Trouble understanding a part of Rudin's construction of the reals - the additive inverse of a real I was studying Rudin's Principle of Mathematical Analysis. In his construction of the reals there is a part I am having trouble comprehending. It concerns the additive inverse of the real. 
I understand that for a real number $\alpha$ , which is a cut, the additive inverse is given by the cut $\beta$ - which is the set of all rational numbers $p$ with the property that , there exists a rational $r > 0$ such that $-p-r \notin \alpha $ .
Now we need to show that $\alpha + \beta$ is indeed $0$ .
For this we need to show that the cut given by $\alpha + \beta$ is a subset of the cut given by $0$ and the cut given by $0$ is a subset of $\alpha + \beta$ . 
I am having trouble understanding proof of the second part viz. the cut given by $0$ is a subset of $\alpha + \beta$ .
As is customary let us denote the cut representing the real $0$ as $0^*$ .
Here is how Ruding approaches the proof :- 
Pick a rational $v \in 0^*$ put $w = -v/2$ . Then , $w>0$ and there is an integer $n$ such that $nw \in \alpha$ but $(n+1)w \notin \alpha$ . This is the statement I am having trouble proving rigorously. 
Rudin states that this follows from the fact that $\mathbb Q$ has the Archimedean property. Now I appreciate that $\mathbb Q$ has the Archimedean property and that  it can be proven without recourse to a least upper bound property which $\mathbb Q$ conspicuously lacks.
But I am unable to prove the existence of $n$ rigorously.
I am having trouble doing so because the set of rationals(cut) $\alpha$ does not have a largest element. 
Here is how I could somewhat proceed :- 
Consider the set of integers(let us call it $I$), $m$ such that $mw \in \alpha$ . This set is non-empty . As we know $\alpha$ is non empty and therefore it has at least one rational say $q$ . Now since $\mathbb Q$ has the Archimedean property there exists an integer $m_1$ such that $m_1 w < q $ . Thus the set of integers $I$ is non empty . Similarly we can show that the set $I$ is bounded above. Because $\alpha$ is bounded above. 
The largest element of $I$ is the $n$ we are looking for. And $(n+1) w$ won't be in $\alpha$. 
Is my proof ok ? I am having second thoughts about it. 
 A: 
there is an integer $n$ such that $nw\in \alpha$ but $(n+1)w \notin \alpha$. 

The proof is based on the induction axiom (also known as the principle of mathematical induction), which is one of Peano axioms. Rudin does not discuss the Peano axioms because he takes natural numbers as understood.  
The induction axiom can be succinctly stated as: every nonempty subset of $\mathbb N$ has the least element. Using the properties of addition/subtraction, we can generalize this to: $\mathbb Z$ has the least upper bound property, as well as the greatest lower bound property. Furthermore, both l.u.b. and g.l.b of a set (if they exist) are contained in the set. Indeed, if $n$  is an upper bound for $A\subset \mathbb Z$ and $n\notin A$, then $n-1$ is also an upper bound for $A$, hence $n$ is not the least upper bound.
Now back to proof. Since $\alpha$ is not all of $\mathbb Q$ there is a rational number $y$ that does not belong to $\alpha$. By the Archimedean property there exists a positive integer $b$ such that $bw>y$. Therefore, $bw\notin \alpha$. In a similar way we can show that there exists an integer $a$ such that $aw\in \alpha$. 
Now consider the set $A=\{m\in\mathbb Z: mw\in \alpha\}$. This set is nonempty because it contains $a$. It has an upper bound $b$. Therefore, there exists $n = \sup A$. By the above, $n\in A$. We have $nw\in\alpha$ and $(n+1)w\notin \alpha$, as required. $\Box$
