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Find $\alpha\in \mathbb{Q}$, such that $v_2(\alpha-1/3)\ge 2$, $v_3(\alpha-1/2)\ge 3$ and $|\alpha-1|_\infty<1/2$, where $v_p$ is the $p$-adic exponential valuation and $|\cdot|_\infty$ is the usual absolute value.

Thanks in advance!

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Is this homework? What have you tried? –  Alon Amit Apr 8 '12 at 6:45
@Alon Amit, thanks for the comment. I have tried to do some 3-adic and 2-adic expansion for 1/2 and 1/3, respctively, then I may find α by the Chinese Remainder Theorem. But I am confused with the expansions, since what I have got is quite different from what I need. –  Qiang Zhang Apr 8 '12 at 7:38

2 Answers 2

up vote 2 down vote accepted

The first two conditions are equivalent to $v_2(3\alpha-1)\ge2$ and $v_3(2\alpha-1)\ge3$. That is,

$$3\alpha-1\equiv 0 \mod 2^2\Bbb Z \qquad and \qquad 2\alpha-1 \equiv 0 \mod 3^3\Bbb Z.$$

Rewriting each side (note $2^{-1}\equiv 14\mod 27$), we obtain

$$\begin{cases} \alpha\equiv -1 \mod 4 \\ \alpha \equiv 14 \mod 27. \end{cases}$$

By CRT we have $\alpha=95 \mod 108$. We need $|p/q-1|_\infty <1/2$ while $pq^{-1} \equiv 95 ~(108)$. The two nontrivial factors of $95$ are $5$ and $19$; we have $5^{-1}\equiv 65$ and $19^{-1}\equiv 91 \mod 108$. The latter is close to $5+108=113$ in the $|\cdot|_\infty$ metric, so we check that $|113/91-1|_\infty<1/2$ indeed holds.

This gives $\alpha=113/91$ as one solution.

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@Qiang: If you want to do it by hand it'd be the extendend Euclidean algorithm. Otherwise cheat and consult google for an online applet. :) –  anon Apr 8 '12 at 16:22
Anon, a thousand thanks for your answer! I want to know if there some tricks in solving the equation such as $19x\equiv1 \mod 108$? P.S. I thought about the weak approximation theorem later on today, feeling the problem would be solved by the constructive proof. But I didn't really do it after I read your answer. It's a really great answer! –  Qiang Zhang Apr 8 '12 at 16:25
Anon, I got it. Thanks. It seems that your solution is actually $\alpha=113/91$. I'm sorry I cannot @you on this computer. –  Qiang Zhang Apr 8 '12 at 16:37
@Qiang: Oops, typo. Fixed, thanks. Don't worry, comments on a person's answer always ping that person, so @anon would be redundant here. –  anon Apr 8 '12 at 16:50

Hint: If $\alpha = \frac{a}{b}$ then

$\alpha-\frac{1}{2} = \frac{2a-b}{2b}$

$\alpha-\frac{1}{3} = \frac{3a-b}{3b}$

You need to make the first fraction divisible by a high power of 3, and the second divisible by a high power of 2. It's easiest to make $b$ relatively prime to 6 so you don't have to worry about the denominators. Pick such a $b$. Can you find an $a$ such that $2a-b$ is divisible by 27? Can you find an $a$ that makes $3a-b$ divisible by 4? Can you find an $a$ that does both? Finally, can you make $a$ close enough to $b$ so that $\frac{a}{b}$ is not too far from 1?

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Alon, thanks for your hint! I will try that. Later on today, I found that the weak approximation theorem will be useful here. –  Qiang Zhang Apr 8 '12 at 16:08

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