The fraction 355/113 is famously close to π. According to continued fraction theory, what makes it so special compared to other fractions with denominator ≤ 113?
AIt is a convergent of π's continued fraction expansion, so no fraction with a smaller denominator is closer to π
BIt was computed by truncating the decimal expansion of π to 3 digits in numerator and denominator
CIt is the unique fraction of that denominator that equals π when rounded to 6 decimal places
DIt minimizes the sum of numerator and denominator while staying within 0.001 of π
355/113 is a convergent of π's continued fraction expansion. A fundamental theorem states that convergents are the best rational approximations: no fraction with a smaller denominator lies closer to the real number. This is much stronger than just being accurate — it means 355/113 is optimal among all fractions p/q with q ≤ 113.
Question 2 Multiple Choice
A student is told that √2 has a periodic continued fraction expansion. They conclude that π must also have a periodic expansion since both are irrational. What theorem directly refutes this reasoning?
AOnly quadratic irrationals — numbers of the form (p + √q)/r — have eventually periodic continued fraction expansions
BOnly algebraic numbers of any degree can have periodic continued fraction expansions
CPeriodic expansions are only possible for rational numbers, not irrationals
DThe periodicity of √2 is a special coincidence with no general theorem behind it
EAny irrational with bounded partial quotients has a periodic expansion
Lagrange's theorem states that a real number has an eventually periodic continued fraction if and only if it is a quadratic irrational — a number of the form (p + √q)/r with p, q, r integers. π is transcendental (not even algebraic), so its continued fraction cannot be periodic. Irrationality alone is insufficient.
Question 3 True / False
The continued fraction expansion of every rational number terminates in finitely many steps.
TTrue
FFalse
Answer: True
True. Expanding a rational number p/q via the continued fraction algorithm is equivalent to running the Euclidean algorithm on p and q. Since the Euclidean algorithm always terminates (remainders strictly decrease and are non-negative integers), the continued fraction expansion of any rational number terminates in finitely many steps.
Question 4 True / False
Any irrational number with a periodic continued fraction expansion is expected to be transcendental.
TTrue
FFalse
Answer: False
False. It is exactly the opposite: any irrational with a periodic continued fraction is a quadratic irrational (a root of a degree-2 polynomial with integer coefficients), which is algebraic, not transcendental. Transcendental numbers like e and π cannot have periodic continued fraction expansions.
Question 5 Short Answer
Why do the convergents of a continued fraction give the best rational approximations to a real number, rather than just good ones?
Think about your answer, then reveal below.
Model answer: Convergents are optimal in the sense that no fraction with a smaller denominator lies closer to the target number. The algorithm that generates them — mimicking the Euclidean algorithm — discards only information about how much the approximation can improve, keeping only the partial quotients that give the largest improvement per unit of denominator size. This means a convergent beats every fraction p/q with q smaller than its own denominator.
The key is the combination of two facts: (1) the error of each convergent pₙ/qₙ satisfies |x − pₙ/qₙ| < 1/(qₙ·qₙ₊₁), and (2) any non-convergent fraction with denominator between qₙ and qₙ₊₁ has larger error than pₙ/qₙ. Together these prove optimality — being a convergent is both necessary and sufficient for best-approximation status among fractions with that denominator.