Find p-adic valuation and p-norm

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SUMMARY

This discussion focuses on calculating the p-adic valuation and p-norm for the fractions \( x = \frac{3}{686} \) and \( x = \frac{56}{12} \) using primes \( p = 7 \) and \( p = 5 \). For \( x = \frac{3}{686} \), the calculations yield \( w_7\left(\frac{3}{686}\right) = -3 \) and \( |x|_7 = \frac{1}{343} \). For \( x = \frac{56}{12} \), the results are \( w_7\left(\frac{56}{12}\right) = 1 \) and \( |x|_7 = \frac{1}{7} \). Both calculations for \( p = 5 \) yield \( w_5\left(\frac{3}{686}\right) = 0 \) and \( w_5\left(\frac{56}{12}\right) = 0 \), confirming the correctness of the computations.

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evinda
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Hi! (Nerd)

I have to find the values $w_p(x)$ and $|x|_p$ for $x=\frac{3}{686}$ and $x=\frac{56}{12}$ for $p=7,5$.

That's what I have tried:

  • $$x=\frac{3}{686}=\frac{3}{2 \cdot 7^3}=3 \cdot 2^{-1} \cdot 7^{-3} $$

    $$p=7: x=7^{-3} \cdot 3 \cdot 2^{-1} \mapsto 7^3=343= \left |\frac{3}{686} \right|_7$$

    $$w_7 \left ( \frac{3}{686}\right)=-3$$

    $$p=5: x=5^0 \cdot 7^{-3} \cdot 3 \cdot 2^{-1} \mapsto 5^{-0}=1=\left |\frac{3}{686} \right |_5$$

    $$w_5 \left ( \frac{3}{686}\right)=0$$
    $$$$
  • $$x=\frac{56}{12}=2 \cdot 7 \cdot 3^{-1} $$

    $$p=7: x=7^1 \cdot 2 \cdot 3^{-1} \mapsto 7^{-1}=\frac{1}{7}=\left |\frac{56}{12} \right |_7$$

    $$w_7 \left ( \frac{56}{12}\right)=1$$

    $$p=5: x=5^0 \cdot 2 \cdot 7 \cdot 3^{-1} \mapsto 5^{-0}=1=\left |\frac{56}{12} \right |_5$$

    $$w_5 \left ( \frac{56}{12}\right)=0$$

Could you tell me if it is right or if I have done something wrong? (Thinking)
 
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Looks good to me. (Yes)
 
mathbalarka said:
Looks good to me. (Yes)

Nice! Thanks a lot! (Smile)
 

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