Problems from Euclidian distance

Development:

1. $d({\displaystyle \frac{1}{5}},1)=|1-{\displaystyle \frac{1}{5}}|=|{\displaystyle \frac{5-1}{5}}|={\displaystyle \frac{4}{5}}$

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

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

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

Solution:

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

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