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I need to calculate the 5 adic expansion of $\frac{1}{45}$. Since i cannot compute it normally, i expand $\frac{1}{45}$ into $\frac{1}{5}*\frac{1}{9}$.

I calculated the 5 adic expansion of $\frac{1}{9}$, but i still cannot calculate the expansion of $\frac{1}{5}$ Please give me some advice if possible.

user108680
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    Doesn't multiplication by $1/5$ just shift the places of the "digits" one to the right? – coffeemath Nov 23 '13 at 05:17
  • Nope, it's supposed to have some negative power of p. Not sure though, that's why i'm asking.

    Are you sure about the shifting to the right? Please explain it to me if possible. I'm really in the dark regarding this stuff.

    thank you

     – user108680 Nov 23 '13 at 05:42
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    If the 5-adic expansion of $x$ is $a+5b+25c+125d+\dots$, then the 5-adic expansion of $x/5$ is $a/5+b+5c+25d+\dots$. – Gerry Myerson Nov 23 '13 at 05:48
  • user108680 Yes, I got it backward, and shift is to the left as Gerry Myerson's comment makes clear. I forgot that the powers go in increasing order in p-adic expansions. – coffeemath Nov 23 '13 at 11:47

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The $5$-adic expansion of $5^{-1}$ is simply $5^{-1}$. Seems you are overthinking things.

Just like how multiplying real numbers by powers of $10$ shifts their decimal expansions to the left or right as appropriate, and just like how multiplying a Laurent series $\sum a_nx^n$ by powers of $x$ shift their coefficients to the left or right, so too will multiplying a $p$-adic number by powers of $p$ shift its $p$-adic digital representations, in the same exact manner as you would expect.

Since $9^{-1}=\overline{210234}21024_5$, we get $45^{-1}=\overline{210234}2102.4_5$, or

$$\frac{1}{45}=4\cdot5^{-1}+2\cdot5^0+0\cdot5^1+1\cdot5^2+2\cdot5^3+\cdots$$

anon
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  • Thank you for your help, seems like i really am over-thinking about this. Wish i can thank each and every of you. – user108680 Nov 23 '13 at 06:32
  • can you tell me how to calculate the expansion? My method is also working for me but it's really long and hard. – user108680 Nov 23 '13 at 09:26
  • @user108680 You said you calculated the expansion of $\frac{1}{9}$? – anon Nov 23 '13 at 09:36
  • Anyway, to calculate the $p$-adic expansion of a rational $a/b$, you want to multiply numerator and denominator by a $c$ for which the denominator is of the form (plus or minus) $p^s-1$ (consider Euler's theorem for $p$ mod $b$), in which case you can use the geometric series formula. I explain in more detail here. In this particular case, however, for speed and convenience I just computed $9^{-1}$ mod $5^{20}$, wrote the result in base $5$ then found out where the repetition was occurring. – anon Nov 23 '13 at 09:39
  • Well, i can only do it up to about 5 or 6 digits, i have never get more than 10 like you: 21023421024 – user108680 Nov 24 '13 at 05:45
  • Ah okay. The method I outlined allows the whole digital expansion to be computed by hand (since it's repeating). But like I said, I just used modular arithmetic (on Wolfram Alpha) and eyeballed the expansion to guess where the repetition would be for the sake of convenience. – anon Nov 24 '13 at 08:34