Re: 7-adic distance in Q7

$\displaystyle 305_{10} = 614_7$

So, you are looking for $\displaystyle k = \sum_{n\ge 0}d_n 7^n$ such that $\displaystyle k^2 = 614_7$. So, suppose $\displaystyle d_0=2$ (it can be 2 or 5). Then $\displaystyle 2d_0d_1 \equiv 1 \pmod{7}$, so $\displaystyle 4d_1 \equiv 1 \pmod{7}$. That gives $\displaystyle d_1 = 2$, and a carry of 1. So, $\displaystyle 1+2d_0d_2 + d_1^2 = 4d_2 + 5 \equiv 6 \pmod{7}$. This gives $\displaystyle d_2=2$ as you suggested. Again, the carry is 1. Then, $\displaystyle 1+2d_0d_3 + 2d_1d_2 = 4d_3+9 \equiv 0 \pmod{7}$, so $\displaystyle d_3 = 3$ and the carry is 3. Next, $\displaystyle 3+2d_0d_4 + 2d_1d_3 + d_2^2 = 3+4d_4 + 12 + 4 = 19+4d_4 \equiv 0 \pmod{7}$. So, $\displaystyle d_4 = 4$, and the carry is 5. So, I don't know how you calculated the square root of 305, but I got a very different result.

Re: 7-adic distance in Q7

yes ur right, i got some mistakes of my caculation. now, i got the new representation for root 305 should be right: 2.22345... or

5.4432....

and my question is how to caculate the distance between -1/3

and root of 305 in Q7 (i.e. w.r.t the norm|.|p) Hints for the question is:

compute first few terms of 7-adic expansion of -1/3 and root 305 .

Now , i dont know how to consider the distance..

Re: 7-adic distance in Q7

The easiest way to check if a solution is correct is to use evaluate its geometric sum: $\displaystyle \sum_{n \ge 0}2\cdot 7^n = 2\sum_{n\ge 0}7^n = 2\dfrac{1}{1-7} = 2\left(\dfrac{1}{-6}\right) = -\dfrac{1}{3}$. So, you found the correct representation for $\displaystyle -1/3$. So, $\displaystyle \left|-\dfrac{1}{3} - \sqrt{305} \right|_7$ is the minimum distance between the possible square roots of $\displaystyle 305$. So, if $\displaystyle k = \ldots 43222._7$ and $\displaystyle -k = \ldots 34555._7$, and both of them squared give $\displaystyle 614_7 = 305_{10}$, the 7-adic distance is $\displaystyle \min \left\{ \left|\ldots 22222._7 - \ldots 43222._7 \right|_7, \left| \ldots 22222._7 - \ldots 34555._7 \right|_7 \right\} = \min \left\{7^{-3},7^0 \right\}$

Re: 7-adic distance in Q7

sorry, do you mean if i continue to do the method for more digits of 2.22345... or 5.4432.... then i can get k and -k as ur post?? but how could u ensure that for 2.22345... , u can only get 5 on the digits after 4？？？

Re: 7-adic distance in Q7

You are writing $\displaystyle 2.22345...$. That is not a notation I am familiar with. I am used to $\displaystyle \ldots 43222._7$ which means $\displaystyle 2 + 2\cdot 7 + 2\cdot 7^2 + 3\cdot 7^3 + 4\cdot 7^4 + \cdots$. I suspect they mean the same thing. I am using only digits to the left of the decimal place. You are using digits to the right of the decimal place. I learned that $\displaystyle 2.2_7 = 2\cdot 7^{-1} + 2\cdot 7^0$. Anyway, using your notation, you want $\displaystyle \min\left( \left|2.22222 - 2.22345\right|_7, \left|2.22222 - 5.44321 \right|_7\right)$. Note that $\displaystyle 5.44321 = -2.22345$.

Re: 7-adic distance in Q7

Thats clear for me now. thanks a lot

Re: 7-adic distance in Q7

How did you calculate the 7-adic expansion of square root 305? i honestly cannot do it

Re: 7-adic distance in Q7

Quote:

Originally Posted by

**lahuxixi** How did you calculate the 7-adic expansion of square root 305? i honestly cannot do it

me too

Re: 7-adic distance in Q7

Calculate it digit-by-digit. I went through it in post #2.

Re: 7-adic distance in Q7

I don't need the answer to the question of this thread. I want to ask you about the method of calculating the expansion for other cases. Thank you

Re: 7-adic distance in Q7

Quote:

Originally Posted by

**SlipEternal** $\displaystyle 305_{10} = 614_7$

what does this mean?