Rational Bounds and Whole Bounds

Theorem 248 (Landau 154). If X > Y, or X = Y, or X < Y, then X* > Y*, or X* = Y*, or X* < Y*, respectively, and conversely.

If X > Y, then Y is lower for X* and upper for Y*, so X* > Y*. If X = Y, clearly X* = Y*. The third follows from the first. The converse holds because the cases are exclusive and exhaustive.

Theorem 249 (Landau 155). (X + Y)* = X* + Y*. (X − Y)* = X* − Y* if X > Y. (X•Y)* = X*•Y*. (X/Y)* = X*/Y*.

Proof: I) A lower ratio for X* + Y* is the sum of a ratio less than X and a ratio less than Y, so it is less than X + Y. Conversely, if U < X + Y, then U/(X + Y) < 1, and U = X•(U/(X + Y)) + Y•(U/(X + Y)) is the sum of a ratio less than X and a ratio less than Y. II) If X > Y, then X = (X − Y) + Y, so X* = (X − Y)* + Y* by I). III) A lower ratio for X*•Y* is the product of a ratio less than X and one less than Y, so it is less than X•Y. Conversely, if U < X•Y, take U1 with U < U1 < X•Y (Theorem 184). Then U/U1 < 1 and U1/Y < X, and U = (U1/Y)•(Y•(U/U1)) is the product of a ratio less than X and one less than Y. IV) X = (X/Y)•Y, so X* = (X/Y)* • Y* by III).

Theorem 250 (Landau 156). The whole bounds answer one to one to the unit and the numbers, 1* to the unit and (A′)* to the successor of what answers to A*. Order, sum and product agree. So Axioms 1 to 3 hold of whole bounds.

By Theorems 248 and 249, as in Theorem 206.

Theorem 251 (Landau 157). The rational bounds are those that have a least upper ratio X, and the bound is then X.

For X* the least upper ratio is X. Conversely, if ξ has a least upper ratio X, then every lower ratio is less than X and every upper ratio is not less than X, so ξ = X*.

Theorem 252 (Landau 158). X is a lower ratio for ξ if and only if X < ξ, and an upper ratio if and only if X ≧ ξ.

If X is lower for ξ, then since X is upper for X*, X < ξ. If X is the least upper ratio for ξ, then X = ξ by Theorem 251. If X is upper but not least, take an upper X1 < X; it is lower for X*, so X > ξ.

Theorem 253 (Landau 159). If ξ < η, there is a ratio Z with ξ < Z < η.

Take an upper ratio X for ξ that is lower for η, and a greater lower ratio Z for η. By Theorem 252, ξ ≦ X < Z < η.

Theorem 254 (Landau 160). Every ratio Z > ξ•η can be written Z = X•Y with X ≧ ξ and Y ≧ η.

Proof: Let ζ be the lesser of 1 and (Z − ξ•η)/((ξ + η) + 1). By Theorem 253 take ratios Z1 and Z2 with ξ < Z1 < ξ + ζ and η < Z2 < η + ζ. Then Z1•Z2 < (ξ + ζ)•(η + ζ) = (ξ + ζ)•η + (ξ + ζ)•ζ ≦ (ξ + ζ)•η + (ξ + 1)•ζ = ξ•η + ((ξ + η) + 1)•ζ ≦ ξ•η + (Z − ξ•η) = Z. Let X = Z/Z2 and Y = Z2. Then X > (Z1•Z2)•(1/Z2) = Z1 > ξ, and Y > η.

Theorem 255 (Landau 161). For each ζ, the equation ξ•ξ = ζ has exactly one solution.

Proof: I) At most one: if ξ1 > ξ2 then ξ1•ξ1 > ξ2•ξ2 by Theorem 241. II) Call a ratio X lower when X•X < ζ, and upper otherwise. This gives a bound. (1) If X < 1 and X < ζ, then X•X < X < ζ, so X is lower. If X ≧ 1 and X ≧ ζ, then X•X ≧ X ≧ ζ, so X is upper. (2) If X•X < ζ and Y < X, then Y•Y < X•X < ζ. (3) Let X•X < ζ. Take a ratio Z less than the lesser of 1 and (ζ − X•X)/(X + (X + 1)). Then X + Z > X, and (X + Z)•(X + Z) = (X + Z)•X + (X + Z)•Z < (X•X + Z•X) + (X + 1)•Z = X•X + (X + (X + 1))•Z < X•X + (ζ − X•X) = ζ. Call this bound ξ. If ξ•ξ > ζ, take a ratio Z with ξ•ξ > Z > ζ (Theorem 253). Being lower for ξ•ξ, Z = X1•X2 with X1 and X2 lower for ξ. If X is the greater of them, Z ≦ X•X < ζ, which is impossible. If ξ•ξ < ζ, take Z with ξ•ξ < Z < ζ. By Theorem 254, Z = X1•X2 with X1 ≧ ξ and X2 ≧ ξ. If X is the lesser of them, X is upper for ξ, so Z ≧ X•X ≧ ζ, which is impossible.

Remark. The sorting used here escapes a circle. The square root of ζ is not defined as what lies between the ratios less and greater than itself. Ratios are sorted by a test that does not mention it: whether the square is less than ζ.

Theorem 256 (Landau 162). There is an irrational bound.

Proof: The solution ξ of ξ•ξ = 2 (Theorem 255) is irrational. Otherwise ξ is a ratio. Let a, b be the least numbers in that ratio (Theorem 72); they are prime to one another (Theorem 61). Then the square on a is to the square on b as 2 is to the unit. The squares on a and b are prime to one another (Theorem 66), so they are the least in their ratio (Theorem 60). But 2 and the unit are also the least in that ratio. So the square on a is 2. This is impossible: the square on the unit is the unit, and the square on any number is not less than 4 (Theorem 38).

This is the first case of Theorem 292, which proves by the same steps that the side of any number that is not square is irrational.