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.

  • $\begingroup$ Point (b) in the definition of cut is incomplete $\endgroup$
    – NazimJ
    Feb 20, 2020 at 22:35
  • $\begingroup$ Use the same properties of addition in $\Bbb Q$. $\endgroup$
    – Berci
    Feb 21, 2020 at 0:03
  • 1
    $\begingroup$ Try to prove these properties and let us know which steps are bothering you. The proofs are almost trivial. $\endgroup$
    – Paramanand Singh
    Feb 21, 2020 at 2:35
  • $\begingroup$ @Berci said i can use the same properties of addition in Q, could you elaborate on that? $\endgroup$
    – BOI DOI
    Feb 21, 2020 at 2:35

1 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


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