Best Rational Approximation Calculator

Rational Approximation

Every irrational number (like π or √2) can be approximated by a fraction p/q. The challenge is finding the best fraction, the one that gives the smallest error for the smallest denominator.

Continued Fractions

Any real number can be written as a continued fraction: x = a₀ + 1/(a₁ + 1/(a₂ + 1/(a₃ + …)))

Written as [a₀; a₁, a₂, a₃, …]

The convergents (truncations of this expansion) give the best rational approximations, better than any other fraction with a smaller denominator.

Famous Examples

Why 355/113 is remarkable

355/113 = 3.14159292…, which matches π to six decimal places using only three digits top and bottom. What makes it stand out is the jump. The convergent before it, 333/106, is already good at 0.00265%, but 355/113 is roughly 300 times better while adding only 7 to the denominator. That leap happens because the next term in π’s continued fraction expansion is 292, an unusually large number, and a large term means the truncation before it was already very close. The Chinese mathematician Zu Chongzhi found 355/113 around 480 AD and called it the 密率, the “close ratio”. Europe did not match it for another thousand years.

Convergents are not the whole story

A convergent is the best approximation in a strict sense: no fraction with a smaller denominator gets closer. But when you cap the denominator at some awkward value, the winner is often a semiconvergent, formed by taking a partial step toward the next convergent.

Cap the denominator at 100 and ask for π. The convergents available are 3/1 and 22/7, since 333/106 is already too big. Yet 311/99 fits comfortably and is about seven times closer than 22/7. It is a semiconvergent, sitting fourteen steps of the way from 22/7 toward 333/106.

This calculator checks the semiconvergents as well, which is why it can return a fraction you will not find in a table of convergents.

The Stern-Brocot Tree

Every positive fraction appears exactly once in this infinite binary tree, in lowest terms. Walking down it by repeatedly taking the mediant (a+c)/(b+d) of the two fractions bracketing your target generates exactly the convergents and semiconvergents above, which is why the two methods always agree.