Are there methods for approximating square roots of numbers in two different $p$-adic valuations. The squares mod $3$ are $\square =\{ 0,1\}$. Therefore, by Hensel's lemma $\sqrt{2} \notin \mathbb{Q}_3$. Therefore, could I find a rational number $\frac{a}{b} \in \mathbb{Q}$ such that:

$$ \left|\,\frac{a}{b} - \sqrt{2}\,\right|_3 < \epsilon $$

I want to know how good these approximations are. Over the real numbers we could use Dirichlet's approximation theorem to find infinitely many rationals such that:

$$ \left|\,\frac{a}{b} - \sqrt{2}\,\right| < \frac{1}{b^2}$$

Is there a $p$-adic version of Dirichlet's approximation theorem. Since $\sqrt{2}$ is not a 3-adic rational, can I find finitely many fractions $\frac{a}{b} \in \mathbb{Q}$ such that:

$$ \left|\,\frac{a}{b} - \sqrt{2}\,\right|_3 < \frac{1}{|b|_3^2} = \frac{1}{3^m}$$

since all $3$-adic valuations are powers of $3$.


Since as you recognize, $\sqrt2\notin\Bbb Q_3$, $\sqrt2$ is not $3$-approximable by rational numbers. What you want is to know the $3$-distance between $\sqrt2$ and $\Bbb Q_3$.

Indeed, let $a$ be any $3$-adic integer, i.e. $a\in\Bbb Z_3$. Now $\vert a-\sqrt2\vert_3=\vert a+\sqrt2\vert_3$ follows from the uniqueness of the extension of the valuation on a complete field to any algebraic extension. Consequently, $\vert a-\sqrt2\vert_3^2=\vert a^2-2\vert_3$. But $a^2$ is congruent to either $0$ or $1$ modulo $3$, and in either case, $\vert a^2-2\vert_3=1$.

The moral? That $\big\vert\frac mn-\sqrt2\big\vert_3\ge1$ always. (I leave it to you to verify that the distance from $\sqrt2$ to an element of $\Bbb Q_3$ that is not in $\Bbb Z_3$ is always greater than $1$.)

  • 1
    $\begingroup$ why did I even say that in the first place? I think even nonsense statements like mine have a place in number theory :-) $\endgroup$ – cactus314 Feb 10 '18 at 16:13

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