Problems from p-adic distance

Development:

1. Using the distance definition we have $$d_5(3,-{\displaystyle \frac{1}{8}})=|3-{\displaystyle \frac{-1}{8}}{|}_5=|{\displaystyle \frac{3\cdot 8+1}{8}}{|}_5=|{\displaystyle \frac{25}{8}}{|}_5$$

   According to the previous notations we have that $a=25$ and $b=8$ with $m=1$ and $n=8$ and also $r=2$ and $s=0$. Then: $$|{\displaystyle \frac{25}{8}}{|}_5=5^{s-r}=5^{0-2}={\displaystyle \frac{1}{25}}$$

   And we have $$d_5(3,-{\displaystyle \frac{1}{8}})={\displaystyle \frac{1}{25}}$$

2. Using the distance definition we have $$d_2({\displaystyle \frac{1}{3}},{\displaystyle \frac{3}{5}})=|{\displaystyle \frac{1}{3}}-{\displaystyle \frac{3}{5}}{|}_2=|{\displaystyle \frac{1\cdot 5-3\cdot 3}{3\cdot 5}}{|}_2=|{\displaystyle \frac{-4}{15}}{|}_2$$

   According to the previous notations we have that $a=-4$ and $b=15$ with $m=-1$ and $n=15$ and also $r=2$ and $s=0$. Then: $$|{\displaystyle \frac{-4}{15}}{|}_2=2^{s-r}=2^{0-2}={\displaystyle \frac{1}{4}}$$

   Therefore $$d_2({\displaystyle \frac{1}{3}},{\displaystyle \frac{3}{5}})={\displaystyle \frac{1}{4}}$$

3. Using the distance definition we have $$d_{13}({\displaystyle \frac{22}{17}},{\displaystyle \frac{1}{12}})=|{\displaystyle \frac{22}{17}}-{\displaystyle \frac{1}{12}}{|}_{13}=|{\displaystyle \frac{12\cdot 22-1\cdot 17}{17\cdot 12}}{|}_{13}=|{\displaystyle \frac{247}{204}}{|}_{13}$$

   According to the previous notations we have $a=247$ and $b=204$ with $m=19$ and $n=204$ and also $r=1$ and $s=0$. Then: $$|{\displaystyle \frac{247}{204}}{|}_{13}={13}^{s-r}={13}^{0-1}={\displaystyle \frac{1}{13}}$$

   And consequently $$d_{13}({\displaystyle \frac{22}{17}},{\displaystyle \frac{1}{12}})={\displaystyle \frac{1}{13}}$$

Solution:

1. $d_5(3,-{\displaystyle \frac{1}{8}})={\displaystyle \frac{1}{25}}$
2. $d_2({\displaystyle \frac{1}{3}},{\displaystyle \frac{3}{5}})={\displaystyle \frac{1}{4}}$
3. $d_{13}({\displaystyle \frac{22}{17}},{\displaystyle \frac{1}{12}})={\displaystyle \frac{1}{13}}$

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