Question

# To prove: D^+ subseteq Q^+ Given information: text{Each} x in D text{is identified with} [x,e] text{in Q}

Transformation properties
To prove: $$\displaystyle{D}^{+}\subseteq{Q}^{+}$$
Given information:
$$\displaystyle\text{Each}\ {x}\ \in{D}\ \text{is identified with}\ {\left[{x},{e}\right]}\text{in Q}$$

2020-11-28

Formula used:
1) An ordered field is an ordered integral domain that is also a field.
2) In the quotient field Q of an ordered integral domain D, defined $$\displaystyle{Q}^{+}$$ by
$$Q^+=\left\{[a,b]\mid ab \in D^+\right\}$$
3) Well-ordered $$\displaystyle{D}^{{+}}:$$
If D is an ordered integral domain in which the set $$\displaystyle{D}^{{+}}$$ of positive elements is well-ordered, then e is the least element of $$\displaystyle{D}^{{+}}.$$
Proof:
$$\displaystyle\text{Let}\ {x}\in{D}^{+},\text{then}\ {x}\in{D}$$
$$\displaystyle\text{So},{\left[{x},{e}\right]}\text{in Q where e is the least element of}\ {D}^{+}$$
$$\displaystyle\text{Now}\ {e}\in{D}^{+}\text{and}\ {x}\in{D}^{+}\text{implies that}\ {x}{e}\in{D}^{+}$$
$$\displaystyle\text{Therefore},\ {\left[{x},{e}\right]}\in{Q}^{+}\subseteq{Q}$$
$$\displaystyle\text{So},\ {x}\in{Q}^{+}$$
$$\displaystyle\text{Hence},{D}^{+}\subseteq{Q}^{+}$$