Proof.
Write
![$ \alpha = c/d$](img1753.png)
with
![$ c,d\in\mathbf{Z}$](img460.png)
and
![$ d>0$](img1108.png)
.
First suppose
![$ d$](img179.png)
is exactly divisible by a power of
![$ N$](img5.png)
,
so for some
![$ r$](img38.png)
we have
![$ N^r\mid d$](img1754.png)
but
![$ \gcd(N,d/N^r)=1$](img1755.png)
.
Then
If
![$ N^s$](img1757.png)
is the largest power of
![$ N$](img5.png)
that divides
![$ c$](img352.png)
, then
![$ e=s-r$](img1758.png)
,
![$ a=c/N^s$](img1759.png)
,
![$ b=d/N^r$](img1760.png)
satisfy the conclusion of the lemma.
By unique factorization of integers, there is a smallest multiple
of
such that
is exactly divisible by
. Now apply the
above argument with
and
replaced by
and
.
Proof.
The first two properties of Definition
16.2.1 are
immediate. For the third, we first prove that if
![$ \alpha,\beta\in\mathbf{Q}$](img1777.png)
then
Assume, without loss, that
![$ \ord _N(\alpha) \leq \ord _N(\beta)$](img1779.png)
and
that both
![$ \alpha$](img11.png)
and
![$ \beta$](img1780.png)
are nonzero.
Using Lemma
16.2.5 write
![$ \alpha=N^e(a/b)$](img1781.png)
and
![$ \beta=N^f(c/d)$](img1782.png)
with
![$ a$](img163.png)
or
![$ c$](img352.png)
possibly negative. Then
Since
![$ \gcd(N,bd)=1$](img1784.png)
it follows that
![$ \ord _N(\alpha+\beta)\geq e$](img1785.png)
.
Now suppose
![$ x,y,z\in\mathbf{Q}$](img1775.png)
. Then
so
hence
![$ d_N(x,z) \leq \max(d_N(x,y), d_N(y,z))$](img1788.png)
.