Question

Verify whether commutative property is satisfied for addition, subtraction, multiplication and division of the following pairs of rational numbers.
(i) 4 and $\dfrac{2}{5}$
(ii) $\dfrac{{ - 3}}{7}\,{\text{and }}\dfrac{{ - 2}}{7}$

Answer 1

Hint – In order to solve this problem we need to add, subtract, divide and multiply the numbers and then again do that after swapping those numbers. If the numbers obtained are the same then it is commutative other-wise not.

Complete step-by-step answer:
(i) 4 and $\dfrac{2}{5}$
Addition:
$ \Rightarrow 4 + \dfrac{2}{5} = \dfrac{{22}}{5}$
After swapping the numbers,
$ \Rightarrow \dfrac{2}{5} + 4 = \dfrac{{22}}{5}$
The answer is the same in both cases so addition is commutative between 4 and $\dfrac{2}{5}$.
Subtraction:
$ \Rightarrow 4 - \dfrac{2}{5} = \dfrac{{18}}{5} $
After swapping the numbers,
$ \Rightarrow \dfrac{2}{5} - 4 = \dfrac{{ - 18}}{5}$
The answer is different so subtraction is not commutative between 4 and $\dfrac{2}{5}$.
Multiplication:
$ \Rightarrow \dfrac{2}{5}(4) = \dfrac{8}{5}$
After swapping the numbers,
$ \Rightarrow (4)\dfrac{2}{5} = \dfrac{8}{5}$
Hence the answer is the same in both the cases so multiplication is commutative between 4 and $\dfrac{2}{5}$.
Division:
$ \Rightarrow \dfrac{{\dfrac{2}{5}}}{4} = \dfrac{2}{{20}} = \dfrac{1}{{10}}$
After swapping the numbers,
$ \Rightarrow \dfrac{4}{{\dfrac{2}{5}}} = \dfrac{{20}}{2} = 10$
The numbers obtained are not the same so the division is not commutative between 4 and $\dfrac{2}{5}$.
(ii) $\dfrac{{ - 3}}{7}\,{\text{and }}\dfrac{{ - 2}}{7}$
Addition:
$ \Rightarrow \dfrac{{ - 3}}{7} + \dfrac{{ - 2}}{7} = \dfrac{{ - 5}}{7}$
After swapping the numbers,
$ \Rightarrow \dfrac{{ - 2}}{7} + \dfrac{{ - 3}}{7} = \dfrac{{ - 5}}{7}$
The answer is same in both case so addition is commutative between $\dfrac{{ - 3}}{7}\,{\text{and }}\dfrac{{ - 2}}{7}$.
Subtraction:
$ \Rightarrow \dfrac{{ - 3}}{7} - \dfrac{{ - 2}}{7} = \dfrac{{ - 1}}{7}$
After swapping the numbers,
$ \Rightarrow \dfrac{{ - 2}}{7} - \dfrac{{ - 3}}{7} = \dfrac{1}{7}$
The answer is different so subtraction is not commutative between $\dfrac{{ - 3}}{7}\,{\text{and }}\dfrac{{ - 2}}{7}$.
Multiplication:
$ \Rightarrow \dfrac{{ - 2}}{7}\left( {\dfrac{{ - 3}}{7}} \right) = \dfrac{6}{7}$
After swapping the numbers,
$ \Rightarrow \left( {\dfrac{{ - 3}}{7}} \right)\dfrac{{ - 2}}{5} = \dfrac{6}{7}$
Hence the answer is same in both the cases so multiplication is commutative between $\dfrac{{ - 3}}{7}\,{\text{and }}\dfrac{{ - 2}}{7}$.
Division:
$ \Rightarrow \dfrac{{\dfrac{{ - 2}}{7}}}{{\dfrac{{ - 3}}{7}}} = \dfrac{2}{3}$
After swapping the numbers,
$ \Rightarrow \dfrac{{\dfrac{{ - 3}}{7}}}{{\dfrac{{ - 2}}{7}}} = \dfrac{3}{2}$
The numbers obtained are not same so division is not commutative between $\dfrac{{ - 3}}{7}\,{\text{and }}\dfrac{{ - 2}}{7}$.

Note – To solve such problems we need to know that if two numbers are showing commutative properties then the answer obtained by swapping the two numbers will be the same. Addition and multiplication is always commutative whether the number is real or imaginary.