p-adic distances

Question

We take $\mathbb{Q}_p$ to be the completion of $\mathbb{Q}$ with respect to $|\cdot|_p$. If $x=\sum_{j=k}^{\infty} a_jp^j$ is some element in $\mathbb{Q}_p$, then how exactly does $|\cdot|_p$ extend? Does one define $|x|_p=\sum_{j\geq k}p{-j}$ ? This doesn't quite seem right to me.

Answer 1

If $x=\sum_{j=k}^{\infty} a_jp^j$ defines a $p$-adic number, with $a_j \in \{0,1,\dots,p-1\}$ for all $j$ and $a_k\ne 0$, then $|x|_p = p^{-k}$.

This also works for the usual positive rationals...

If $$ x = \frac{14}{9} $$ write $$ \frac{14}{9} = \frac{2}{3^2}+\frac{1}{3^1} + \frac{1}{3^0} $$ so $|x|_3 = 9$ because the first nonzero term is the $3^{-2}$ term.

Answer 2

Remember that the $p$-adic numbers satisfy the strengthened triangle inequality: $|a+b|\le\max\bigl(|a|,|b|\bigr)$, and you prove almost immediately that if one of $|a|$, $|b|$ is smaller than the other, the absolute value of the sum is equal to the greater of the two absolute values.

Similarly, $\big|\sum_ia_i\big|\le\max_i\bigl(|a_i|\bigr)$, and if one of the separate absolute values is bigger than all the others, the absolute value of the sum is equal to that maximum absolute value. This extends to infinite sums as well, so that if you have a sum $\sum_0^\infty a_ip^i$ with all the $a_i$’s being integers in the range $0,1,\dots p-1$, then the absolute value of the sum is the absolute value of the first nonzero term in your sum.

And now I can not resist the impulse to sermonize: if you write your infinite sum $\sum_0^\infty a_ip^i$ in the form $\dots a_3a_2a_1a_0;$ with the understanding that this is $p$-ary expansion, then hand computations can be done by exactly the method you learned in elementary school, though you may find it easiest to prepare yourself with a $p$-ary multiplication table beforehand.