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 $\ds r = p^{k_1} \frac {m_1} {n_1} = p^{k_2} \frac {m_2} {n_2}$, 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.

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Hence, we reached a contradiction.

$\blacksquare$

Also see

• P-adic Valuation of Rational Number is Well Defined

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.2$: Normed and Banach spaces. Normed spaces