Intervals in Real Numbers

Say which of the following affirmations are true or false:

  1. $\dfrac{1}{\sqrt{5}}$ belongs to the interval $\Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$
  2. $\sqrt{2}$ belongs to the interval $\Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$
  3. $\dfrac{1}{\sqrt{7}}$ belongs to the interval $\Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$

  1. As $\sqrt{2} < \sqrt{5} < \sqrt{7}$, we have that $\dfrac{1}{\sqrt{5}} \in \Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$

  2. $\sqrt{2} > \dfrac{1}{\sqrt{7}}$, and therefore $\sqrt{2} \notin \Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$

  3. The interval $\Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$ is closed and bounded, so that the endpoints belong to it.

  1. True.
  2. False.
  3. True.

Calculate:

  1. The center and the radius of the interval $[-\sqrt{5},2].$
  2. The endpoints of the interval of center $-\dfrac{1}{3}$ and radius $1$.

  1. The center of an interval is: $$C=\dfrac{a+b}{2}=\dfrac{2-\sqrt{5}}{2}=1-\dfrac{\sqrt{5}}{2}$$ and the radius is: $$d(a,C)=d \Big(-\sqrt{5},1-\dfrac{\sqrt{5}}{2}\Big)=\Big|1-\dfrac{\sqrt{5}}{2}+\sqrt{5}\Big|=1+\dfrac{\sqrt{5}}{2}$$

  2. The lower endpoint is: $a=C-r=-\dfrac{1}{3}-1=-\dfrac{4}{3},$ and the uppe endpoint is: $b=C+r=-\dfrac{1}{3}+1=\dfrac{2}{3}.$

  1. $C=1-\dfrac{\sqrt{5}}{2}$ and $r=1+\dfrac{\sqrt{5}}{2}$
  2. $a=-\dfrac{4}{3}$ and $b=\dfrac{2}{3}.$
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