Theorem 339 (Landau 205). Let it be determined of any real number whether it is of the first kind or of the second, such that:
(1) some number is of the first kind and some number is of the second;
(2) every number of the first kind is less than every number of the second.
Then there is exactly one real number Ξ such that every number less than Ξ is of the first kind and every number greater than Ξ is of the second.
Proof: A) There is at most one. Let Ξ₁ < Ξ₂ both serve, and let M = (Ξ₁ + Ξ₂)/2. From Ξ₁ + Ξ₁ < Ξ₁ + Ξ₂ < Ξ₂ + Ξ₂ (Theorem 322) and Theorem 337, Ξ₁ < M < Ξ₂. Since M > Ξ₁, M is of the second kind. Since M < Ξ₂, M is of the first kind. This is impossible.
B) There is one. Note first, from (2): any number less than a number of the first kind is of the first kind, and any number greater than a number of the second kind is of the second kind. Three cases.
I) Some positive number P is of the first kind.
Sort the ratios as follows. A ratio X is lower if X is of the first kind and some number of the first kind is greater than X. Otherwise X is upper. This is a sorting in the sense of Definition 43:
(a) There is a lower ratio. Any lower ratio X for P is less than P, so it is of the first kind and P is a greater number of the first kind.
(b) There is an upper ratio. Let Q be of the second kind. Q > P, so Q is positive. Any upper ratio X for Q is not less than Q, so it is of the second kind, and so upper.
(c) Every lower ratio is less than every upper ratio. Let X be lower and Y upper. If Y is of the second kind, X < Y by (2). If Y is of the first kind, then no number of the first kind is greater than Y, so X ≦ Y, and X is not Y.
(d) No lower ratio is greatest. Let X be lower, and Λ of the first kind with Λ > X. By Theorem 253 there is a ratio X′ with X < X′ < Λ. X′ is of the first kind and Λ is greater, so X′ is lower. Let ξ be the bound of this sorting and Ξ = +ξ.
Let Θ < Ξ. If Θ ≦ 0, then Θ < P and Θ is of the first kind. If Θ > 0, take a ratio X with Θ < X < Ξ (Theorem 253). X is a lower ratio for ξ, so of the first kind, and so Θ is.
Let Θ > Ξ. Take a ratio X with Ξ < X < Θ. X is an upper ratio for ξ. So either X is of the second kind, and then Θ is; or X is of the first kind and no number of the first kind is greater, and then Θ, being greater, is of the second kind.
II) Every positive number is of the second kind, and some negative number is of the second kind.
Call Θ of the first kind′ when −Θ is of the second kind, and of the second kind′ when −Θ is of the first kind. By Theorem 317, conditions (1) and (2) hold for the kinds′, and some positive number is of the first kind′. By I) there is Ξ′ for the kinds′. Then Ξ = −Ξ′ serves, by Theorem 317 again.
III) Every positive number is of the second kind and every negative number is of the first kind. Then Ξ = 0 serves.
These three cases are all: either some positive number is of the first kind, or none is; and if none is, either some negative number is of the second kind, or none is. Whether Ξ itself is of the first or the second kind does not matter and is not settled by the theorem.
What the theorem says. Ξ is the relative maximum of the first kind and the relative minimum of the second: nothing of the first kind exceeds it, nothing of the second falls short of it, and nothing lies between. This is the completeness of the real numbers, and it is reached without a set of real numbers and without a division of any whole. The proof needs only: a rule of two kinds, the ratios one at a time, and Definition 43.