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  1. MHB Master
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    #1
    Hey!!

    I want to find the $4$-adic expansion of $\frac{1}{5}$. I have done the following:


    $\displaystyle{\frac{1}{5}\equiv a_0\pmod 4 \Rightarrow a_0\equiv \frac{1}{5}\pmod 4}$
    We multiply by $5$ and we get $\displaystyle{5a_0\equiv 1\pmod 4}$.
    The only residue of division by $4$ that solves this is $a_0=1$.


    Then we have $\displaystyle{\frac{1}{5}-1\equiv 4a_1 \pmod {4^2} \Rightarrow -\frac{4}{5}\equiv 4a_1\pmod {4^2}}$.
    We divide by $4$ and we get $\displaystyle{-\frac{1}{5}\equiv a_1\pmod {4}}$.
    We multiply then by $5$ and we get $\displaystyle{-1\equiv 5a_1\pmod {4}\Rightarrow 5a_1\equiv -1 \pmod 4 \Rightarrow 5a_1\equiv 3\pmod 4 \Rightarrow a_1\equiv 3\pmod 4}$. The only residue of division by $4$ that solves this is $a_1=3$.


    Then we have $\displaystyle{\frac{1}{5}-1-3\cdot 4\equiv 4^2a_2 \pmod {4^3} \Rightarrow -\frac{4}{5}-3\cdot 4\equiv 4^2a_2\pmod {4^3}}$.
    We multiply by $4$ and we get $\displaystyle{-\frac{1}{5}-3\equiv 4a_2\pmod {4^2}\Rightarrow -\frac{16}{5}\equiv 4a_2 \pmod {4^2}}$, we multiply then by $5$ and we get $\displaystyle{-16\equiv 20a_2\pmod {4^2} \Rightarrow 0\equiv 4a_2 \pmod {4^2} \Rightarrow 4a_2\equiv 0\pmod {4^2}}$. We divide by $4$ and we get $\displaystyle{a_2\equiv 0\pmod {4}}$. The only residue of division by $4$ that solves this is $a_2=0$.


    Then we have $\displaystyle{\frac{1}{5}-1-3\cdot 4\equiv 4^3a_3 \pmod {4^4} \Rightarrow -\frac{4}{5}-3\cdot 4\equiv 4^3a_3\pmod {4^4}}$.
    We divide by $4$ and we get $\displaystyle{-\frac{1}{5}-3\equiv 4^2a_3\pmod {4^3}\Rightarrow -\frac{16}{5}\equiv 4^2a_3 \pmod {4^3}}$, then we divide by $4^2$ and we get $\displaystyle{ -\frac{1}{5}\equiv a_3 \pmod 4}$. Then we multiply by $5$ and we get $\displaystyle{ -1\equiv 5a_3 \pmod 4\Rightarrow 5a_3\equiv -1 \pmod 4 \Rightarrow a_3\equiv 3 \pmod 4 }$. The only residue of division by $4$ that solves this is $a_3=1$.


    Then we have $\displaystyle{\frac{1}{5}-1-3\cdot 4-4^3\equiv 4^4a_4 \pmod {4^5} \Rightarrow -\frac{4}{5}-3\cdot 4-4^3\equiv 4^4a_4\pmod {4^5}}$.
    We divide by $4$ and get $\displaystyle{-\frac{1}{5}-3-4^2\equiv 4^3a_4\pmod {4^4}\Rightarrow -\frac{16}{5}-4^2\equiv 4^3a_4 \pmod {4^4}}$, then we divide by $4^2$ and get $\displaystyle{ -\frac{1}{5}-1\equiv 4a_4 \pmod 4^2 \Rightarrow -\frac{6}{5}\equiv 4a_4 \pmod {4^2}}$. We multiply then by $5$ and get $\displaystyle{ -6\equiv 20a_4 \pmod {4^2}\Rightarrow 20a_4\equiv -6 \pmod {4^2} \Rightarrow 4a_4\equiv 10 \pmod {4^2} }$.
    This doesn't have a solution, right?

    Where have I done something wrong?

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    #2
    Quote Originally Posted by mathmari View Post
    I want to find the $4$-adic expansion of $\frac{1}{5}$. I have done the following:

    .
    .
    .

    Then we have $\displaystyle{\frac{1}{5}-1-3\cdot 4\equiv 4^3a_3 \pmod {4^4} \Rightarrow -\frac{4}{5}-3\cdot 4\equiv 4^3a_3\pmod {4^4}}$.
    We divide by $4$ and we get $\displaystyle{-\frac{1}{5}-3\equiv 4^2a_3\pmod {4^3}\Rightarrow -\frac{16}{5}\equiv 4^2a_3 \pmod {4^3}}$, then we divide by $4^2$ and we get $\displaystyle{ -\frac{1}{5}\equiv a_3 \pmod 4}$. Then we multiply by $5$ and we get $\displaystyle{ -1\equiv 5a_3 \pmod 4\Rightarrow 5a_3\equiv -1 \pmod 4 \Rightarrow {\color{red}a_3\equiv 3 \pmod 4 }}$. The only residue of division by $4$ that solves this is $\color{red}a_3=\Huge 3$.


    Then we have $\displaystyle{\frac{1}{5}-1-3\cdot 4-4^3\equiv 4^4a_4 \pmod {4^5} \Rightarrow -\frac{4}{5}-3\cdot 4-4^3\equiv 4^4a_4\pmod {4^5}}$.
    We divide by $4$ and get $\displaystyle{-\frac{1}{5}-3-4^2\equiv 4^3a_4\pmod {4^4}\Rightarrow -\frac{16}{5}-4^2\equiv 4^3a_4 \pmod {4^4}}$, then we divide by $4^2$ and get $\displaystyle{ -\frac{1}{5}-1\equiv 4a_4 \pmod 4^2 \Rightarrow -\frac{6}{5}\equiv 4a_4 \pmod {4^2}}$. We multiply then by $5$ and get $\displaystyle{ -6\equiv 20a_4 \pmod {4^2}\Rightarrow 20a_4\equiv -6 \pmod {4^2} \Rightarrow 4a_4\equiv 10 \pmod {4^2} }$.
    This doesn't have a solution, right?

    Where have I done something wrong?
    It looks as though the coefficients will be alternately $3$ and $0$ (after the initial $1$).

  3. MHB Apprentice

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    #3
    Show that 1/5 is ...


    Using () to represent the repeated digits under the vinculum and everything base 4.
    Then show that 1/5 is ...
    (03)1.(0)

    Step 1: Calculate 4/5 (base 4)




    Step 2: Calculate -4/5 (4-adic)
    Notice that
    if n = ...3030.3030....
    then 100 n = n
    Therefore n=0
    and
    -4/5 = (03)0.(0)


    Step 3: Calculate 1/5 (4-adic)
    using 1/5 = 1+(-4/5)
    1/5 = (03)1.(0)


    Note: Quaternary and 4-adic number expansions have different metric spaces. However, they share the same arithmetic for the + and * operations.
    Last edited by RLBrown; December 18th, 2016 at 10:29.

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