Definition 68 (Landau 52). The sum Ξ + Θ is the operation of doing Ξ and then Θ. For positive Ξ and Θ it is the sum of Definition 49. Otherwise, with ξ = |Ξ| and η = |Θ|:
If both are negative, Ξ + Θ = −(ξ + η).
If Ξ is positive and Θ negative, Ξ + Θ is +(ξ − η) when ξ > η, is 0 when ξ = η, and is −(η − ξ) when ξ < η.
If Ξ is negative and Θ positive, Ξ + Θ = Θ + Ξ.
If Θ = 0, Ξ + Θ = Ξ. If Ξ = 0, Ξ + Θ = Θ.
These cases are not stipulated. They are what the two operations yield when done one after the other on any bound great enough for both to apply. Adding ξ and then removing a lesser η leaves ξ − η added (Theorem 234); adding ξ and then removing a greater η leaves η − ξ removed.
Theorem 309 (Landau 175). Ξ + Θ = Θ + Ξ. (Commutative Law of Addition)
Proof: For positive Ξ and Θ, Theorem 224. If both are negative, −(ξ + η) = −(η + ξ) by Theorem 224. If one is positive and one negative, by the third case of Definition 68. If either is 0, by the last case.
Definition 69 (Landau 53). −Ξ is 0 if Ξ = 0, and is |Ξ| if Ξ is negative. For positive Ξ it is already defined. −Ξ is the contrary of Ξ: it undoes what Ξ does.
Theorem 310 (Landau 176). If Ξ > 0, Ξ = 0, Ξ < 0, then −Ξ < 0, −Ξ = 0, −Ξ > 0 respectively, and conversely.
Definitions 59 and 69 and Theorem 303.
Theorem 311 (Landau 177). −(−Ξ) = Ξ.
If Ξ is +ξ, then −Ξ is −ξ and −(−ξ) is |−ξ| = ξ. If Ξ is 0, obvious. If Ξ is −ξ, then −Ξ is ξ and −ξ is Ξ.
Theorem 312 (Landau 178). |−Ξ| = |Ξ|. Definitions 60 and 69.
Theorem 313 (Landau 179). Ξ + (−Ξ) = 0.
If Ξ = 0, obvious. Otherwise one of Ξ, −Ξ is positive and the other negative, with the same absolute value, so the sum is 0 by Definition 68.
Theorem 314 (Landau 180). −(Ξ + Θ) = −Ξ + (−Θ).
Proof: By Theorem 309 and the symmetry of the claim, it is enough to take these cases. (1) Θ = 0. Both sides are −Ξ.
(2) Both positive. −Ξ and −Θ are negative, and their sum is −(ξ + η), which is −(Ξ + Θ).
(3) Both negative. Ξ + Θ = −(ξ + η), so −(Ξ + Θ) = ξ + η = −Ξ + (−Θ).
(4) Ξ positive, Θ negative. Then −Ξ is −ξ and −Θ is η. If ξ > η: Ξ + Θ = ξ − η, and −Ξ + (−Θ) = η + (−ξ) = −(ξ − η). If ξ = η: both sides are 0. If ξ < η: Ξ + Θ = −(η − ξ), and −Ξ + (−Θ) = η + (−ξ) = η − ξ.
Definition 70 (Landau 54). Ξ − Θ = Ξ + (−Θ). It is the difference of Ξ and Θ.
For positive Ξ > Θ this agrees with Definition 50, by Definition 68.
Theorem 315 (Landau 181). −(Ξ − Θ) = Θ − Ξ.
−(Ξ + (−Θ)) = −Ξ + (−(−Θ)) = −Ξ + Θ = Θ + (−Ξ), by Theorems 314, 311 and 309.
Theorem 316 (Landau 182). Ξ − Θ is positive, 0, or negative according as Ξ > Θ, Ξ = Θ, Ξ < Θ.
Proof: The three cases on each side exclude one another and exhaust the possibilities (Theorem 301), so it is enough to prove each case from left to right.
If Ξ = Θ: Theorem 313.
If Ξ > Θ: (1) Both positive: −Θ is negative with |−Θ| < |Ξ|, so Ξ + (−Θ) is +(ξ − η). (2) Ξ positive, Θ = 0: Ξ − Θ = Ξ. (3) Ξ positive, Θ negative: −Θ is positive, and a sum of positives is positive. (4) Ξ = 0, Θ negative: Ξ − Θ = −Θ, positive. (5) Both negative, with ξ < η: −Θ is +η, and η + (−ξ) is +(η − ξ).
If Ξ < Θ: then Θ > Ξ, so Θ − Ξ is positive, and Ξ − Θ = −(Θ − Ξ) is negative by Theorems 315 and 310.
Theorem 317 (Landau 183). If Ξ > Θ, Ξ = Θ, Ξ < Θ, then −Ξ < −Θ, −Ξ = −Θ, −Ξ > −Θ respectively.
(−Θ) − (−Ξ) = −Θ + Ξ = Ξ − Θ, by Theorems 311 and 309. Then Theorem 316.
Theorem 318 (Landau 184). Every real number is the difference of two positive numbers.
If Ξ > 0: Ξ = (Ξ + 1) − 1. If Ξ = 0: Ξ = 1 − 1. If Ξ < 0: Ξ = 1 − (|Ξ| + 1), since 1 < |Ξ| + 1 and (|Ξ| + 1) − 1 = |Ξ|.
Theorem 319 (Landau 185). For positive ξ₁, ξ₂, η₁, η₂: (ξ₁ − ξ₂) + (η₁ − η₂) = (ξ₁ + η₁) − (ξ₂ + η₂).
Proof: (1) Let ξ₁ > ξ₂, with ξ₁ = ξ₂ + α.
If η₁ > η₂, with η₁ = η₂ + β: ξ₁ + η₁ = (ξ₂ + η₂) + (α + β), so both sides are α + β.
If η₁ = η₂: ξ₁ + η₁ = (ξ₂ + η₂) + α, so both sides are α.
If η₁ < η₂, with η₂ = η₁ + β: the left side is α + (−β). When α > β, with α = β + γ: ξ₁ + η₁ = (ξ₂ + η₂) + γ, and both sides are γ. When α = β: ξ₁ + η₁ = ξ₂ + η₂, and both sides are 0. When α < β, with β = α + γ: ξ₂ + η₂ = (ξ₁ + η₁) + γ, and both sides are −γ.
(2) Let ξ₁ = ξ₂. The left side is η₁ − η₂. The right side is (ξ₁ + η₁) − (ξ₁ + η₂), which is η₁ − η₂ in each of the cases η₁ > η₂, η₁ = η₂, η₁ < η₂, by Theorems 225 and 228.
(3) Let ξ₁ < ξ₂. By (1) with the indices exchanged, (ξ₂ − ξ₁) + (η₂ − η₁) = (ξ₂ + η₂) − (ξ₁ + η₁). Taking the contrary of each side, by Theorems 314 and 315, gives the claim.
Theorem 320 (Landau 186). (Ξ + Θ) + Λ = Ξ + (Θ + Λ). (Associative Law of Addition)
Proof: By Theorem 318, Ξ = ξ₁ − ξ₂, Θ = η₁ − η₂, Λ = ζ₁ − ζ₂ with all six positive. By Theorem 319 twice,
(Ξ + Θ) + Λ = ((ξ₁ + η₁) + ζ₁) − ((ξ₂ + η₂) + ζ₂),
Ξ + (Θ + Λ) = (ξ₁ + (η₁ + ζ₁)) − (ξ₂ + (η₂ + ζ₂)).
These are the same by Theorem 225.
Theorem 321 (Landau 187). For given Ξ and Θ, the equation Θ + Φ = Ξ has exactly one solution, Φ = Ξ − Θ.
Proof: It is a solution: Θ + (Ξ − Θ) = (Ξ + (−Θ)) + Θ = Ξ + ((−Θ) + Θ) = Ξ + 0 = Ξ. It is the only one: if Θ + Φ = Ξ, then Ξ − Θ = (−Θ) + (Θ + Φ) = ((−Θ) + Θ) + Φ = Φ.
So subtraction is now always possible. In Theorem 234 it needed ξ > η. The restriction belonged to bounds, not to the operations on them.
Theorem 322 (Landau 188). Ξ + Λ > Θ + Λ, Ξ + Λ = Θ + Λ, Ξ + Λ < Θ + Λ according as Ξ > Θ, Ξ = Θ, Ξ < Θ.
(Ξ + Λ) − (Θ + Λ) = (Ξ + Λ) + ((−Λ) + (−Θ)) = Ξ − Θ, by Theorems 314, 320 and 313. Then Theorem 316.
Theorem 323 (Landau 189). If Ξ > Θ and Λ > Φ, then Ξ + Λ > Θ + Φ.
Ξ + Λ > Θ + Λ and Θ + Λ = Λ + Θ > Φ + Θ = Θ + Φ, by Theorem 322. Then Theorem 305.
Theorem 324 (Landau 190). If Ξ ≧ Θ and Λ > Φ, or Ξ > Θ and Λ ≧ Φ, then Ξ + Λ > Θ + Φ.
With equality, Theorem 322. Otherwise Theorem 323.
Theorem 325 (Landau 191). If Ξ ≧ Θ and Λ ≧ Φ, then Ξ + Λ ≧ Θ + Φ.
Obvious with two equalities. Otherwise Theorem 324.