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Convergents are best rational approximations of the first kind

Statement

Let α be an irrational real number with convergents pn/qn, and let n1. If r,sZ with s>0 and sαr<qnαpn, then sqn+1. Consequently, no rational number with denominator at most qn approximates α more closely than pn/qn.

Facts & Assumptions

Given: An irrational real number α, an index n1, its convergents pn/qn, and integers r,s with s>0.

[F1]

Since n1, the corollary Convergents are reduced fractions applied at the index n1 shows that the vectors (qn,pn),(qn1,pn1) form a Z-basis of Z2.

[F2]

The complete-quotient algorithm produces a unique integer an+1 with an+1αn+1<an+1+1, and every later complete quotient satisfies αn+1>1 (Complete quotients in the continued-fraction algorithm).

[F3]

If αn+1 is the next complete quotient, then α=αn+1pn+pn1αn+1qn+qn1. (Complete-quotient tail formula).

[F4]

The convergent denominators satisfy q1=0,q0=1,qn+1=an+1qn+qn1. (Convergents of a regular continued fraction)

[F5]

The convergent errors satisfy αpnqn=(1)nqn(αn+1qn+qn1),αpnqn<1qnqn+1. (Convergent error bound).

Proof

technique · direct
1.1

From [F3] and [F5] one obtains. [F3, F5, algebra] qnαpn=(1)nαn+1qn+qn1,qn1αpn1=αn+1(qnαpn). So the consecutive errors have opposite signs and satisfy qn1αpn1=αn+1qnαpn.

F3F5algebra
2.1

By the basis statement in [F1], there are unique integers u,v with. [F1, step 1.1, algebra] (sr)=u(qnpn)+v(qn1pn1). Subtracting r from sα gives sαr=u(qnαpn)+v(qn1αpn1)=(uvαn+1)(qnαpn) by step 1.1.

F1step 1.1algebra
3.1

Assume sαr<qnαpn. Step 2.1 gives. [step 2.1, F2, algebra] uvαn+1<1. If v<0, then u<0 as well, because otherwise uvαn+11+αn+1>1; but then s=uqn+vqn1<0, impossible. If v=0, then u<1, so u=0 and again s=0, impossible. Therefore v>0. Since u is an integer and [F2] gives αn+1>an+1, the inequality above implies u>vαn+11>van+11, hence uvan+1.

step 2.1F2algebra
4.1

Now. [step 3.1, F4, algebra] s=uqn+vqn1v(an+1qn+qn1)=vqn+1qn+1, which is the first claim. Because n1, fact [F2] gives an+11, and [F4] then gives qn+1=an+1qn+qn1>qn. For the consequence, suppose sqn and αrs<αpnqn. Then sαr=sαrs<sαpnqnqnαpnqn=qnαpn, contradicting the first claim because s<qn+1. Thus no denominator at most qn gives a closer rational approximation.

step 3.1F2F4algebra

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