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Question

# To prove: [a, b] > [c, d] text{if and only if} abd^2 - cdb^{2} in D^+. Given information: text{Acoording to the definition of "greater than,"} > text{is defined in Q by} [a, b] > [c, d] text{if and only if} [a, b] - [c, d] in Q^{+}

Transformation properties
To prove: $$\displaystyle{\left[{a},\ {b}\right]}\ {>}\ {\left[{c},\ {d}\right]}\ \text{if and only if}\ {a}{b}{d}^{{2}}\ -\ {c}{d}{b}^{{{2}}}\ \in{D}^{+}$$.
Given information:
$$\displaystyle\text{Acoording to the definition of "greater than,"}\ {>}\ \text{is defined in Q by}\ {\left[{a},\ {b}\right]}\ {>}\ {\left[{c},\ {d}\right]}\ \text{if and only if}\ {\left[{a},\ {b}\right]}\ -\ {\left[{c},\ {d}\right]}\in{Q}^{{+}}$$

2021-02-19

Formula used:
1) Greatest than definition:
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
$$\displaystyle{x}\ {>}\ {y}\ \text{if and only if}\ {x}\ -\ {y}\ \in{D}^{+}$$.
2) Property of greater that (>):
$$\displaystyle\text{If}\ {x}\ {>}\ {y}\ \text{and}\ {z}\ {>}\ {0},\text{then}\ {x}{z}\ {>}\ {y}{z}$$
Proof:
$$\displaystyle\text{Let}\ {\left[{a},{b}\right]}\ {>}\ {\left[{c},{d}\right]}$$
$$\displaystyle\text{Then}\ {\left[{a},{b}\right]}-{\left[{c},{d}\right]}\in\ {Q}^{+}$$
$$\displaystyle\text{So},{\left[{a},{b}\right]}\ +\ {\left[-{c},{d}\right]}\in{Q}^{+}$$
$$\displaystyle\text{Therefore},{\left[{a}{d}\ -\ {b}{c},{b}{d}\right]}\in{Q}^{+}$$
$$\displaystyle\text{This implies that}{\left({a}{d}\ -\ {b}{c}\right)}{b}{d}\in{D}^{+}$$
$$\displaystyle\text{So},{a}{d}{b}^{{2}}\ -\ {c}{d}{b}^{{2}}\in{D}^{+}$$
Conversely,
$$\displaystyle\text{Let}\ {a}{b}{d}^{{2}}\ -\ {c}{d}{b}^{{2}}\in{D}^{+}$$
$$\displaystyle\text{Therefore},{\left({a}{d}\ -\ {b}{c}\right)}{b}{d}\in{D}^{+}$$
$$\displaystyle\text{So},{\left[{a}{d}\ -\ {b}{c},{b}{d}\right]}\in{Q}^{+}\text{as}\ {b}{d}\ne{q}{0}$$
$$\displaystyle\text{Then}{\left[{a},{b}\right]}\ +\ {\left[-{c},{d}\right]}\in{Q}^{+}$$
$$\displaystyle\text{Implies that}{\left[{a},{b}\right]}\ -\ {\left[{c},{d}\right]}\in{Q}^{+}$$
$$\displaystyle\text{Hence},{\left[{a},{b}\right]}\ {>}\ {\left[{c},{d}\right]}$$
$$\displaystyle\text{Therefore},{\left[{a},{b}\right]}\ {>}\ {\left[{c},{d}\right]}\text{if and only if}\ {a}{b}{d}^{{2}}\ -\ {c}{d}{b}^{{2}}\in{D}^{+}$$