Question

Verify that commutative property for the addition of the following set of rational numbers.
1) \[\dfrac{2}{3}\]and \[\dfrac{1}{2}\]
2) \[\dfrac{2}{5}\] and \[\dfrac{2}{3}\]

Answer 1

Hint:
Here in this question we will use the basic concept of the commutative property. We will apply commutative property on the numbers. Then we will find the value of LHS of the equation then we will find the value of the RHS of the equation. If the LHS and RHS are equal then the commutative property is verified.

Complete step by step solution:
Commutative property states that if \[a\] and are two numbers, then \[a + b = b + a\]. It is an additive property.
1) Given rational numbers are \[\dfrac{2}{3}\]and \[\dfrac{1}{2}\].
Now according to the commutative property we know that
\[\dfrac{2}{3} + \dfrac{1}{2} = \dfrac{1}{2} + \dfrac{2}{3}\]
Now we will take the LHS of the equation and calculate the value of LHS. Therefore, we get
\[LHS = \dfrac{2}{3} + \dfrac{1}{2} = \dfrac{{\left( {2 \times 2} \right) + \left( {1 \times 3} \right)}}{6} = \dfrac{{4 + 3}}{6} = \dfrac{7}{6}\]
Now we will take RHS of the equation and calculate the value of RHS. Therefore, we get
\[RHS = \dfrac{1}{2} + \dfrac{2}{3} = \dfrac{{\left( {1 \times 3} \right) + \left( {2 \times 2} \right)}}{6} = \dfrac{{3 + 4}}{6} = \dfrac{7}{6}\]
We can clearly see that the LHS is equals to the RHS of the equation i.e. \[LHS = RHS\].
Hence the commutative property is verified.
2) Given rational numbers are \[\dfrac{2}{5}\] and \[\dfrac{2}{3}\].
Now according to the commutative property we know that
\[\dfrac{2}{5} + \dfrac{2}{3} = \dfrac{2}{3} + \dfrac{2}{5}\]
Now we will take the LHS of the equation and calculate the value of LHS. Therefore, we get
\[LHS = \dfrac{2}{5} + \dfrac{2}{3} = \dfrac{{\left( {2 \times 3} \right) + \left( {2 \times 5} \right)}}{{15}} = \dfrac{{6 + 10}}{{15}} = \dfrac{{16}}{{15}}\]
Now we will take RHS of the equation and calculate the value of RHS. Therefore, we get
\[RHS = \dfrac{2}{3} + \dfrac{2}{5} = \dfrac{{\left( {2 \times 5} \right) + \left( {2 \times 3} \right)}}{{15}} = \dfrac{{10 + 6}}{{15}} = \dfrac{{16}}{{15}}\]
We can clearly see that the LHS is equals to the RHS of the equation i.e. \[LHS = RHS\].

Hence the commutative property is verified.

Note:
Here we should know the basic commutative property. We have to remember that for a property to be applied its LHS must be equals to the RHS.
The rational number is the number which is in fraction. The three main types of fraction are proper fractions, improper fractions and mixed fractions.
Proper fractions are a fraction having the numerator less, or lower in degree, than the denominator. The value of proper fraction after simplification is always less than 1.
Improper Fraction is a fraction where the numerator is greater than or equals to the denominator, then it is known as an improper fraction. After the simplification of an improper fraction results in the value which is equal or greater than 1, but not less than 1.
Mixed Fraction is the combination of a natural number and fraction. It is basically an improper fraction. After the simplification of a mixed fraction results in the value which is always greater than 1.