Distancia euclídea entre dos números reales
Calcula
- $d\Big(\dfrac{1}{5},1\Big)$
- $d\Big(-\dfrac{1}{3},\dfrac{1}{5} \Big)$
- $d\Big(-\dfrac{1}{5},-6 \Big)$
- todos los números reales $x$ tales que $d(x,-1)=\dfrac{1}{3}$
$d\Big(\dfrac{1}{5},1\Big)=\Big|1-\dfrac{1}{5}\Big|=\Big|\dfrac{5-1}{5}\Big|=\dfrac{4}{5}$
$d\Big(-\dfrac{1}{3},\dfrac{1}{5} \Big)=\Big|\dfrac{1}{5} - \Big(-\dfrac{1}{3}\Big)\Big|=\Big|\dfrac{1}{5}+\dfrac{1}{3}| = \dfrac{3+5}{15}=\dfrac{8}{15}$
$d\Big(-\dfrac{1}{5},-6 \Big)= \Big|-6-\Big(-\dfrac{1}{5}\Big)\Big|= \Big|-6+\dfrac{1}{5}\Big|=\Big|\dfrac{-30+1}{5}\Big|=\dfrac{29}{5}$
Si $d(x,-1) < \dfrac{1}{3}$, entonces $|-1-x| < \dfrac{1}{3}$, es decir, $-\dfrac{1}{3} < -1-x < \dfrac{1}{3}$, que sumando $1$ a todos los términos de la desigualdad tenemos que: $$-\dfrac{1}{3} +1 < -x < \dfrac{1}{3} + 1$$ $$\dfrac{2}{3} < -x < \dfrac{4}{3}$$ Multiplicando todos los términos de la igualdad por $-1$ obtenemos que $$-\dfrac{2}{3} > x > -\dfrac{4}{3}$$
- $\dfrac{4}{5}$
- $\dfrac{8}{15}$
- $\dfrac{29}{5}$
- Todos los números reales tales que $-\dfrac{2}{3} > x > -\dfrac{4}{3}$