Euclidian distance between two real numbers

Calculate

  1. $d\Big(\dfrac{1}{5},1\Big)$
  2. $d\Big(-\dfrac{1}{3},\dfrac{1}{5} \Big)$
  3. $d\Big(-\dfrac{1}{5},-6 \Big)$
  4. all real numbers $x$ such that $d(x,-1)=\dfrac{1}{3}$

  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 $$-\dfrac{2}{3} > x > -\dfrac{4}{3}$$

  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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