See Andrew Baker's p-adic notes:
For a non-Archimedean norm $N$ it is true that "$N(x + y) \leq \max\{N(x), N(y)\}$, with equality if $N(x) \neq N(y)$."
Having trouble proving this. Please give a hint.
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Brian M. Scott
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Daniel Donnelly
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If not, suppose without loss of generality that $N(x)<N(y)$ and that $N(x+y)<N(y)$, and try to derive a contradiction. HINT: $y=(x+y)+(-x)$
Note that you quoted the notes incorrectly: you had ‘$N(x+y)\le\max\{N(x),N(y)\}$, with equality if and only if $N(x)\ne N(y)$’ instead of the correct ‘$N(x+y)\le\max\{N(x),N(y)\}$, with equality if $N(x)\ne N(y)$’. I corrected the quoted part when I edited the question.
Brian M. Scott
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To prove the required statement following this hint, do we need to know that $N(-x)=N(x)$? How is that clear from the definitions given? – G Tony Jacobs May 22 '17 at 18:34
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@GTonyJacobs No, only need to note that $N(-x)\geq 0$ – T C Jun 17 '18 at 16:44