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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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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.

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  • $\begingroup$ 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? $\endgroup$ – G Tony Jacobs May 22 '17 at 18:34
  • $\begingroup$ @GTonyJacobs No, only need to note that $N(-x)\geq 0$ $\endgroup$ – chí trung châu Jun 17 '18 at 16:44

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