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Question

Insert five
rational numbers between:


(i) $$\dfrac{-4}{5}$$ and $$\dfrac{-2}{3}$$


(ii) $$\dfrac{-1}{2}$$ and $$\dfrac{2}{3}$$


Solution

(i) $$\dfrac{-4}{5}$$
and $$\dfrac{-2}{3}$$


LCM of $$5,3=15$$


$$\therefore
\dfrac{-4}{5}=\dfrac{-4 \times 3}{5 \times 3}=\dfrac{-12}{15} $$


$$\text {
and } \dfrac{-2}{3}=\dfrac{-2 \times 5}{3 \times 5}=\dfrac{-10}{15}$$


$$\because$$
Between -10 and $$-12,$$ there are is one rational.


$$\therefore$$
Multiplying numerator and denominator by $$(5+1)=6$$


$$\therefore \dfrac{-12}{15}=\dfrac{-12 \times 6}{15 \times
6}=\dfrac{-72}{90} $$


$$\text {
and } \dfrac{-10}{15}=\dfrac{-10 \times 6}{15 \times 6}=\dfrac{-60}{90}$$


$$\therefore
5$$ rational numbers between $$\dfrac{-72}{90}$$ and $$\dfrac{-60}{90}$$


$$\dfrac{-71}{90},
\dfrac{-70}{90}, \dfrac{-69}{90}, \dfrac{-68}{90}, \dfrac{-67}{90}$$


$$\dfrac{-71}{90},
\dfrac{-7}{9}, \dfrac{-23}{30}, \dfrac{-34}{45}, \dfrac{-67}{90}$$


$$(i i)-\dfrac{1}{2}$$
and $$\dfrac{2}{3}$$


LCM of $$2,3,=6$$


$$\therefore
\dfrac{-1}{2}=\dfrac{-1 \times 3}{2 \times 3}=\dfrac{-3}{6}$$ and


$$\dfrac{2}{3}=\dfrac{2
\times 2}{3 \times 2}=\dfrac{4}{6}$$


Now 5
rational numbers between


$$-\dfrac{1}{2} \text { and } \dfrac{2}{3} $$


$$\dfrac{-2}{6},
\dfrac{-1}{6}, \dfrac{1}{6}, \dfrac{2}{6}, \dfrac{3}{6}$$


Mathematics

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