Question

Insert fiverational 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 $$(5+1)=6$$$$\therefore \dfrac{-12}{15}=\dfrac{-12 \times 6}{15 \times6}=\dfrac{-72}{90}$$$$\text {and } \dfrac{-10}{15}=\dfrac{-10 \times 6}{15 \times 6}=\dfrac{-60}{90}$$$$\therefore5$$ 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 5rational 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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