Order and absolute value encode distance inequalities
Objective: Why does division preserve both inequality directions?
Absolute value is distance from zero, and the triangle and reverse-triangle inequalities turn geometric separation into algebraic error control.
Objects and quantifiers: For real x, |x|=x when x≥0 and |x|=−x when x<0; distance is d(x,y)=|x−y|. The result to establish is: For all real x,y, ||x|−|y||≤|x−y|≤|x|+|y|. State who chooses each tolerance, what may depend on it, and where one bound must work for every point.
The worked question is: Solve |2x−3|<5 and express the result as an interval. Begin with “Rewrite as −5<2x−3<5.” and finish with “Divide by positive 2 to obtain −1<x<4.” Then compare the quantified result with this boundary: Squaring an inequality without first controlling signs can reverse or destroy equivalence.
For all real x,y, ||x|−|y||≤|x−y|≤|x|+|y|.
From x=(x−y)+y, triangle inequality gives |x|≤|x−y|+|y|.
Thus |x|−|y|≤|x−y|; exchange x,y for the opposite difference.
Combining yields the reverse inequality, while ordinary triangle inequality gives the upper bound.
Solve |2x−3|<5 and express the result as an interval.
- Rewrite as −5<2x−3<5.
- Add 3 to obtain −2<2x<8.
- Divide by positive 2 to obtain −1<x<4.
Result: The solution set is (−1,4).