Dedekind Cuts and showing that they satisfy the additive axioms

Question

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 $s<r$, then s is in α;

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

Answer 1

Once you start I think it you can get the hang of it. So let's get some of it started:

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