Absolute value of a real number

Calculate:

  1. $|-0,75|$
  2. $\Big| \dfrac{1}{5}-\dfrac{1}{3}\Big|$
  3. $\Big| \dfrac{3\sqrt{8}}{9-\sqrt{6}}\Big|$

  1. $|-0.75|=max(0.75, -0.75)=0.75$

  2. $\Big| \dfrac{1}{5}-\dfrac{1}{3}\Big|=\Big|\dfrac{3-5}{15}\Big|=\Big|\dfrac{-2}{15}\Big|=\dfrac{2}{15}$

  3. $\Big| \dfrac{3\sqrt{8}}{9-\sqrt{6}}\Big|=\dfrac{|3\sqrt{8}|}{|9-\sqrt{6}|}=\dfrac{|3|\cdot|\sqrt{8}|}{9-\sqrt{6}}=\dfrac{3\sqrt{8}}{9-\sqrt{6}} $

  1. $0,75$
  2. $\dfrac{2}{15}$
  3. $\dfrac{3\sqrt{8}}{9-\sqrt{6}}$

Say if the following inequalities are true or false:

  1. $\Big|\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{5}\Big| < 0,6$
  2. $\Big|\dfrac{1}{3}\Big(\dfrac{1}{7}+\dfrac{2}{5}\Big)\Big| < \dfrac{19}{35}$
  1. $\Big|\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{5}\Big|=\Big|\dfrac{15-20+6}{30}\Big|=\Big|\dfrac{1}{30}\Big|=\dfrac{1}{30}$

$0,6=\dfrac{6}{10}=\dfrac{3}{5}$, and to be able to compare both fractions, we do $\dfrac{3}{5}=\dfrac{18}{30}$, then it is easy to see that $\dfrac{1}{30} < \dfrac{18}{30} $

  1. We calculate first: $$\Big|\dfrac{1}{3}\Big(\dfrac{1}{7}+\dfrac{2}{5}\Big)\Big| = \Big|\dfrac{1}{3}\Big|\cdot \Big|\dfrac{1}{7}+\dfrac{2}{5}\Big|=\dfrac{1}{3}\cdot \Big|\dfrac{5+14}{35}\Big|= \dfrac{1}{3}\cdot\dfrac{19}{35}$$ and as $\dfrac{1}{3} < 1$, we have that $\dfrac{1}{3}\cdot\dfrac{19}{35} < \dfrac{19}{35}.$
  1. True.
  2. True.
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