Problems from Euclidian distance

Development:

1. $$d\Big(\dfrac{1}{5},1\Big)=\Big|1-\dfrac{1}{5}\Big|=\Big|\dfrac{5-1}{5}\Big|=\dfrac{4}{5}$$

2. $$d\Big(-\dfrac{1}{3},\dfrac{1}{5} \Big)=\Big|\dfrac{1}{5} - \Big(-\dfrac{1}{3}\Big)\Big|=\Big|\dfrac{1}{5}+\dfrac{1}{3}| = \dfrac{3+5}{15}=\dfrac{8}{15}$$

3. $$d\Big(-\dfrac{1}{5},-6 \Big)= \Big|-6-\Big(-\dfrac{1}{5}\Big)\Big|= \Big|-6+\dfrac{1}{5}\Big|=\Big|\dfrac{-30+1}{5}\Big|=\dfrac{29}{5}$$

4. If $$d(x,-1) < \dfrac{1}{3}$$, then $$|-1-x| < \dfrac{1}{3}$$, that is to say $$-\dfrac{1}{3} < -1-x < \dfrac{1}{3}$$, that adding $$1$$ to all the terms of the inequality we have that: $$$-\dfrac{1}{3} +1 < -x < \dfrac{1}{3} + 1$$$ $$$\dfrac{2}{3} < -x < \dfrac{4}{3}$$$ Multiplying all the terms of to equality by $$-1$$ we obtain $$$-\dfrac{2}{3} > x > -\dfrac{4}{3}$$$

Solution:

1. 1. $$\dfrac{4}{5}$$
   2. $$\dfrac{8}{15}$$
   3. $$\dfrac{29}{5}$$
   4. All the real numbers such that $$-\dfrac{2}{3} > x > -\dfrac{4}{3}$$

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