P-adic Norm is Well Defined
Theorem
P-adic norm $\norm {\, \cdot \,}_p$ is well defined.
Proof
Aiming for a contradiction, suppose $\norm {\, \cdot \,}_p$ is not well defined.
Then, given $r \in \Q$, for two equivalent representations of $r$ $\norm {r}_p$ will yield two different results.
Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$.
Let $\displaystyle r = p^{k_1} \frac {m_1} {n_1} = p^{k_2} \frac {m_2} {n_2}$, with $k_1 \ne k_2$.
Suppose $k_2 < k_1$.
Then:
- $p^{k_1 - k_2} m_1 n_2 = m_2 n_1$
Therefore:
- $p \divides m_2 n_1$
Since $p$ is prime, it cannot be expressed as a product of selected divisors of both $m_2$ and $n_1$.
Hence $p \divides m_2$ or $p \divides n_1$.
This is a contradiction.
Thus, $k_1 \le k_2$.
Similarly, assuming $k_1 < k_2$ leads to a contradiction.
Hence, $k_1 \ge k_2$.
Since $k_1 \ne k_2$, both $k_1$ and $k_2$ have to be such that:
- $k_1 < k_2$
- $k_1 > k_2$
are satisfied.
No integers satisfy this.
Hence, we reached a contradiction.
$\blacksquare$
Also see
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.2$: Normed and Banach spaces. Normed spaces