p-adic distance between two real numbers

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