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I'm trying to prove that p-adic space is a metric space(also a ultrametric space), but I find it difficult to prove the triangle inequality. So if one can prove the strong triangle inequality, then the triangle inequality will be easily solved. Does anyone know the proof of the Strong Triangle Inequality or any helpful hint?

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Why set-theory tag? I think metric-spaces could be appropriate tag for this question. –  Martin Sleziak Oct 2 '12 at 8:14
    
@Martin: I agree. –  Brian M. Scott Oct 2 '12 at 8:16
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Are you looking at the $p$-adic numbers, or the $p$-adic integers? And which of the various definitions do you have available to use? Do you have the $p$-adic norm? –  Brian M. Scott Oct 2 '12 at 8:17
    
@BrianM.Scott I just encountered the definition of p-adic norm in a chapter of number theory in A First Course in Abstract Algebra with Applications by Joseph Rotman. So basicly the p-adic norm and p-adic metric are available for me. –  Pan Yan Oct 2 '12 at 8:25
    
This is an exercise at the end of the section on the fundamental theorem? Have you proved yet that $\|a+b\|_p\le\max\{\|a\|_p,\|b\|_p\}$? –  Brian M. Scott Oct 2 '12 at 8:36
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1 Answer

up vote 4 down vote accepted

The case in which $a=0$ or $b=0$ is easy, so assume that $a\ne 0\ne b$. Let $a=\pm p^mr$ and $b=\pm p^ns$, where $m$ and $n$ are integers, and $r$ and $s$ are rationals whose numerators and denominators in lowest terms are not divisible by $p$. (This is the representation used by Rotman; I just collapsed all of the factors not involving $p$ into single rationals $r$ and $s$.)

By definition $\|a\|_p=p^{-m}$ and $\|b\|_p=p^{-n}$. Without loss of generality assume that $m\le n$. Then $p^m\le p^n$, so $p^{-m}\ge p^{-n}$, and $\max\{\|a\|_p,\|b\|_p\}=p^{-m}$. Thus, we must show that $\|a+b\|_p\le p^{-m}$.

Now $a+b=\pm p^mr\pm p^ns=\pm p^m\left(r\pm p^{n-m}s\right)$, where $n-m\ge 0$. Now let’s write $r$ and $s$ as fractions in lowest terms, say as $r=\frac{i}j$ and $s=\frac{k}\ell$, where $p$ does not divide $i,j,k$, or $\ell$. Then

$$r\pm p^{n-m}s=\frac{i}j\pm\frac{p^{n-m}k}\ell=\frac{i\ell\pm p^{n-m}kj}{j\ell}\;.\tag{1}$$

The denominator of this last fraction clearly has no factors of $p$. The numerator has no factors of $p$ if $n-m>0$ (why?). Thus, if $n-m>0$, $\pm p^m(r\pm p^{n-m}s)$ expresses $a+b$ as a product of a power of $p$ and a rational whose numerator and denominator in lowest terms have no factors of $p$, and by definition $\|a+b\|_p=p^{-m}$, which is fine.

If $n-m=0$, the numerator in the last fraction of $(1)$ might have factors of $p$, but if so, then $a+b=\pm p^{m'}t$ for some $m'>m$, where $t$ in lowest terms has no factors of $p$ in either numerator or denominator, and $\|a+b\|_p=p^{-m'}<p^{-m}$, which is also fine.

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Great answer! Thanks. –  Pan Yan Oct 2 '12 at 12:43
    
One more question: if $x$ and $y$ are non-equal real numbers, then either $x<y$ or $x>y$. But I read from a website linkthat there is no linear ordering of the p-adic numbers. I am a little confused about that, because I think each p-adic numbers can be written into decimals which can be compared with order. –  Pan Yan Oct 2 '12 at 13:46
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