Dedekind cut problem in rudin; Problem 20, Chapter 1, Rudin

[saim_]

I need help with the last part of Problem 20, Chapter 1 of Rudin. Here's the problem:

"With reference to the Appendix, suppose that property (III) were omitted from the definition of a cut. Keep the same definitions of order and addition. Show that the resulting ordered set has the least-upper-bound property, that addition satisfies axioms (A1) to (A4) (with a slightly different zero-element!) but that (A5) fails."

I'm just having trouble proving why A5 would fail without property (III).

Note: The appendix, referred to here, contains construction of reals using Dedekind cuts; property (III) is the requirement that a cut have no largest number; properties A1 to A4 are field axioms of closure, associativity, commutativity and existence of identify for addition. A5 is the property of existence of an additive inverse.

 

[micromass]

Hi saim!

Let $\mathbb{Q}^-$ be the set of all negative rationals (excluding zero). You've probably found out that $\mathbb{Q}^-\cup\{0\}$ is the identity for addition. Maybe you can show that $\mathbb{Q}^-$ does not have an additive inverse...

 

[HallsofIvy]

I thought micromass had gone off the rails until I realized he was talking about the situation in which axiom A5 fails!

 

[Khichdi lover]

Hi saim!

Let $\mathbb{Q}^-$ be the set of all negative rationals (excluding zero). You've probably found out that $\mathbb{Q}^-\cup\{0\}$ is the identity for addition. Maybe you can show that $\mathbb{Q}^-$ does not have an additive inverse...

here is my attempt at a proof :-
let us assume that $\mathbb{Q}^-$ has an additive inverse. Let us call it β.
Then $\mathbb{Q}^-$ + β = $\mathbb{Q}^-\cup\{0\}$

Then there is a p $\in$ $\mathbb{Q}^-$ and a q $\in$ β such that
p + q = 0. (since 0 is an element of $\mathbb{Q}^-\cup\{0\}$).

But p is negative , thus q has to be positive.
Now , there exists a rational q1 such that 0<q1<q .

Further , $\mathbb{Q}^-$ contains the rational -q1.
By definition of addition ,the set $\mathbb{Q}^-$ + β
must contain an element
s = (-q1) + q.
But s is positive. Thus the set $\mathbb{Q}^-$ + β also has some positive rationals as it's elements, and is hence not the additive identity.

Thus there doesn't exist an additive inverse for $\mathbb{Q}^-$ .

*I wanted to know if this is a valid/proper way of proving what was asked in Rudin. I am new to analysis so wanted to better my proving skills. Sorry for posting in an old thread, but what I wanted to ask was being discussed here. Plus I was struck at precisely the point at which the hint is provided by the PF member - micromass.*