You visited us 2 times! Enjoying our articles? Unlock Full Access!

Associative Property of Rational Numbers

Verify associ...

Question

Verify associativity of addition of rational numbers i.e.,
$(x + y) + z = x + (y + z)$

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

Open in App

Solution

Given: $x = - 2, y = \frac{3}{5}, z = \frac{- 4}{3}$

LH.S.: $(x + y) + z = (- 2 + \frac{3}{5}) + (\frac{- 4}{3})$
$= (\frac{- 2 \times 5}{1 \times 5} + \frac{3}{5}) - \frac{4}{3}$
$= (\frac{- 10}{5} + \frac{3}{5}) - \frac{4}{3}$
$= \frac{- 7}{5} - \frac{4}{3} = \frac{- 7 \times 3}{5 \times 3} - \frac{4 \times 5}{3 \times 5}$
$= \frac{- 21}{15} - \frac{20}{15}$
$= \frac{- 41}{15}$
R.H.S: $x + (y + z) = - 2 + (\frac{3}{5} + \frac{- 4}{3})$
$= - 2 + (\frac{3 \times 3}{5 \times 3} - \frac{4 \times 5}{3 \times 5})$
$= - 2 + (\frac{9}{15} - \frac{20}{15})$
$= - 2 - \frac{11}{15}$
$= \frac{- 2 \times 15}{1 \times 15} - \frac{11}{15}$
$= \frac{- 30}{15} - \frac{11}{15} = \frac{- 41}{15}$

Thus, $(x + y) + z = x + (y + z)$
Hence, law of associativity is verified.

Suggest Corrections

Similar questions

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

x = \frac{1}{2}, y = \frac{2}{3}, z = - \frac{1}{5}

x = \frac{- 2}{5}, y = \frac{4}{3}, z = \frac{- 7}{10}

x = \frac{- 7}{11}, y = \frac{2}{- 5}, z = \frac{- 3}{22}

x = - 2, y = \frac{3}{5}, z = \frac{- 4}{3}

x \times (y \times z) = (x \times y) \times z

x = 1, y = \frac{- 1}{2}, z = \frac{1}{4}

x = \frac{2}{3}, y = \frac{- 3}{7}, z = \frac{1}{2}

x \times (y + z) = x \times y + x \times z

x = - 3, y = \frac{9}{2}, z = \frac{- 7}{9}

x \times (y + z) = x y + x z

x = - \frac{1}{2}, y = \frac{3}{4}, z = \frac{1}{4}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program