# Verify commutativty of addition of rational numbers for each of the following pairs of rotional numbers:

Question:

Verify commutativty of addition of rational numbers for each of the following pairs of rotional numbers:

(i) $\frac{-11}{5}$ and $\frac{4}{7}$

(ii) $\frac{4}{9}$ and $\frac{7}{-12}$

(iii) $\frac{-3}{5}$ and $\frac{-2}{-15}$

(iv) $\frac{2}{-7}$ and $\frac{12}{-35}$

(v) 4 and $\frac{-3}{5}$

(vi) $-4$ and $\frac{4}{-7}$

Solution:

$C$ ommutativity of the addition of rational numbers means that if $\frac{\mathrm{a}}{\mathrm{b}}$ and $\frac{\mathrm{c}}{\mathrm{d}}$ are two rational numbers, then $\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{c}}{\mathrm{d}}+\frac{\mathrm{a}}{\mathrm{b}}$.

(i) We have $\frac{-11}{5}$ and $\frac{4}{7}$.

$\therefore \frac{-11}{5}+\frac{4}{7}=\frac{-11 \times 7}{5 \times 7}+\frac{4 \times 5}{7 \times 5}=\frac{-77}{35}+\frac{20}{35}=\frac{-77+20}{35}=\frac{-57}{35}$

$\frac{4}{7}+\frac{-11}{5}=\frac{4 \times 5}{7 \times 5}+\frac{-11 \times 7}{5 \times 7}=\frac{20}{35}+\frac{-77}{35}=\frac{20-77}{35}=\frac{-57}{35}$

$\therefore \frac{-11}{5}+\frac{4}{7}=\frac{4}{7}+\frac{-11}{5}$

Hence verified.

(ii) We have $\frac{4}{9}$ and $\frac{7}{-12}$.

$\therefore \frac{4}{9}+\frac{-7}{12}=\frac{4 \times 4}{9 \times 4}+\frac{-7 \times 3}{12 \times 3}=\frac{16}{36}+\frac{-21}{36}=\frac{16-21}{36}=\frac{-5}{36}$

$\frac{-7}{12}+\frac{4}{9}=\frac{-7 \times 3}{12 \times 3}+\frac{4 \times 4}{9 \times 4}=\frac{-21}{36}+\frac{16}{36}=\frac{-21+16}{36}=\frac{-5}{36}$

$\therefore \frac{4}{9}+\frac{-7}{12}=\frac{-7}{12}+\frac{4}{9}$

Hence verified.

(iii) We have $\frac{-3}{5}$ and $\frac{-2}{-15}$ or $\frac{-3}{5}$ and $\frac{2}{15}$.

$\therefore \frac{-3}{5}+\frac{2}{15}=\frac{-9}{15}+\frac{2}{15}=\frac{-9+2}{15}=\frac{-7}{15}$

$\frac{2}{15}+\frac{-3}{5}=\frac{2}{15}+\frac{-9}{15}=\frac{2-9}{15}=\frac{-7}{15}$

$\therefore \frac{-3}{5}+\frac{-2}{-15}=\frac{-2}{-15}+\frac{-3}{5}$

Hence verified.

(iv) We have $\frac{2}{-7}$ and $\frac{12}{-35}$.

$\therefore \frac{-2}{7}+\frac{-12}{35}=\frac{-2 \times 5}{7 \times 5}+\frac{-12}{35}=\frac{-10-12}{35}=\frac{-22}{35}$

$\frac{12}{-35}+\frac{2}{-7}=\frac{-12}{35}+\frac{-2 \times 5}{7 \times 5}=\frac{-12-10}{35}=\frac{-22}{35}$

$\therefore \frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

Hence verified.

(v) We have 4 and $\frac{-3}{5}$.

$\therefore 4+\frac{-3}{5}=\frac{4 \times 5}{1 \times 5}+\frac{-3}{5}=\frac{20-3}{5}=\frac{17}{5}$

$\frac{-3}{5}+4=\frac{-3}{5}+\frac{4 \times 5}{1 \times 5}=\frac{-3+20}{5}=\frac{17}{5}$

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

Hence verified.

(vi) We have $-4$ and $\frac{4}{-7}$.

$\therefore \frac{-4}{1}+\frac{-4}{7}=\frac{-4 \times 7}{1 \times 7}+\frac{-4}{7}=\frac{-28-4}{7}=\frac{-32}{7}$

$\frac{-4}{7}+\frac{-4}{1}=\frac{-4}{7}+\frac{-4 \times 7}{1 \times 7}=\frac{-4-28}{7}=\frac{-32}{7}$

$\therefore-4+\frac{4}{-7}=\frac{4}{-7}-4$

Hence verified.