Problems from p-adic distance

Development:

1. Using the distance definition we have $$$d_5\Big(3,-\dfrac{1}{8} \Big)=\Big|3-\dfrac{-1}{8}\Big|_5 = \Big|\dfrac{3\cdot8+1}{8}\Big|_5 = \Big|\dfrac{25}{8}\Big|_5$$$

   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: $$$\Big|\dfrac{25}{8}\Big|_5=5^{s-r}=5^{0-2}=\dfrac{1}{25}$$$

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

2. Using the distance definition we have $$$d_2 \Big(\dfrac{1}{3},\dfrac{3}{5} \Big)=\Big|\dfrac{1}{3}-\dfrac{3}{5}\Big|_2 = \Big|\dfrac{1\cdot5-3\cdot3}{3\cdot5}\Big|_2 = \Big|\dfrac{-4}{15}\Big|_2$$$

   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: $$$\Big|\dfrac{-4}{15}\Big|_2=2^{s-r}=2^{0-2}=\dfrac{1}{4}$$$

   Therefore $$$d_2\Big(\dfrac{1}{3},\dfrac{3}{5} \Big)=\dfrac{1}{4}$$$

3. Using the distance definition we have $$$d_{13}\Big(\dfrac{22}{17},\dfrac{1}{12} \Big)=\Big|\dfrac{22}{17}-\dfrac{1}{12}\Big|_{13} = \Big|\dfrac{12\cdot22-1\cdot17}{17\cdot12}\Big|_{13} = \Big|\dfrac{247}{204}\Big|_{13}$$$

   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: $$$\Big|\dfrac{247}{204}\Big|_{13}=13^{s-r}=13^{0-1}=\dfrac{1}{13}$$$

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

Solution:

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

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