# 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) $x=\frac{-7}{11}, y=\frac{2}{-5}, z=\frac{-3}{22}$(iv) $x=-2, y=\frac{3}{5}, z=\frac{-4}{3}$

To do:

We have to verify associativity of addition of given rational numbers.

Solution:

Associative property explains that the addition and multiplication of numbers are possible regardless of how they are grouped.

Therefore,

(i) $(x+y)+z=(\frac{1}{2}+\frac{2}{3})+\frac{-1}{5}$

$=(\frac{1\times3+2\times2}{6})+\frac{-1}{5}$            (LCM of 2 and 3 is 6)

$=(\frac{3+4}{6})+\frac{-1}{5}$

$=\frac{7}{6}+\frac{-1}{5}$

$=\frac{7\times5+(-1)\times6}{30}$            (LCM of 5 and 6 is 30)

$=\frac{35-6}{30}$

$=\frac{29}{30}$

$x+(y+z)=\frac{1}{2}+(\frac{2}{3}+\frac{-1}{5})$

$=\frac{1}{2}+(\frac{2\times5+(-1)\times3}{15})$            (LCM of 3 and 5 is 15)

$=\frac{1}{2}+(\frac{10-3}{15})$

$=\frac{1}{2}+\frac{7}{15}$

$=\frac{1\times15+7\times2}{30}$            (LCM of 2 and 15 is 30)

$=\frac{15+14}{30}$

$=\frac{29}{30}$

Hence verified.

(ii) $(x+y)+z=(\frac{-2}{5}+\frac{4}{3})+\frac{-7}{10}$

$=(\frac{-2\times3+4\times5}{15})+\frac{-7}{10}$            (LCM of 5 and 3 is 15)

$=(\frac{-6+20}{15})+\frac{-7}{10}$

$=\frac{14}{15}+\frac{-7}{10}$

$=\frac{14\times2+(-7)\times3}{30}$            (LCM of 15 and 10 is 30)

$=\frac{28-21}{30}$

$=\frac{7}{30}$

$x+(y+z)=\frac{-2}{5}+(\frac{4}{3}+\frac{-7}{10})$

$=\frac{-2}{5}+(\frac{4\times10+(-7)\times3}{30})$            (LCM of 3 and 10 is 30)

$=\frac{-2}{5}+(\frac{40-21}{30})$

$=\frac{-2}{5}+\frac{19}{30}$

$=\frac{-2\times6+19\times1}{30}$            (LCM of 5 and 30 is 30)

$=\frac{-12+19}{30}$

$=\frac{7}{30}$

$(x+y)+z=x+(y+z)$

Hence verified.

(iii) $(x+y)+z=(\frac{-7}{11}+\frac{2}{-5})+\frac{-3}{22}$

$=(\frac{-7\times5+(-2)\times11}{55})+\frac{-3}{22}$            (LCM of 11 and 5 is 55)

$=(\frac{-35+(-22)}{55})+\frac{-3}{22}$

$=\frac{-57}{55}+\frac{-3}{22}$

$=\frac{-57\times2+(-3)\times5}{110}$            (LCM of 55 and 22 is 110)

$=\frac{-114-15}{110}$

$=\frac{-129}{110}$

$x+(y+z)=\frac{-7}{11}+(\frac{2}{-5}+\frac{-3}{22})$

$=\frac{-7}{11}+(\frac{-2\times22+(-3)\times5}{110})$            (LCM of 5 and 22 is 110)

$=\frac{-7}{11}+(\frac{-44-15}{110})$

$=\frac{-7}{11}+\frac{-59}{110}$

$=\frac{-7\times10+(-59)\times1}{110}$            (LCM of 11 and 110 is 110)

$=\frac{-70-59}{110}$

$=\frac{-129}{110}$

$(x+y)+z=x+(y+z)$

Hence verified.

(iv) $(x+y)+z=(-2+\frac{3}{5})+\frac{-4}{3}$

$=(\frac{-2\times5+3\times1}{5})+\frac{-4}{3}$            (LCM of 1 and 5 is 5)

$=(\frac{-10+3}{5})+\frac{-4}{3}$

$=\frac{-7}{5}+\frac{-4}{3}$

$=\frac{-7\times3+(-4)\times5}{15}$            (LCM of 5 and 3 is 15)

$=\frac{-21-20}{15}$

$=\frac{-41}{15}$

$x+(y+z)=-2+(\frac{3}{5}+\frac{-4}{3})$

$=-2+(\frac{3\times3+(-4)\times5}{15})$            (LCM of 5 and 3 is 15)

$=-2+(\frac{9-20}{15})$

$=-2+\frac{-11}{15}$

$=\frac{-2\times15+(-11)\times1}{15}$            (LCM of 1 and 15 is 15)

$=\frac{-30-11}{15}$

$=\frac{-41}{15}$

$(x+y)+z=x+(y+z)$

Hence verified.

Updated on: 10-Oct-2022

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