[MATH]a < b \implies \exists \ c \text { such that } c > 0 \text { and } a + c = b \implies c = b - a.[/MATH]
Will you accept that as a theorem?
[MATH]\text {Given:} a < b \implies \exists \ c \text { such that } c > 0 \text { and } c = b - a\\
\text {ASSUME, for purposes of contradiction that } - a < - b\\
\therefore \exists \ d \text { such that } d > 0 \text { and } - a + d = - b \implies\\
(-1)(-a + d) = (-1)(b) \implies a - d = b \implies - d = b - a \implies \\
- d = c > 0 \implies d < 0, \text { which contradicts } d \ge 0, \implies
d \not > 0.[/MATH][MATH]\text {ASSUME, for purposes of contradiction that } - a = - b \implies \\
(-1)(-a) = (-1)((-b) \implies a = b, \text { which contradicts } a < b, \implies \\
-a \not = - b.[/MATH][MATH]- a \not \le - b \implies - a > - b.[/MATH]
What is this saying in practical terms.
[MATH]- 4 < - 2 \implies 4 > 2.[/MATH]
When you multiply or divide both sides of an inequality by a negative number, the inequality reverses.