Note: this question was updated (2) after GNiklasch's answer was posted, and taking Gro-Tsen's comment into account. The initial question (1) dealt with $\mathbb{Q}_3$.

Original post (1). Let's try to solve the equation $x^2 - 2 = 0$ with $x = \frac{a}{b} \in \mathbb{Q}$. We can't have $x^2 \neq 2$, so the best we can do is minimize $|x^2 - 2|$. Let's try to find an approximation that works over two different completions. Can we have this?

$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_3 \ll \frac{1}{a} $$

I'm trying to write an $S$-adic approximate solution over two places $S = \{ 3, \infty\}$ and $x = \frac{a}{b} \mapsto (\frac{a}{b}, \frac{a}{b}) \in \mathbb{Q} \times \mathbb{Q} \subset \mathbb{R} \times \mathbb{Q}_3 $. What are the correct exponents?

Edit (2). Perhaps i need to find a problem statement that has a solution. gro-tsen suggests I change $p=3$ to $p=7$ so that $\sqrt{2}\in \mathbb{Q}_7$.

$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_7 \ll \frac{1}{a} $$

possibly I can leave the places the same and change the thing I'm approximating. $\sqrt{7}\in \mathbb{Q}_3$ so perhaps I can find a rational number $\frac{a}{b}\in \mathbb{Q}$ such that

$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 7 |_3 \ll \frac{1}{a} $$

Excuse me while I try to state an instance of weak approximation that's not vacuous.

  • $\begingroup$ Do you mean $1/b$, at least in the first of the inequalities? The second inequality is even less clear to me (I know how to motivate just the first one). $\endgroup$ Mar 8, 2018 at 12:49
  • 13
    $\begingroup$ Assuming you meant $\mathbb{Q}_7$ instead of $\mathbb{Q}_3$, so that $2$ can indeed be approximated by $7$-adic rationals, one thing you can do is start with a rational that is "fairly" close to $\sqrt{2}$ in $\mathbb{R}$ and $\mathbb{Q}_7$, like $5/4$, and iterate $x \mapsto \frac{x}{2}+\frac{1}{x}$ (Newton's method). This will give good approximations for $\sqrt{2}$ in both $\mathbb{R}$ and $\mathbb{Q}_7$, like $6449/4560$. $\endgroup$
    – Gro-Tsen
    Mar 8, 2018 at 15:01
  • $\begingroup$ Nice problem! I gave a full solution below. $\endgroup$
    – GH from MO
    Mar 9, 2018 at 2:15

2 Answers 2


Here is a full solution for the modified problem, inspired by Gro-Tsen's valuable comment.

1. There are infinitely many rational numbers $a/b\in\mathbb{Q}$ in lowest terms such that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{1}{b}\qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7\ll\frac{1}{b}.$$ To see this, we work in $\mathbb{Z}[\sqrt{2}]$, the ring of integers of $\mathbb{Q}(\sqrt{2})$. In this ring, the rational prime $(7)$ splits as $(3+\sqrt{2})(3-\sqrt{2})$, while the totally positive units are $(3+2\sqrt{2})^m$. For any positive integer $n$, we can choose the positive integer $m$ so that \begin{align*}a+b\sqrt{2}&=(3+2\sqrt{2})^m(3+\sqrt{2})^n\asymp 7^n,\\ a-b\sqrt{2}&=(3-2\sqrt{2})^m(3-\sqrt{2})^n\asymp 1.\end{align*} This is because $a^2-2b^2=7^n$ holds regardless of $m$. By basic arithmetic in $\mathbb{Z}[\sqrt{2}]$, the integers $a$ and $b$ are relatively prime to each other and to $7$ as well. Moreover, $a$ and $b$ are positive and of size $\asymp 7^n$. It follows that $$\left|\frac{a^2}{b^2}-2\right|_\infty=\frac{7^n}{b^2}\asymp\frac{1}{b} \qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7=7^{-n}\asymp\frac{1}{b}.$$

2. By modifying the above argument, we can achieve that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{c}{b}\qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7\ll\frac{c^{-1}}{b}$$ for any constant $c>0$. This is essentially best possible, because it is straightforward to see that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\left|\frac{a^2}{b^2}-2\right|_7\geq\frac{1}{|b^2|_\infty|b^2|_7}\geq\frac{1}{b^2}.$$


Squares of elements of the $3$-adic integers $\mathbb{Z}_3$ are congruent to $0$ or $1$ modulo $3$, thus they are all at $3$-adic distance $1$ from $2$.

Squares of elements of $\mathbb{Q}_3 \setminus \mathbb{Z}_3$ are $3$-adically even further away from the target.

Thus in $\mathbb{Q}_3$, you cannot approximate a square root of $2$ at all well - just like you can't approximate a square root of $-1$ at all well in the reals.

  • $\begingroup$ can you find a place where simultaneous approximation can work? $\endgroup$ Mar 8, 2018 at 18:08
  • $\begingroup$ can you find a place where simultaneous approximation can work? $\endgroup$ Mar 8, 2018 at 18:10
  • 1
    $\begingroup$ Doesn't Gro-Tsen's comment on your question help? For rational approximations to work, the $p$-adic completion needs to contain a square root of $2$ to begin with - in other words, $2$ must be a quadratic residue mod (odd) $p$. $\endgroup$
    – GNiklasch
    Mar 8, 2018 at 18:20
  • $\begingroup$ i apologize. I'm asking for tge rate at which Gro-Tsen's suggestion works. Obviously I can find $x\in \mathbb{Q}$ that work to arbitrary accuracy in both places (or not at all since. I didn't choose the primes correctly). So now I'm asking how quickly. $\endgroup$ Mar 8, 2018 at 19:20

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