Аксиомы вещественных чисел

Следствие 1

Единственность нуля.

Доказательство

Пусть нуля два θ₁ и θ₂. Тогда θ₁ = θ₁ + θ₂ = θ₂ + θ₁ = θ₂

Следствие 2

−(−x) = x

Доказательство

−(−x) = θ + (−(−x)) = x + (−x) + (−(−x)) = x + θ = x

Следствие 3

θ ⋅ x = θ, ∀ x ∈ R

Доказательство

x = x ⋅ 1 = x(1 + θ) = x + x ⋅ θ = x + θ ⋅ x → θ ⋅ x = θ

Следствие 4

(–1)⋅(–1) = 1

Доказательство

θ = (1 + (−1))⋅(1 + (−1)) = 1⋅1 + (−1)⋅1 + 1⋅(−1) + (−1)⋅(−1) = 1 + (−1) + (−1) + (−1)⋅(−1) =
= (−1) + (−1)⋅(−1) → (−1)⋅(−1) = 1