Prove why convergents are close and reduced

Try

Follow the successive truncations of [1; 2, 2, 2,…], the continued fraction for the square root of two. They begin one, three-halves, seven-fifths and seventeen-twelfths. Compute their squares to determine which lie above the square root. The approximants alternate sides, and adjacent fractions are unusually close. We want a proof that explains both patterns without relying on a decimal display of the irrational target.

Predict

Predict whether multiplying all previous numerators and denominators is necessary when another coefficient arrives. It is enough to retain the last two of each. Set p at index minus two to zero, p at minus one to one, q at minus two to one, and q at minus one to zero. For index n≥0 define p_n=a_n p_(n−1)+p_(n−2) and q_n=a_n q_(n−1)+q_(n−2). Backward substitution, or multiplying the two-by-two matrices with rows (a_n, 1),(1, 0), proves that p_n/q_n is the truncated fraction.

Experiment

Fill the last row independently and compute each adjacent cross-product.

Index	Coefficient	Numerator	Denominator	p²−2q²
0	1	1	1	−1
1	2	3	2	1
2	2	7	5	−1
3	2	17	12	1

For example, 7·2−3·5=−1. The difference between seven-fifths and three-halves is therefore exactly −1/10. These are signed differences, so taking absolute values too early would hide alternation.

Understand

Write D_n=p_n q_(n−1)−p_(n−1)q_n. Substitution of the recurrences cancels the two terms containing a_n and yields D_n=−D_(n−1). At n=0 the initial values give D_0=−1; hence D_n=(−1)^(n−1). Any common divisor of numerator and denominator divides this determinant, proving every convergent reduced. For an irrational α let t be its complete quotient after coefficient a_n, so t>a_(n+1)>0. The exact tail formula is α=(p_n t+p_(n−1))/(q_n t+q_(n−1)). Subtracting p_n/q_n gives absolute error 1/[q_n(q_n t+q_(n−1))], which is less than 1/(q_n q_(n+1)). The denominators eventually strictly increase and tend to infinity, proving convergence.

Apply

The signed numerator in the same subtraction proves alternation around α. The error is eventually less than 1/q_n² because q_(n+1)≥q_n. That inequality is an upper bound, not an exact error or a guarantee for arbitrary fractions. A useful geometric comparison is the neighboring-fraction lemma: if a/b< c/d and bc−ad=1, any fraction strictly between them has denominator at least b+d. Indeed, the positive integer differences br−as and cs−dr combine to give s=d(br−as)+b(cs−dr). This explains why adjacent convergents leave little room for a competing fraction with a small denominator.

Continue

Use the recurrence to approximate a measurement with a denominator budget, then report both the rational candidate and the theorem that bounds its error. Do not silently replace that theorem by a general claim of best approximation in every possible metric. Stronger best-approximation and Legendre criteria can be proved separately. The bound established here already supplies infinitely many close fractions and will be enough to prove that Pell equations have nontrivial positive solutions.

Return to the number theory unit to choose a targeted retry. Continue with build and classify solutions of pell equations.