P-adic Norm is Norm

= Theorem =

The p-adic measure of distance forms a norm on $$\Q_p$$.

= Proof =

We prove in succession each of the three properties of norms.

Property 1
From the definition of the p-adic function $$|*|_p$$, we have the first property of a norm.

Property 2
For either $$x=0$$ or $$y=0$$ property 2 reduces to property 1, so suppose $$x,y \neq 0$$ are p-adic numbers. Then $$ord_p (xy) = ord_p (x) + ord_p (y)$$, so $$|xy|_p = \tfrac{1}{p^{ord_p(x)+ord_p(y)}} = \tfrac{1}{p^{ord_p(x)}} \tfrac{1}{p^{ord_p(y)}} = |x|_p |y|_p$$.

Property 3
Finally, for either $$x=0$$, $$y=0$$, or $$x+y=0$$, property 3 is true by definition, so $$x,y, x+y$$ are all non-zero. First, assume $$x,y \in \Q$$. Let $$x=a/b, y=c/d$$ be in lowest terms. Then $$x+y = \tfrac{ad+bc}{bd}$$ and $$ord_p(x+y)=ord_p(ad+bc)-ord_p(b)-ord_p(d)$$. The highest power of $$p$$ dividing the sum of two numbers is at least the minimum of the highest power dividing the first and the highest power dividing the second. Therefore,

$$ord_p(x+y) \geq min(ord_p(ad), ord_p(bc))-ord_p(b)-ord_p(d) \ $$

$$= min(ord_p(a)+ord_p(d),ord_p(b)+ord_p(c))-ord_p(b)-ord_p(d) \ $$

$$= min(ord_p(a) - ord_p(b), ord_p(c)-ord_p(d)) \ $$

$$= min(ord_p(x),ord_p(y)) \ $$

Therefore,

$$|x+y|_p = p^{-ord_p(x+y)} \leq max(p^{-ord_p(x)},p^{-ord_p(y)}) = max(|x|_p,|y|_p) \leq |x|_p+|y|_p \ $$