Query regarding proof of Existence Theorem (1.19, pg-17, Baby Rudin) I was reading the proof of Existence Theorem (Statement (1.19)- There exists an ordered field $\mathbb{R}$ which has the Least Upper Bound Property. Moreover, $\mathbb{R}$ contains $\mathbb{Q}$ as a subfield), which is given in Rudin's Principles of Mathematical Analysis (pg-8 -  statement, pg-17 - proof).
I got stuck at finding multiplicative inverse in Step - 6. Also, how do we prove for rest of the $\mathbb{R}$ after proving the multiplicative properties for $\mathbb{R^+}$?
And in Step - 9, I did not understand the "isomorphism" part (We know that isomorphism is a one-one, onto homomorphism. Homomorphism is fine because the arithmetic and order is preserved, but what about one-one and onto?) and how did they conclude that $\mathbb{Q}$ is a subfield of $\mathbb{R}$ and not the other way round? (of course, it is so; but how did they prove?)
 A: I'm not sure I understand your question about the isomorphism, but the author says that "$\mathbb{Q}$ is isomorphic to the ordered field $\mathbb{Q}^*$", the latter being a subset of $\mathbb{R}$, not the entire set. Anyway, here is an attempt for $(M1)-(M5)$. I hope I haven't made any catastrofic mistakes.
If $\alpha \in \mathbb{R}^+$, and $\beta \in \mathbb{R}^+$, we define $\alpha \cdot \beta$ to be the set of all $p$ such that $p<rs$ for some choice of $r \in \alpha$, $s \in \beta$, $r>0$, $s>0$. We define $1^*$ to be the set of all $q<1$. It is clear that 
$1^*$ is a cut. We verify that the axioms for multiplication hold in $\mathbb{R}$, with $1^*$ playing the role of $1$.
$(M1)$ We have to show that $\alpha \cdot \beta$ is a cut. It is clear that $\alpha \cdot \beta$ is а nonempty subset of $\mathbb{Q}$. Take $г' \notin \alpha$, $s' \notin \beta$. Then $r's'>rs$ for all choices of $r \in \alpha$, $s \in \beta$, $r>0$, $s>0$. Thus $r's' \notin \alpha \cdot \beta$. It follows that $\alpha \cdot \beta$ has property $(I)$. Pick $p \in \alpha \cdot \beta$. Then $p<rs$, with $r \in \alpha$, $s \in \beta$, $r>0$, $s>0$. If $q<p$, then $q/s<r$, so $q/s \in \alpha$, and $q=(q/s)s<rs$ so that $q \in \alpha \cdot \beta$. Thus $(II)$ holds. Choose $t \in \alpha$ so that $t>r>0$. Then $rs<ts$ and $rs \in \alpha \cdot \beta$. Thus $(III)$ holds.
$(M2)$ $\alpha \cdot \beta$ is the set of all $rs$, with $r \in \alpha$, $s \in \beta$, $r>0$, $s>0$. By the same definition, $\beta \cdot \alpha$ is the set of all $sr$. Since $rs=sr$ for all $r \in \mathbb{Q}$, $s \in \mathbb{Q}$, we have $\alpha \cdot \beta = \beta \cdot \alpha$. 
$(M3)$ As above, this follows from the associative law in $\mathbb{Q}$. 
$(M4)$ If $p \in \alpha \cdot 1^*$, then $p<rs$ for some choice of $r \in \alpha$ and $s \in 1^*$, $r>0$, $s>0$. Then $p<rs<r$, hence $p \in \alpha$. Thus $\alpha \cdot 1^* \subset \alpha$. To obtain the opposite inclusion, pick $p \in \alpha$, and pick $r,s \in \alpha$, $s>r>p>0$. Then $p/r \in 1^*$, $0<p/s<p/r$, and $p=s(p/s)<s(p/r)$ so that $p \in \alpha \cdot 1^*$. Thus $\alpha \subset \alpha \cdot 1^*$. We conclude that $\alpha \cdot 1^*=\alpha$. 
$(M5)$ Fix $\alpha \in \mathbb{R}^+$. Let $\beta^+$ be the set of all $p$ with the following property: There exists $r>1$ such that $(1/p)(1/r) \notin \alpha$. In other words, some rational number smaller than $1/p$ fails to be in $\alpha$. We show that $\beta = \beta^+ \cup \mathbb{Q}^- \in \mathbb{R}^+$ and that $\alpha \cdot \beta =1^*$.
If $s \notin \alpha$ and $p=(1/s)(1/2)$, then $(1/p)(1/2) \notin \alpha$, hence $p \in \beta^+$. So $\beta$ is not empty. If $q \in \alpha$ and $q>0$ then $0<1/q$ and $1/q \notin \beta^+$. So $\beta \neq \mathbb{Q}$. Hence $\beta$ satisfies $(I)$. 
Pick $p \in \beta^+$, and pick $r>1$, so that $(1/p)(1/r) \notin \alpha$. If $0<q<p$, then $(1/q)(1/r)>(1/p)(1/r)$, hence $(1/q)(1/r) \notin \alpha$. Thus $q \in \beta^+$, and $(II)$ holds.
Put $s=(r+1)/2$, $t=p(r/s)$. Then $s>1$, $t>p$, and $(1/t)(1/s)=(1/p)(1/r)$, so that $(1/t)(1/s)  \notin \alpha$, and $t \in \beta^+$. Hence $\beta$ satisfies $(III)$. We have proved that $\beta \in \mathbb{R}^+$. 
If $r \in \alpha$ and $s \in \beta$, $r>0$, $s>0$, then $1/s \notin \alpha$, $r<1/s$, hence $rs<1$. Thus $\alpha \cdot \beta \subset 1^*$. To prove the opposite inclusion, pick $u \in 1^*$, $u>0$, put $v=[(1+u)/2]^2$, $w=2/(1+u)$. Then $u<v$, $w>1$, and there is an integer $n$ such that $w^n \in \alpha$ but $w^{n+1} \notin \alpha$. Put $p=1/w^{n+2}$. Then $p \in \beta$, since $(1/p)(1/w)=w^{n+1} \notin \alpha$ and, $v=1/w^2=w^np \in \alpha \cdot \beta$, so that $u \in \alpha \cdot \beta$. Thus $1^* \subset \alpha \cdot \beta$. We conclude that $\alpha \cdot \beta =1^*$. 
