# Dedekind Cuts and showing that they satisfy the additive axioms

Let α,β be cuts and let α+β={r+s|r∈αands∈β}. How can I show that for all cuts in R with the addition defined here, we can satisfy the additive axioms of commutativity, closure, identity, inverse, and associativity? (A1)-(A5).

We have defined a cut to be:

A subset α of Q (the rationals) is said to be a cut if:

a.) the set α =/ null set and α =/ Q.

b.) if r is in α and s is in Q satisfies $$s, then s is in α;

c.) if r is in α, then there exists s in Q with s>r and s in α.

• Point (b) in the definition of cut is incomplete Feb 20, 2020 at 22:35
• Use the same properties of addition in $\Bbb Q$. Feb 21, 2020 at 0:03
• Try to prove these properties and let us know which steps are bothering you. The proofs are almost trivial. Feb 21, 2020 at 2:35
• @Berci said i can use the same properties of addition in Q, could you elaborate on that? Feb 21, 2020 at 2:35

Commutativity and Associativity are almost identical to prove, so I will do Commutativity: $$\alpha + \beta = \{ r+s : r \in \alpha,s\in\beta \} = \{ s+r : r \in \alpha,s\in\beta \} = \beta + \alpha$$
For closure, consider cuts $$\alpha$$ and $$\beta$$. You have to show that the set $$\gamma = \{ r+s : r \in \alpha,s\in\beta\}$$ satisfies all 3 conditions of the definition of cut. This is purely an excercise in verifying the definition. If you get stuck leave a comment about where/why
Identity, my hint is to consider the cut $$\theta=\{ r \in \mathbb{Q}:r < 0 \}$$. Obviously you have to prove that $$\theta$$ is even a cut in the first place
Inverse, I will leave to you. For any cut $$\alpha$$, find a $$\beta := \alpha^{-1}$$ such that $$\alpha + \beta = \theta$$. It might help to remember that in $$\mathbb{Q}$$, the additive inverse of $$r$$ is $$-r$$ and vice versa