

InterviewSolution
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Verify associativity of addition of rational numbers i.e., \((x+y)+z=x+(y+z)\) when: (i) \(x=\frac{1}{2}\,,y=\frac{2}{3}\;,z=\frac{-1}{5}\)(ii) \(x=\frac{-2}{5}\;,y=\frac{4}{3}\,,z=\frac{-7}{10}\)(iii) \(x=-2\;,y=\frac{3}{5}\;,z=\frac{-4}{3}\) |
Answer» i) As the property states \((x+y)+z=x+(y+z)\) We use value as \((\frac{1}{2}+\frac{2}{3})+\frac{-1}{5}\\=\frac{1}{2}+(\frac{2}{3}+\frac{-1}{5})\) Let us consider L.H.S \((\frac{1}{2}+\frac{2}{3})+\frac{-1}{5}\\\) L.C.M for 2 and 3 is 6 \(\frac{(1\times3)}{(2\times3)}+\frac{(2\times2)}{(3\times2)}\\=\frac{3}{6}+\frac{4}{6}\\=\frac{7}{6}\\=\frac{7}{6}+\frac{-1}{5}\) (Since the denominators are same we add it) \(\frac{(7\times5)}{(6\times5)}+\frac{(-1\times6)}{(5\times6)}\\=\frac{35}{30}+\frac{-6}{30}\\=\frac{(35+(-6))}{30}\\=\frac{(35-6)}{30}\\=\frac{29}{30}\)(Since denominators are same we add it) Let us consider R.H.S \(=\frac{1}{2}+(\frac{2}{3}+\frac{-1}{5})\) L.C.M for 3 and 5 is 15 \(\frac{2}{3}+\frac{-1}{5}\\=\frac{(2\times5)}{(3\times5)}+\frac{(-1\times3)}{(5\times3)}\\=\frac{10}{15}+\frac{-3}{15}\) \(\frac{10}{15}+\frac{-3}{15}\\=\frac{(10-3)}{15}\\=\frac{7}{15}\) (Since the denominators are same we add it) L.C.M for 2 and 15 is 30 \(\frac{1}{2}+\frac{7}{15}\\=\frac{(1\times15)}{(2\times15)}+\frac{(7\times2)}{(15\times2)}\\=\frac{15}{30}+\frac{14}{30}\\=\frac{29}{30}\) (Since denominator is same we add it) \(\therefore\) L.H.S = R.H.S (associativity of addition of rational number is verified. ii) As the property states \((x+y)+z=x+(y+z)\) We use value as \((\frac{-2}{5}+\frac{4}{3})+\frac{-7}{10}\\=\frac{-2}{5}+(\frac{4}{3 }+\frac{-7}{10})\) Let us consider L.H.S \((\frac{-2}{5}+\frac{4}{3})+\frac{-7}{10}\\ \) LCM for 5 and 3 is 15 \(\frac{(-2\times3)}{(5\times3)}+\frac{(4\times5)}{(3\times5)}\\=\frac{-6}{15}+\frac{20}{15}\\=\frac{-6}{15}+\frac{20}{15}\\=\frac{(-6+20)}{15}\\=\frac{14}{15}\) ( Since denominators are same we add it) Now, \(\frac{14}{15}+\frac{-7}{10}\) L.C.M for 15 and 30 is 30 \(\frac{(14\times2)}{(15\times2)}+\frac{(-7\times3)}{(10\times3)}\\=\frac{28}{30}+\frac{-21}{30}\\=\frac{(28+(-21))}{30}\\=\frac{(28-21)}{30}\\=\frac{7}{30}\) (Since the denominators are same we add it) L.C.M for 3 and 10 is 30 \(\frac{4}{3}+\frac{-7}{10}\\=\frac{(4\times10)}{(3\times10)}+\frac{(-7\times3)}{(10\times3)}\\=\frac{40}{30}+\frac{-21}{30}\\=\frac{40}{30}+\frac{-21}{30}\\=\frac{(40-21)}{30}\\=\frac{19}{30}\) ( Since the denominators are same we add it) Now, \(\frac{-2}{5}+\frac{19}{30}\) LCM for 5 and 30 is 30 \(\frac{-2}{5}+\frac{19}{30}\\=\frac{(-2\times6)}{(5\times6)}+\frac{(19\times1)}{(30\times1)}\\=\frac{-12}{30}+\frac{19}{30}\\=\frac{(-12+19)}{30}\\=\frac{7}{30}\) (Since denominators are same we add it) \(\therefore\) L.H.S = R.H.S (associativity of addition of rational number is verified. iii) As the property states \((x+y)+z=x+(y+z)\) We use value as \((\frac{-2}{1}+\frac{3}{5})+\frac{-4}{3}=\frac{-2}{1}+(\frac{3}{5}+\frac{-4}{3})\) Let us consider L.H.S \((\frac{-2}{1}+\frac{3}{5})+\frac{-4}{3}\) LCM for 1 and 5 is 5 \(\frac{(-2\times5)}{(1\times5)}+\frac{(3\times1)}{(5\times1)}\\=\frac{-10}{5}+\frac{3}{5}\\=\frac{(-10+3)}{5}\\=\frac{-7}{5}\) Now, \(\frac{-7}{5}+\frac{-4}{3}\) LCM for 5 and 3 is 15 \(\frac{(-7\times3)}{(5\times3)}+\frac{(-4\times5)}{(3\times5)}\\=\frac{-21}{15}+\frac{-20}{15}\\=\frac{(-21+(-20))}{15}\\=\frac{-21-20}{15}\\=\frac{-41}{15}\) (Since the denominators are same) Let us consider R.H.S \(\frac{-2}{1}+(\frac{3}{5}+\frac{-4}{3})\) LCM for 5 and 3 is 15 \(\frac{-3}{5}+\frac{-4}{3}\\=\frac{(3\times3)}{(5\times3)}+\frac{(-4\times5)}{(3\times5)}\\=\frac{9}{15}+\frac{-20}{15}\\=\frac{(9-20)}{15}\\=\frac{-11}{15}\) (Since denominators are same) Now, \(\frac{-2}{1}+\frac{-11}{15}\\=\frac{(-2\times15)}{(1\times15)}+\frac{(-11\times1)}{(15\times1)}\\=\frac{-30}{15}+\frac{-11}{15}\\=\frac{-41}{15}\) \(\therefore\) L.H.S = R.H.S (associativity of addition of rational number is verified. |
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