This is always true. If, say, $|x|>|y|$, then the ultrametric inequality applied to $x=(x+y)-y$ forces $|x+y|=|x|$; applying it to $x=(x-y)+y$ similarly gives $|x-y|=|x|$.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 136 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-136/1/b/ii/solution