Real Analysis: Proof that Multiplicative Inverse is a cut This requires knowledge of Dedekind cuts. 
Let us define the cut $\beta$
$\beta = \mathbb{Q}_- \cup \lbrace 0\rbrace \cup \lbrace t: 0<t<r, \frac{1}{r} \notin \alpha \rbrace $
with the property 
$ \alpha \beta = 1^*$ 
where $1^*$ is the multiplicative identity, i.e. $1^* = \lbrace p: p<1 \rbrace$
I don't know how to prove $\beta$ is a cut (let alone that it actually behaves like a multiplicative inverse), and I can't find anything online to help me figure it out. So, for $\beta$, how do I prove 


*

*$\beta \neq \mathbb{Q}, \emptyset$ (i.e. $\beta$ consists of neither all the rationals nor the empty set). 

*$\beta$ is closed downward: I.e. for any rational $r\in\beta$, there exists $s\in\beta$ such that $s<r$.

*$\beta$ has no largest element: I.e. for any rational $r\in\beta$, there exists $t\in\beta$ such that $r<t$.
I literally can't even prove that $\beta$ is non-empty. Now, maybe if I get some help here I'll be able to prove the rest, but I doubt that. 
Thanks.  
 A: We'll assume that $\alpha > 0^*$ since otherwise $\beta$ as defined is all of $\mathbb{Q}$. We show that then $\beta$ as defined is a cut, by proving it has the three properties you've mentioned.
A cut is defined as a set $\alpha \subset \mathbb{Q}$ with the following properties:
I. $\alpha$ is not empty, $\alpha \neq \mathbb{Q}$
II. If $p \in \alpha, q \in \mathbb{Q}$ and $q<p$ then $q \in \alpha$
III. If $p \in \alpha$, then $p<r$ for some $r \in \alpha$
By $\alpha > 0^*$, the set $\alpha$ contains $0^*$ as a proper subset which implies $\alpha \setminus 0^*$ is non-empty.
Recalling that $0^*$ is defined as all negative rationals, $\alpha$ then contains an element $x \geq 0$. By property (III) of cuts, $\alpha$ contains some $v \gt x \geq 0$.
By $\alpha \neq \mathbb{Q}$, there exists some $w \in \mathbb{Q}$ for which $w' \notin \alpha$ for all $w'>w$. If there were no such $w$, by (II),  we would have $\alpha = \mathbb{Q}$.
By the archimedean property of $\mathbb{Q}$, $nv>w$ for some positive integer $n$ and it follows that $nv \notin \alpha$.
Then since $0 < \frac{1}{(n+1)v} < \frac{1}{nv}$ and $nv \notin \alpha$, we have $\frac{1}{(n+1)v} \in \beta$, so $\beta$ is non-empty.
Now consider arbitrary $r$ such that $\frac{1}{v} < r$. Equivalently, we have $0 < \frac{1}{r} < v$ and by (II), $\frac{1}{r} \in \alpha$. So, it follows that $\frac{1}{v} \notin \beta$ and $\beta \neq \mathbb{Q}$. So property (I) is satisfied.
Now let $x \in \beta$, and $0< y < x$. Then by definition, there exists some $x<r$ such that $\frac{1}{r} \notin \alpha$ and clearly $y < r$, so property (II) is satisfied.
Finally, let $x \in \beta$ such that $x > 0$, then  by definition there exists an $r$ where $0 < x< r$ such that $\frac{1}{r} \notin \alpha$, and by the fact that $\mathbb{Q}$ is dense-in-itself, there exists some $y$ such that $x < y < r$ and so $y \in \beta$ and $x < y$. So property (III) is satisfied.
