a0
    for _ in range(num_terms - 1):
        m = d * a - m
        d = (D - m * m) // d
        a = (a0 + m) // d
        coeffs.append(a)
    return coeffs


# uses: cf_sqrt()
import math

for D in [2, 3, 5, 7, 13]:
    print(f"sqrt({D:2d}): {cf_sqrt(D, 12)}")

sqrt( 2): [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
sqrt( 3): [1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1]
sqrt( 5): [2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]
sqrt( 7): [2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1]
sqrt(13): [3, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1]

math.isqrt(D) (Python 3.8+) returns \(\lfloor\sqrt{D}\rfloor\) using integer arithmetic alone – no floating-point approximation ever occurs. The output reveals an immediate pattern: every non-square positive integer \(D\) produces a repeating block. We will examine why in Section 8.4.

For arbitrary floating-point numbers, a simpler loop works for the first ten or so coefficients before rounding errors accumulate:

import math

def cf_float(x, max_terms=10):
    """Approximate CF from a float (loses precision after ~12 steps)."""
    coeffs = []
    for _ in range(max_terms):
        a = math.floor(x)
        coeffs.append(a)
        frac = x - a
        if frac < 1e-9:
            break
        x = 1.0 / frac
    return coeffs


# uses: cf_float()
import math

print("pi:", cf_float(math.pi, 8))
print("e: ", cf_float(math.e, 10))

pi: [3, 7, 15, 1, 292, 1, 1, 1]
e:  [2, 1, 2, 1, 1, 4, 1, 1, 6, 1]

The pi output [3, 7, 15, 1, 292, 1, 1, 1] matches the known values exactly. The e output reveals a striking pattern: [2, 1, 2, 1, 1, 4, 1, 1, 6, 1] – every third term is an even number that grows by 2. A transcendental number with such clean structure is quite unusual; the full pattern is \(e = [2;\; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, \ldots]\).

8.3 Convergents: Best Rational Approximations

MathsNeedNotApply
The Pi Approximation Tier List
youtu.be/S0QtJ3kRezM

Ranks π approximations from worst to best — showing why convergents from the continued fraction are provably the best possible rational approximations.

Truncating a continued fraction at position \(n\) gives a rational approximation called the \(n\)-th convergent. The convergents of \([a_0;\, a_1, a_2, \ldots]\) satisfy a two-step recurrence:

\[h_{-1} = 1, \quad h_0 = a_0, \quad h_n = a_n\, h_{n-1} + h_{n-2}\]

\[k_{-1} = 0, \quad k_0 = 1, \quad k_n = a_n\, k_{n-1} + k_{n-2}\]

The \(n\)-th convergent is \(p_n/q_n = h_n/k_n\). Notice how similar this is to the Fibonacci recurrence in Section 6.1: when every \(a_n = 1\), the numerators and denominators are exactly the Fibonacci numbers. We will return to this connection in Section 8.5.

# uses: cf_float()
from fractions import Fraction
import math

def convergents(coeffs):
    """Convergents of CF with given coefficients.
    Returns a list of Fraction objects (exact arithmetic)."""
    h_prev, h_curr = 1, coeffs[0]
    k_prev, k_curr = 0, 1
    result = [Fraction(h_curr, k_curr)]
    for a in coeffs[1:]:
        h_prev, h_curr = h_curr, a * h_curr + h_prev
        k_prev, k_curr = k_curr, a * k_curr + k_prev
        result.append(Fraction(h_curr, k_curr))
    return result


pi_cf = cf_float(math.pi, 7)      # [3, 7, 15, 1, 292, 1, 1]
pi_convs = convergents(pi_cf)
for c in pi_convs:
    err = abs(float(c) - math.pi)
    print(f"{str(c):13s} {float(c):.9f}  {err:.2e}")

3             3.000000000  1.42e-01
22/7          3.142857143  1.26e-03
333/106       3.141509434  8.32e-05
355/113       3.141592920  2.67e-07
103993/33102  3.141592653  5.78e-10
104348/33215  3.141592654  3.32e-10
208341/66317  3.141592653  1.22e-10

fractions.Fraction is a Python standard-library class (no SymPy needed) that stores a fraction as an exact ratio of two integers. All arithmetic on Fraction objects is exact: Fraction(1,3) + Fraction(1,6) gives Fraction(1,2), never 0.4999…, which makes it ideal for verifying convergent computations.

The convergents alternate above and below the true value: \(p_0/q_0 < \pi < p_1/q_1\), then \(p_2/q_2 < \pi < p_3/q_3\), and so on, closing in from both sides. The jump from \(p_3/q_3 = 355/113\) to \(p_4/q_4 = 103993/33102\) is tiny in absolute error but requires a denominator 300 times larger – because the next coefficient is 292.

Research Example: Do Convergents Alternate Above and Below \(\pi\)?

Each new convergent overshoots or undershoots π and then flips to the other side — they zero in from alternating directions. The gap between 355/113 and the next convergent should look absurdly small compared to the others. Let us check.

# uses: convergents()
#| label: fig-cfrac-closeup
#| fig-cap: "Convergents of π alternate above (red) and below (blue) the true value. 355/113 sits so close that it takes a denominator 300× larger to improve on it."
import math
import matplotlib.pyplot as plt

RED   = '#d62728'
GREEN = '#2ca02c'
BLUE  = '#1f77b4'

labels_plot = ['3/1', '22/7', '333/106', '355/113', '103993/33102']
conv_vals = [3, 22/7, 333/106, 355/113, 103993/33102]
errors = [v - math.pi for v in conv_vals]
fig, ax = plt.subplots(figsize=(8, 4))
ax.axhline(0, color=RED, linewidth=2, label='π (true value)')
above_color, below_color = GREEN, BLUE
for i, (err, lbl) in enumerate(zip(errors, labels_plot)):
    c = above_color if err > 0 else below_color
    ax.plot([i, i], [0, err], color=c, linewidth=3)
    ax.plot(i, err, 'o', color=c, markersize=9)
    offset = abs(err) * 0.15
    ax.text(i, err + (offset if err > 0 else -offset), lbl, ha='center',
            va='bottom' if err > 0 else 'top', color=c, fontsize=8, fontweight='bold')
ax.set_xticks(range(5))
ax.set_xticklabels([f'n={i}\nq={[1,7,106,113,33102][i]}' for i in range(5)], fontsize=8)
ax.set_ylabel("Convergent − π")
ax.set_title("Convergents Bracket π — Closing In from Alternating Sides", fontsize=11)
ax.legend()
plt.tight_layout(); plt.show()

Red above, blue below — the alternation is exact, every time. And that near-zero bar at 355/113 shows why ancient astronomers used it for centuries: it is the last stop before the denominators explode. You found this with five lines of arithmetic.

More precisely, the convergents satisfy the Best Approximation Theorem (Hardy and Wright 1979): no fraction \(p/q\) with \(q \le q_n\) comes closer to \(\alpha\) than the \(n\)-th convergent \(p_n/q_n\), except \(p_n/q_n\) itself.

We can verify this computationally: search all fractions with denominator at most 113 and confirm none beats 355/113.

# uses: cf_float(), convergents()
import math
from fractions import Fraction

best_err = float('inf')
best_frac = None
for q in range(1, 114):
    p = round(math.pi * q)
    err = abs(p / q - math.pi)
    if err < best_err:
        best_err = err
        best_frac = Fraction(p, q)

print(f"Best approx (denom <= 113): {best_frac}")
print(f"Error:                      {best_err:.4e}")
ref = abs(355 / 113 - math.pi)
print(f"355/113 error:              {ref:.4e}")

Best approx (denom <= 113): 355/113
Error:                      2.6676e-07
355/113 error:              2.6676e-07

The best rational approximation to \(\pi\) with denominator at most 113 is indeed 355/113, confirming the theorem for this case.

8.4 Quadratic Irrationals Are Periodic

Joseph-Louis Lagrange (1736–1813)
Public domain, via Wikimedia Commons

Mathologer
Infinite fractions and the most irrational number
youtu.be/CaasbfdJdJg

Why the golden ratio is the “most irrational” number — its all-ones continued fraction makes every rational approximation as bad as possible.

The cf_sqrt output showed that \(\sqrt{2} = [1;\, \overline{2}]\), \(\sqrt{3} = [1;\, \overline{1, 2}]\), \(\sqrt{5} = [2;\, \overline{4}]\) (the bar means the block repeats forever). Joseph-Louis Lagrange proved in 1770 that a number has an eventually periodic continued fraction if and only if it is a quadratic irrational – the irrational root of a quadratic equation \(ax^2 + bx + c = 0\) with integer coefficients (Lagrange 1770; Hardy and Wright 1979, Thm. 177).

We detect the period by watching for a repeated \((m, d)\) state pair:

import math

def cf_sqrt_period(D):
    """Return (a0, repeating_block) for the CF of sqrt(D)."""
    a0 = math.isqrt(D)
    if a0 * a0 == D:
        return (a0, [])
    seen = {}
    m, d, a = 0, 1, a0
    period = []
    while True:
        m = d * a - m
        d = (D - m * m) // d
        a = (a0 + m) // d
        state = (m, d)
        if state in seen:
            break
        seen[state] = len(period)
        period.append(a)
    return (a0, period)

# uses: cf_sqrt_period()
import math

print(f"{'D':>3}  CF representation         period")
print("-" * 42)
for D in [2, 3, 5, 6, 7, 10, 11, 13, 14]:
    a0, blk = cf_sqrt_period(D)
    cf_str = f"[{a0}; {blk}]"
    print(f"{D:3d}  {cf_str:25s}  {len(blk)}")

  D  CF representation         period
------------------------------------------
  2  [1; [2]]                   1
  3  [1; [1, 2]]                2
  5  [2; [4]]                   1
  6  [2; [2, 4]]                2
  7  [2; [1, 1, 1, 4]]          4
 10  [3; [6]]                   1
 11  [3; [3, 6]]                2
 13  [3; [1, 1, 1, 1, 6]]       5
 14  [3; [1, 2, 1, 6]]          4

Notice that the repeating block always ends with \(2a_0\) (twice the integer part). This is a theorem, not a coincidence.

There is a classical connection to Pell’s equation, which asks for integer solutions to \(x^2 - Dy^2 = \pm 1\). The numerator and denominator of the last convergent before the period repeats always give the smallest positive solution. For example, the convergents of \(\sqrt{2}\) are \(1/1, 3/2, 7/5, 17/12, \ldots\), and indeed: \(1^2 - 2 \cdot 1^2 = -1\), \(3^2 - 2 \cdot 2^2 = 1\), \(7^2 - 2 \cdot 5^2 = -1\), and so on, alternating \(\pm 1\).

# uses: cf_sqrt_period(), convergents()
import math

def pell_solutions(D, n_solutions=5):
    """First n Pell solutions x^2 - D*y^2 = +/- 1."""
    a0, blk = cf_sqrt_period(D)
    if not blk:
        return []
    # extend over several periods to get enough solutions
    full = [a0] + blk * (2 * n_solutions + 4)
    convs = convergents(full)
    results = []
    for c in convs:
        x, y = c.numerator, c.denominator
        val = x * x - D * y * y
        if abs(val) == 1:
            results.append((x, y, val))
            if len(results) >= n_solutions:
                break
    return results

# uses: pell_solutions()
import math

print("Pell solutions for D=2 (x^2 - 2y^2 = +/-1):")
for x, y, v in pell_solutions(2, 6):
    print(f"  {x}^2 - 2*{y}^2 = {v}")

Pell solutions for D=2 (x^2 - 2y^2 = +/-1):
  1^2 - 2*1^2 = -1
  3^2 - 2*2^2 = 1
  7^2 - 2*5^2 = -1
  17^2 - 2*12^2 = 1
  41^2 - 2*29^2 = -1
  99^2 - 2*70^2 = 1

Research Example: Which √D < 100 Has the Longest Continued-Fraction Period?

Period length varies wildly from one D to the next — some square roots repeat after just one coefficient, others take many more. Is there a pattern? Let us compute the period length for every non-square D from 2 to 100 and look for the record-holders.

# uses: cf_sqrt_period()
import math
import matplotlib.pyplot as plt

BLUE = '#1f77b4'
RED  = '#d62728'

ds      = [d for d in range(2, 101) if math.isqrt(d)**2 != d]
periods = [len(cf_sqrt_period(d)[1]) for d in ds]
max_p   = max(periods)

fig, ax = plt.subplots(figsize=(10, 3.5))
colors = [RED if p == max_p else BLUE for p in periods]
ax.bar(ds, periods, color=colors, width=0.8, edgecolor='none')
for d, p in zip(ds, periods):
    if p == max_p:
        ax.text(d, p + 0.15, str(d), ha='center', va='bottom',
                fontsize=7, color=RED, fontweight='bold')
ax.set_xlabel("D")
ax.set_ylabel("Period length")
ax.set_title("CF Period Length for √D (2 ≤ D ≤ 100) — Spikes Appear Without Warning",
             fontsize=10)
plt.tight_layout()
plt.show()

Figure 8.2: CF period lengths for √D, non-square D from 2 to 100. Most periods are short; a handful of D values stand out with the longest period — including the famous D = 61 whose Pell equation solution exceeds a billion.

The bars spike and dip without an obvious formula — yet every value is exact and provable. The red bar at D = 61 is the record-holder: its Pell equation \(x^2 - 61y^2 = 1\) has smallest solution \(x = 1{,}766{,}319{,}049\), a number that Indian mathematician Brahmagupta effectively found in the 7th century (Hardy and Wright 1979) — long before Pell was born.

8.5 The Golden Ratio: Hardest to Approximate

Adolf Hurwitz (1859–1919)
Public domain, Mondadori Publishers, via Wikimedia Commons

Numberphile
The Golden Ratio (why it is so irrational)
youtu.be/sj8Sg8qnjOg

The golden ratio’s continued fraction [1; 1, 1, 1, …] makes it the hardest number to approximate with fractions — visualized with rotating sunflower spirals.

In Section 6.8 we met the golden ratio \(\varphi = (1+\sqrt{5})/2\). Its continued fraction is the simplest possible infinite sequence: \(\varphi = [1;\, 1, 1, 1, \ldots]\) – all ones, forever.

# uses: cf_float()
import math

phi_approx = (1 + math.sqrt(5)) / 2
print("phi (float):", cf_float(phi_approx, 14))

phi (float): [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

Because every partial quotient is 1, the convergents grow as slowly as possible. Computing them with exact arithmetic confirms the Fibonacci connection noted in Section 6.8:

# uses: convergents()
import math

# Phi = [1; 1, 1, 1, ...], 16 terms exactly
phi_convs = convergents([1] * 16)
print("Phi convergents (numerator/denominator):")
for c in phi_convs[:10]:
    print(f"  {c.numerator}/{c.denominator}")

Phi convergents (numerator/denominator):
  1/1
  2/1
  3/2
  5/3
  8/5
  13/8
  21/13
  34/21
  55/34
  89/55

The numerators and denominators are consecutive Fibonacci numbers: \(1/1, 2/1, 3/2, 5/3, 8/5, 13/8, \ldots\) The ratio \(F(n+1)/F(n)\) converges to \(\varphi\) from alternating sides, just as we saw in Section 6.1.

Because \(\varphi\) has all-1 partial quotients, no fraction with a given denominator can approximate it substantially better than the corresponding Fibonacci ratio. This is the content of Hurwitz’s theorem (Hurwitz 1891): for any irrational \(\alpha\), the inequality

\[\left|\alpha - \frac{p}{q}\right| < \frac{1}{\sqrt{5}\, q^2}\]

holds for infinitely many fractions \(p/q\). For the golden ratio, the constant \(\sqrt{5}\) in the denominator cannot be replaced by anything larger: \(\varphi\) is the hardest real number to approximate by rationals.

We can visualize this by plotting the scaled approximation error \(|\alpha - p_n/q_n| \cdot q_n^2\) for the convergents of several numbers. For \(\varphi\) this quantity should hover right at \(1/\sqrt{5} \approx 0.447\); for \(\pi\) it should plunge far below that floor near the 292-coefficient convergent.

Research Example: Does \(\varphi\) Really Hug the Hardest-to-Approximate Boundary?

Hurwitz’s theorem says every irrational can be approximated within \(1/(\sqrt{5}\,q^2)\) infinitely often, and φ is the extreme case — it hugs that floor and never dips below. We can see this directly by plotting the scaled error \(|\alpha - p_n/q_n| \cdot q_n^2\) for φ, √2, and π side by side.

# uses: convergents(), cf_sqrt(), cf_float()
import math
import matplotlib.pyplot as plt

BLUE   = '#1f77b4'
ORANGE = '#ff7f0e'
GREEN  = '#2ca02c'
PURPLE = '#9467bd'

# phi: exact all-1 CF, value approximated from float
phi_val = (1 + math.sqrt(5)) / 2
phi_convs_all = convergents([1] * 20)
phi_errs = [abs(float(c) - phi_val) * c.denominator**2
            for c in phi_convs_all]

# sqrt(2): exact integer CF
s2_convs = convergents(cf_sqrt(2, 18))
s2_val = math.sqrt(2)
s2_errs = [abs(float(c) - s2_val) * c.denominator**2
           for c in s2_convs]

# pi: float CF (accurate for first 8 terms)
pi_convs_all = convergents(cf_float(math.pi, 8))
pi_errs = [abs(float(c) - math.pi) * c.denominator**2
           for c in pi_convs_all]

fig, ax = plt.subplots(figsize=(8, 4))
ax.plot(phi_errs, marker='o', markersize=5, color=PURPLE, linewidth=1.5, label='φ (golden ratio)')
ax.plot(s2_errs,  marker='s', markersize=5, color=BLUE,   linewidth=1.5, label='√2')
ax.plot(pi_errs,  marker='^', markersize=5, color=GREEN,  linewidth=1.5, label='π')
ax.axhline(1 / math.sqrt(5), color=ORANGE, linestyle='--', linewidth=1.5, label='Hurwitz floor 1/√5')
ax.set_yscale('log')
ax.set_xlabel("Convergent index n")
ax.set_ylabel("|α − pₙ/qₙ| · qₙ²")
ax.set_title("φ Hugs the Hurwitz Floor — π Breaks Through It at the 292-Coefficient", fontsize=10)
ax.legend()
plt.tight_layout()
plt.show()

Figure 8.3: Scaled approximation error for φ (circles), √2 (squares), and π (triangles). φ hovers right at the Hurwitz floor 1/√5 ≈ 0.447; π plunges far below at index 4 — the 292 miracle.

φ hugging the green dashed floor while π crashes through it at index 4 — that single dip is the geometric signature of the 292 coefficient. You just reproduced a result from analytic number theory with a for-loop and a plot.

8.6 The Gauss-Kuzmin Distribution

Pick a “random” real number and compute its continued fraction. What fraction of the partial quotients equal 1? Equal 2? Equal 7?

Karl Friedrich Gauss noticed around 1800 that the partial quotients follow a definite probability distribution. This was proved rigorously by Rodion Kuzmin in 1928 and independently by Paul Lévy in 1929 (Kuzmin 1928; Lévy 1929):

\[P(a = k) = \log_2\!\left(\frac{(k+1)^2}{k(k+2)}\right) = \log_2\!\left(1 + \frac{1}{k(k+2)}\right).\]

Here \(\log_2\) means logarithm base 2 (introduced in Section 7.4). The probabilities for small \(k\):

\(k\)	Formula	Probability
1	\(\log_2(4/3)\)	41.5%
2	\(\log_2(9/8)\)	17.0%
3	\(\log_2(16/15)\)	9.3%
4	\(\log_2(25/24)\)	5.9%
5	\(\log_2(36/35)\)	4.1%

Small partial quotients dominate overwhelmingly. The coefficient 1 appears in more than two of every five positions. A telescoping product argument shows that the probabilities sum exactly to 1.

Let us verify this experimentally. The key tool is the Gauss map \(T(x) = \{1/x\}\) (the fractional part of \(1/x\)). Applying \(T\) repeatedly to any \(x \in (0, 1)\) produces one partial quotient at each step and maps the interval back into \((0, 1)\), so the iteration continues indefinitely.

Research Example: Does Partial Quotient 1 Really Appear 41% of the Time?

Gauss conjectured, and Kuzmin proved, that the partial quotients of a random real number follow a precise distribution: coefficient 1 appears about 41% of the time, coefficient 2 about 17%, and so on. Let us generate 3000 random numbers, apply the Gauss map, and see whether the counts match the formula.

import math
import random
import matplotlib.pyplot as plt

BLUE = '#1f77b4'
RED  = '#d62728'

random.seed(42)
counts = {}
n_total = 0
for _ in range(3000):
    x = random.random()        # uniform in (0, 1)
    for _ in range(40):        # apply Gauss map 40 times
        if x < 1e-10:
            break
        x = 1.0 / x            # reciprocal
        a = math.floor(x)      # partial quotient
        x -= a                 # fractional part (back in (0,1))
        counts[a] = counts.get(a, 0) + 1
        n_total += 1

def gk_prob(k):
    return math.log((k + 1)**2 / (k * (k + 2)), 2)

ks = list(range(1, 11))
observed    = [counts.get(k, 0) / n_total for k in ks]
theoretical = [gk_prob(k)                 for k in ks]

fig, ax = plt.subplots(figsize=(8, 4))
width = 0.35
xs = range(len(ks))
ax.bar([x - width / 2 for x in xs], observed,
       width, label='Observed', color=BLUE, edgecolor='white', linewidth=0.6)
ax.bar([x + width / 2 for x in xs], theoretical,
       width, label='Gauss-Kuzmin formula', color=RED, alpha=0.9,
       edgecolor='white', linewidth=0.6)
ax.set_xticks(list(xs))
ax.set_xticklabels([str(k) for k in ks])
ax.set_xlabel("Partial quotient k")
ax.set_ylabel("Frequency")
ax.set_title("Gauss-Kuzmin: Partial Quotient 1 Dominates — It Appears 41% of the Time",
             fontsize=10)
ax.legend()
plt.tight_layout()
plt.show()

Figure 8.4: Observed partial-quotient frequencies (blue) match the Gauss-Kuzmin formula (orange) closely. Coefficient 1 wins by a landslide at 41%.

Blue and orange bars almost indistinguishable — Gauss’s two-century-old conjecture confirmed in seconds by a random simulation. The 1800s had to wait for Kuzmin’s proof; you just watched it happen with three dozen lines of Python.

The Gauss-Kuzmin distribution has two elegant corollaries.

Lévy’s constant. The denominator \(q_n\) of the \(n\)-th convergent grows on average like \(e^{n\beta}\) where \(\beta = \pi^2/(12 \ln 2) \approx 1.1865\). That is, for almost every irrational \(\alpha\), we have \(q_n^{1/n} \to e^\beta \approx 3.276\) as \(n \to \infty\).

Research Example: Do Convergent Denominators Really Grow Like \(e^{n\beta}\)?

Lévy proved that for almost every irrational number the convergent denominators \(q_n\) grow exponentially at rate \(e^\beta\) where \(\beta = \pi^2/(12\ln 2) \approx 1.187\). Let us check whether π’s denominators actually follow this curve — and watch what the 292 coefficient does to the trend.

import math
import matplotlib.pyplot as plt

BLUE   = '#1f77b4'
ORANGE = '#ff7f0e'
GREEN  = '#2ca02c'

beta = math.pi**2 / (12 * math.log(2))
e_beta = math.exp(beta)
pi_cf_long = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14]
qs = [1]; k_p, k_c = 0, 1
for a in pi_cf_long:
    k_p, k_c = k_c, a * k_c + k_p; qs.append(k_c)
ns = list(range(len(qs)))
fig, ax = plt.subplots(figsize=(8, 3.5))
levy_y = [e_beta**n for n in ns]
ax.semilogy(ns, levy_y, '--', color=ORANGE, linewidth=1.8, label=f'e^(nβ), eᵝ ≈ {e_beta:.2f}')
ax.semilogy(ns, qs, 'o-', color=BLUE, markersize=6, linewidth=1.5, label="q_n  (π's convergents)")
ax.annotate('292-coefficient\ncauses huge jump', xy=(4, qs[4]), xytext=(6.5, 5000),
            arrowprops=dict(arrowstyle='->', color=GREEN, lw=1.5), color=GREEN, fontsize=8)
ax.set_xlabel("Convergent index n")
ax.set_ylabel("Denominator q_n")
ax.set_title(f"Lévy's Constant: Denominators Grow Like e^(nβ), eᵝ ≈ {e_beta:.3f}", fontsize=11)
ax.legend()
plt.tight_layout(); plt.show()

Figure 8.5: Denominator growth for π’s convergents (blue) vs. Lévy’s exponential trend e^(nβ) (dashed orange). The 292-coefficient causes a dramatic jump at n = 4.

The blue dots track the orange trend almost perfectly — then the 292 coefficient fires and the denominator rockets upward in one step. Lévy’s constant, proved in 1929, jumps off the page as a single annotated spike. You needed only a list, a loop, and a semilogy call.

Coprime probability. The probability that two randomly chosen positive integers have no common factor is \(6/\pi^2 \approx 60.8\%\) (Hardy and Wright 1979, Thm. 332). This connects the geometry of circles (\(\pi\)) to the arithmetic of primes through the Riemann zeta function at \(s = 2\). (We will meet this connection again in Chapter 9.)

import random
import math

random.seed(0)
n_coprime = sum(
    1 for _ in range(100000)
    if math.gcd(
        random.randint(1, 10**6),
        random.randint(1, 10**6)
    ) == 1
)
frac = n_coprime / 100000
print(f"Observed coprime fraction: {frac:.4f}")
print(f"6/pi^2 =                   {6/math.pi**2:.4f}")

Observed coprime fraction: 0.6088
6/pi^2 =                   0.6079

math.gcd was introduced in Section 6.4. Here it is doing double duty: running 100,000 GCD computations in a loop is fast enough to confirm the theoretical value to three decimal places.

8.7 Further Research Topics

1. The Pell equation survey. For each non-square \(D\) from 2 to 20, use cf_sqrt_period and convergents to find the smallest positive integers \(x, y\) satisfying \(x^2 - Dy^2 = 1\). Record the solution and the period length for each \(D\). Which \(D\) in this range produces the largest fundamental solution? (Hint: try \(D = 13\).) For a harder challenge, extend to \(D \le 100\) and verify that \(D = 61\) gives \(x = 1766319049\), \(y = 226153980\). (Problem proposed by Claude Code.)

2. The pattern in \(e\). Compute the first 30 partial quotients of \(e = 2.71828\ldots\) (use known values if your float loses precision). Confirm or challenge the conjecture \(e = [2;\; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, \ldots]\) where every third term increments by 2. Can you write a formula for the \(n\)-th partial quotient? Search the OEIS for the sequence of partial quotients (see Chapter 8). (Problem proposed by Claude Code.)

3. Coprime probability as a function of scale. Modify the coprime experiment to count pairs \((p, q)\) with \(1 \le p, q \le N\) for \(N = 10, 100, 1000, 10000\). Plot the observed fraction vs \(N\) on a log scale and compare with the horizontal line at \(6/\pi^2\). At what \(N\) does the approximation stabilize to within 1%? (Problem proposed by Claude Code.)

4. Convergent accuracy race. For \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\), and \(\pi\), plot the number of correct decimal digits delivered by the \(n\)-th convergent as a function of \(n\). Which number gives the slowest convergence per convergent? Which gives the fastest? Connect your answer to the size of the typical partial quotients. (Problem proposed by Claude Code.)

5. Large-coefficient treasure hunt. A partial quotient much larger than its neighbors signals an unusually close rational approximation. Compute the first 30 CF coefficients of each of \(\pi, e, \sqrt{2}, \ln 2, \sqrt[3]{2}\), and the golden ratio. Report the largest coefficient found for each number and the fraction it produces. Which number has the most “surprising” near-rational value? (Problem proposed by Claude Code.)

6. Verifying Lévy’s constant (Lévy 1929). Pick any irrational of your choice (e.g., \(\sqrt{7}\)) and compute its first 50 convergents. For each \(n\), compute \(q_n^{1/n}\) where \(q_n\) is the denominator of the \(n\)-th convergent. Does the sequence appear to stabilize? To what value? Compare with \(e^{\pi^2/(12 \ln 2)} \approx 3.2758\). (Problem proposed by Claude Code.)

7. Three-distance theorem. Place \(N\) points at positions \(\lfloor n\alpha \rfloor \bmod 1\) on the interval \([0, 1)\) for \(n = 1, 2, \ldots, N\) and any irrational \(\alpha\). A theorem proved by Vera Sós in 1958 (Sós 1958) (conjectured by Steinhaus) says the \(N\) points divide \([0,1)\) into gaps of at most three distinct lengths, and the gap sizes are controlled by the CF of \(\alpha\). Verify this computationally for \(\alpha = \sqrt{2}\) and \(N = 5, 8, 13, 21, 34\) (the Fibonacci numbers). How do the gap sizes change at each Fibonacci step? (Problem proposed by Claude Code.)

8. Beyond quadratic: CF of \(\sqrt[3]{2}\). Compute the CF of \(\sqrt[3]{2} \approx 1.2599\) to 40 terms using cf_float (or a high-precision library). Does the sequence appear periodic? By Lagrange’s theorem (Lagrange 1770) it should not – only quadratic irrationals are periodic. Is the CF of \(\sqrt[3]{2}\) well-known? Search OEIS for the sequence of partial quoticients. (Problem proposed by Claude Code.)

9. Gauss-Kuzmin chi-squared test. Repeat the Gauss-Kuzmin experiment with 50,000 starting points instead of 3,000. Compute the chi-squared statistic \(\chi^2 = \sum_{k=1}^{10} (O_k - E_k)^2 / E_k\) where \(O_k\) and \(E_k\) are the observed and expected counts. A value below 16.9 indicates the match is within the 95% confidence threshold for 9 degrees of freedom. At what sample size do you first achieve \(\chi^2 < 10\)? (Problem proposed by Claude Code.)

10. Best approximation theorem verification. The theorem states that no fraction \(p/q\) with \(q \le q_n\) approximates \(\alpha\) better than the convergent \(p_n/q_n\). Write a brute-force search that checks all fractions with denominator at most \(q_n\) for a specific irrational \(\alpha\) and confirms that the convergent wins. Test it for \(\alpha = \sqrt{3}\) and the first six convergents. (Problem proposed by Claude Code.)

11. Gosper’s continued fraction arithmetic. Bill Gosper invented an algorithm for adding, multiplying, and taking square roots of continued fractions directly – without ever converting to decimal. Look up Gosper’s 1972 paper (Gosper 1972) (or a modern exposition) and implement cf_add(cf1, cf2) that takes two CF coefficient lists and returns the CF of their sum. Test it on \(\sqrt{2} + 1\), which should give \([2;\, \overline{2}]\). (Problem proposed by Claude Code.)

12. Identifying mystery constants. Use SymPy’s nsimplify function (or look ahead to Chapter 8) to try to identify the following “mystery numbers” from their continued fractions:

    1. \([0;\, 3, 4, 3, 1, 3, 1, 8, \ldots]\) (first 20 terms);
    2. the constant whose CF is \([1;\, 1, 2, 1, 1, 4, 1, 1, 6, \ldots]\) (one-half of something familiar?);
    3. the positive root of \(x^2 = x + 1\) (which you already know – verify by CF). Can you reverse-engineer a number from its partial-quotient fingerprint alone? (Problem proposed by Claude Code.)

13. Calendar design by continued fractions. The true solar year is approximately 365.24219 days, so the fractional part is \(f \approx 0.24219\). Compute the CF of \(f\) with cf_float and find its first several convergents. The Julian calendar uses \(1/4\): add a leap day every four years. The Gregorian calendar uses \(97/400\): skip three leap years every 400 years. Verify in Python which convergent is closer to the true value. What does the next convergent after \(97/400\) suggest? How many years long would a calendar based on that convergent be, and by how many seconds per year would it still drift? This is a perfect example of deliberately choosing a “good enough” approximation over the best one because the denominator fits a human-scale calendar. (Problem proposed by Claude Code.)

14. Music and the twelfth root of 2. A perfect fifth sounds pleasing because two pitches at a \(3:2\) frequency ratio have overlapping overtones. For twelve-tone equal temperament to include a near-perfect fifth, we need \(2^{7/12} \approx 3/2\), which means \(7/12\) must be a good rational approximation to \(\log_2 3 - 1\). Compute the CF of \(\log_2 3\) and find its convergents. Show that the denominators \(1, 2, 5, 12, 41, 53, 306, \ldots\) each give an equal-division scale where the “closest fifth” is especially accurate. For each denominator \(n\) in that list, compute the error \(|2^{k/n} - 3/2|\) where \(k\) is the numerator of the corresponding convergent. Plot this error vs. \(n\) on a log scale. Why does 12 win in practice even though 53 is mathematically better? (Problem proposed by Claude Code.)

15. Lamé’s theorem and the Fibonacci worst case. In 1844 Édouard Lamé proved (Lamé 1844) that the number of steps of the Euclidean algorithm on inputs \((a, b)\) with \(b \le n\) never exceeds \(\log_\varphi(n\sqrt{5})\). The worst-case pair is always consecutive Fibonacci numbers. Verify this experimentally: for each \(k\) from 2 to 30, apply the Euclidean algorithm to \((F_{k+1}, F_k)\) and count the division steps. Plot steps vs. \(k\) and confirm the growth looks like \(k \cdot \log \varphi\). Then write a brute-force search: for each \(n\) from 2 to 200, find the pair \((a, b)\) with \(a, b \le n\) that maximizes the step count. Is the winner always a pair of consecutive Fibonacci numbers? If not, what is it? (Problem proposed by Claude Code.)

16. Khinchin’s constant (Khinchin 1964). One of the strangest theorems in analysis says that for almost every real number \(x\) (exceptions have probability zero), the geometric mean of the first \(n\) partial quotients converges to the same constant \(K \approx 2.6854520010\ldots\) regardless of which \(x\) you pick. Compute the first 80 partial quotients of \(\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \pi, e, \ln 2\), and five random decimals. For each, plot the geometric mean after \(k\) terms as \(k\) goes from 5 to 80. Do the random decimals converge toward \(K\)? What about \(\sqrt{2}\) and \(\varphi\)? (They are exceptional cases — why?) At what \(k\) do the non-exceptional numbers first come within \(0.01\) of \(K\)? (Problem proposed by Claude Code.)

17. The Stern-Brocot tree (Stern 1858). Build the fraction \(3/5\) by starting with neighbors \(0/1\) and \(1/0\) and repeatedly taking the mediant \((a+c)/(b+d)\) of the two nearest bounding fractions. The left/right choices you make form a binary sequence. Now encode the CF of \(3/5\) as a binary sequence: \(a_1\) steps left, \(a_2\) steps right, \(a_3\) steps left, Verify that the two sequences are identical. Then implement stern_brocot_path(p, q) that finds any rational \(p/q\) and returns its path. Use the path to prove algorithmically that every positive rational appears in the tree exactly once. As an extension, navigate to an irrational such as \(\sqrt{2} - 1\) by truncating at 20 levels and see which fraction the tree settles on. (Problem proposed by Claude Code.)

18. Hurwitz’s theorem: how well can you approximate an irrational? Hurwitz proved (Hurwitz 1891) that for any irrational \(\alpha\) there are infinitely many rationals \(p/q\) satisfying \(|\alpha - p/q| < 1/(\sqrt{5}\,q^2)\), and \(\sqrt{5}\) is the sharpest possible constant (it cannot be increased for every irrational). Write code that, for a given \(\alpha\), finds all convergents \(p_n/q_n\) with \(q_n \le 500\) and plots the quantity \(C_n = 1/(q_n^2\,|\alpha - p_n/q_n|)\). Run it for \(\varphi\), \(\sqrt{2}\), \(\sqrt{3}\), \(e\), and \(\pi\). For which number does \(C_n\) always stay below \(\sqrt{5}\)? For which numbers does it appear to approach a larger constant? From your data, can you guess the analogous “best constant” for \(\sqrt{2}\)? (Problem proposed by Claude Code.)

19. Markov numbers and Markov triples. The Markov equation is \(x^2 + y^2 + z^2 = 3xyz\). Its positive-integer solutions are called Markov triples, and the largest element in each triple is a Markov number: \(1, 2, 5, 13, 29, 34, 89, 169, 194, \ldots\) Starting from the root triple \((1, 1, 1)\), any triple \((a, b, c)\) generates three new triples by “flipping” one element: if you replace \(c\) with \(c' = 3ab - c\), you get another solution. Write a BFS or DFS that generates all Markov triples with max element \(\le 1000\) and count them. Can you find two different triples that share the same largest element? (The Markov uniqueness conjecture (Markov 1879), unsolved since 1879, says no such pair exists.) For a deeper dive, look up how Markov numbers connect to the Lagrange spectrum of best-approximation constants. (Problem proposed by Claude Code.)

20. Minkowski’s question-mark function (Minkowski 1904). Define \(?(x)\) for \(x = [0;\, a_1, a_2, a_3, \ldots]\) by \[?(x) \;=\; 2\sum_{i=1}^{\infty}(-1)^{i+1}\,2^{-(a_1+a_2+\cdots+a_i)}.\] Implement this formula and plot \(?(x)\) for \(x \in [0, 1]\) using 200 sample points. The graph looks like a staircase that is continuous and strictly increasing yet has zero derivative almost everywhere — a “devil’s staircase.” Verify numerically: \(?(1/2) = 1/2\); \(?(\varphi - 1) = 2/3\) where \(\varphi\) is the golden ratio; and the image of every quadratic irrational under \(?\) appears to be rational. Test this last claim on \(\sqrt{2}-1\), \(\sqrt{3}-1\), and \((\sqrt{5}-1)/2\). What rational does each map to? Can you prove it from the CF definition of \(?\)? (Problem proposed by Claude Code.)

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