Chapter 5.4, Problem 12E

(See Exercise 10 and 11.) If each $x\in D$ is identified with $\left[x,e\right]$ in $Q,$ prove that $D^{+}\subseteq Q^{+}$. (This means that the order relation defined in Exercise 10 coincides in $D$ with the original order relation in $D$. We say that the ordering in $Q$ is an extension of the ordering in $D$.)

(See Exercise 10.) According to Definition 5.29, $>$ is defined in $Q$ by $\left[a,b\right]>\left[c,d\right]$ if and only if $\left[a,b\right]-\left[c,d\right]\in Q^{+}$. Show that $\left[a,b\right]>\left[c,d\right]$ if and only if $ab{d}^2-cd{b}^2\in D^{+}$.

An ordered field is an ordered integral domain that is also a field. In the quotient field $Q$ of an ordered integral domain $D,$ define $Q^{+}$

by

$Q^{+}=\left\{\left[a,b\right]|ab\in D^{+}\right\}$.

Prove that $Q^{+}$ is a set of positive elements for $Q$ and hence, that $Q$ is an ordered field.

Definition 5.29 Greater than

Let $D$ be an ordered integral domain with $D^{+}$ as the set of positive elements. The relation greater than, denoted by $>,$ is defined on elements $x$ and $y$ of $D$

by

$x>y$ if and only if $x-y\in D^{+}$.

The symbol $>$ is read “greater than.” Similarly, $<$ is read “less than.” We define $x<y$ if and only if $y>x$. As direct consequences of the definition, we have

$x>0$ if and only if $x\in D^{+}$

and

$x<0$ if and only if $-x\in D^{+}$.

The three properties of $D^{+}$ in definition 5.28 translate at once into the following properties of $>$ in $D$.

If $x>0$ and $y>0,$ then $x+y>0$.

If $x>0$ and $y>0,$ then $xy>0$.

For each $x\in D,$ one and only one of the following statements is true:

$x>0,x=0,x<0$.

The other basic properties of $>$ are stated in the next theorem. We prove the first two and leave the proofs of the others as exercises. $$