Lemmas 1–4 · Theorem 298
Let ξ and η be bounds with ξ > η.
The subtraction. Some multiple of η exceeds ξ (Theorem 257), so there is a least number n with n • η > ξ (Theorem 22), and n is not the unit. Let a₀ be the number before n. Then a₀ • η ≦ ξ. If they are equal, the subtraction ends. Otherwise let ξ₂ be the remainder, so that ξ = a₀ • η + ξ₂ with ξ₂ < η. Do the same with η and ξ₂, and so on. Writing ξ₀ for ξ and ξ₁ for η, the step numbered k is
ξₖ = aₖ • ξₖ₊₁ + ξₖ₊₂, with ξₖ₊₂ < ξₖ₊₁,
unless the subtraction ends there. By Theorem 285 and Theorem 286 it ends if and only if ξ and η are commensurable. From here on let them be incommensurable. Then a step is given for any k one may take. This is a rule, a function in the sense of the Specifications. Nothing is completed.
The ratios it yields. Let P₀ = a₀ and Q₀ = 1. Let P₁ = a₀ • a₁ + 1 and Q₁ = a₁. For k ≧ 1 let
Pₖ₊₁ = aₖ₊₁ • Pₖ + Pₖ₋₁ and Qₖ₊₁ = aₖ₊₁ • Qₖ + Qₖ₋₁.
Pₖ/Qₖ is called the ratio of the subtraction at step k. It is even or odd as k is; the one at step 0 counts as even.
Lemma 1. For k ≧ 1: ξ = Pₖ • ξₖ₊₁ + Pₖ₋₁ • ξₖ₊₂ and η = Qₖ • ξₖ₊₁ + Qₖ₋₁ • ξₖ₊₂.
Proof: By Axiom 3. For k = 1: η = a₁ • ξ₂ + ξ₃ = Q₁ • ξ₂ + Q₀ • ξ₃, and ξ = a₀ • (a₁ • ξ₂ + ξ₃) + ξ₂ = (a₀ • a₁ + 1) • ξ₂ + a₀ • ξ₃ = P₁ • ξ₂ + P₀ • ξ₃. If it holds for k, put ξₖ₊₁ = aₖ₊₁ • ξₖ₊₂ + ξₖ₊₃. Then ξ = Pₖ • (aₖ₊₁ • ξₖ₊₂ + ξₖ₊₃) + Pₖ₋₁ • ξₖ₊₂ = Pₖ₊₁ • ξₖ₊₂ + Pₖ • ξₖ₊₃, and likewise for η.
Lemma 2. Let A, B, C, D be numbers with A/B < C/D, and let λ, μ be bounds. Then (A • λ + C • μ)/(B • λ + D • μ) is greater than A/B and less than C/D.
Proof: A • D < B • C. So A • (B • λ + D • μ) = A • B • λ + A • D • μ < A • B • λ + B • C • μ = B • (A • λ + C • μ), which gives the first. And D • (A • λ + C • μ) = A • D • λ + C • D • μ < B • C • λ + C • D • μ = C • (B • λ + D • μ), which gives the second.
Lemma 3. Of Pₖ • Qₖ₋₁ and Pₖ₋₁ • Qₖ, one exceeds the other by the unit: the first is the greater when k is odd, the second when k is even. So two ratios of the subtraction at neighbouring steps differ by 1/(Qₖ₋₁ • Qₖ), and the odd one is the greater.
Proof: By Axiom 3. For k = 1: P₁ • Q₀ = a₀ • a₁ + 1 and P₀ • Q₁ = a₀ • a₁. For the step: Pₖ₊₁ • Qₖ = aₖ₊₁ • Pₖ • Qₖ + Pₖ₋₁ • Qₖ, and Pₖ • Qₖ₊₁ = aₖ₊₁ • Pₖ • Qₖ + Pₖ • Qₖ₋₁. So Pₖ₊₁ • Qₖ and Pₖ • Qₖ₊₁ differ as Pₖ₋₁ • Qₖ and Pₖ • Qₖ₋₁ do: by the unit, with the greater changing sides at each step.
Lemma 4. Qₖ ≧ k for k ≧ 1.
Proof: Q₁ = a₁ ≧ 1. Q₂ = a₂ • Q₁ + Q₀ ≧ 2. And Qₖ₊₁ ≧ Qₖ + Qₖ₋₁ ≧ Qₖ + 1.
Theorem 298 (reciprocal subtraction). Let ξ and η be incommensurable bounds with ξ > η. Then:
(1) Every even ratio of the subtraction is a lower ratio for ξ/η, and every odd one is an upper ratio.
(2) For any ratio R, there are two ratios of the subtraction at neighbouring steps that differ by less than R.
(3) A ratio X is a lower ratio for ξ/η if and only if it is less than some even ratio of the subtraction. It is an upper ratio if and only if it is greater than some odd one.
Proof: (1) P₀/Q₀ = a₀, and ξ = a₀ • η + ξ₂ > a₀ • η, so a₀ < ξ/η. For k ≧ 1, by Lemma 1, ξ/η = (Pₖ • ξₖ₊₁ + Pₖ₋₁ • ξₖ₊₂)/(Qₖ • ξₖ₊₁ + Qₖ₋₁ • ξₖ₊₂). The ratios at steps k − 1 and k differ (Lemma 3), so by Lemma 2, ξ/η lies between them. So if the one at step k − 1 is less than ξ/η, the one at step k is greater, and conversely. The one at step 0 is less. So by Axiom 3 the even ones are less and the odd ones greater. Then Theorem 252.
(2) By Lemmas 3 and 4, for k ≧ 2 the ratios at steps k − 1 and k differ by not more than 1/((k − 1) • k). By Theorem 208 this is less than R for some k.
(3) If X is less than an even ratio of the subtraction, it is less than ξ/η by (1). Conversely, let X < ξ/η. Take a ratio Z with X < Z < ξ/η (Theorem 253), and let R = Z − X. By (2) take an even ratio E and an odd ratio O at neighbouring steps with O < E + R. Then E + R > O > ξ/η > Z = X + R, so E > X. The other half is proved in the same way with the sides exchanged.
What this gives. For any two incommensurable bounds, the sorting of Definition 43 for their ratio is now given by a rule that works on numbers alone, one step at a time: Euclid's own subtraction. The even ratios are lower and the odd upper; neither kind has a last; and they come as near one another as one pleases. That is the relative maximum and minimum of Definition 43, reached from Theorem 41 and Theorem 285.
For the side of 2 and the unit, the first ratios of the subtraction are 1, 3/2, 7/5, 17/12. These are the side and diagonal numbers that Theon of Smyrna reports.