Equivalence of Definitions of P-adic Integer/Definition 2 Implies Definition 1

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Theorem

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the valued field of $p$-adic numbers for some prime $p$.

That is, such that:

$\Q_p$ is the field of $p$-adic numbers

$\norm {\,\cdot\,}_p$ is the $p$-adic norm on $\Q_p$.

Let $x \in \Q_p$ such that the canonical expansion of $x$ contains only positive powers of $p$.

Then:

$\norm x_p \le 1$

Proof

Let the canonical expansion of $x$ contain only positive powers of $p$.

That is:

$x = \ds \sum_{n \mathop = 0}^\infty d_n p^n : \forall n \in \N : 0 \le d_n < p$

Case 1 : $\forall n \in \N : d_n = 0$

Let:

$\forall n \in \N : d_n = 0$

Then:

\(\ds x\)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty 0 * p^n\)

Definition of Canonical P-adic Expansion

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty 0\)

\(\ds \)

\(=\)

\(\ds 0\)

Hence:

\(\ds \norm x_p\)

\(=\)

\(\ds \norm 0_p\)

\(\ds \)

\(=\)

\(\ds 0\)

\(\ds \)

\(<\)

\(\ds 1\)

$\Box$

Case 2 : $\exists n \in \N : d_n > 0$

Let:

$\exists n \in \N : d_n > 0$

Let:

$l = \min \set {i: i \ge 0 \land d_i \ne 0}$

Hence:

$l \ge 0$

Thus:

\(\ds \norm x_p\)

\(=\)

\(\ds p^{-l}\)

P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient

\(\ds \)

\(\le\)

\(\ds p^0\)

\(\ds \)

\(=\)

\(\ds 1\)

$\blacksquare$