Rational Numbers Formula
Before learning rational numbers formulas, let us recall what are rational numbers. A rational number is a fraction whose numerator is an integer and the denominator is a nonzero integer. But all fractions are not rational numbers as a fraction may also have its numerator and/or denominator to be an irrational number(s). Let us learn the rational numbers formulas in detail in the next section. The set of rational numbers is denoted by 'Q' and it includes:
 The set of natural numbers, \(\mathbb{N}\)
 The set of whole numbers, \(\mathbb{W}\)
 The set of integers, \(\mathbb{Z}\)
 fractions of integers where the denominator is not zero.
What Are Rational Numbers Formulas?
Using the definition of a rational number, which we discussed in the previous section, a rational number is of the form \(\dfrac p q\), where \(p\) and \(q\) are integers and \(q \neq 0\). So some examples of rational numbers can be 2, 1, 3/2, 1/3, 0, etc. We operate the rational numbers in just the way as we operate the fractions. Thus, the rational numbers formulas are:

\(\mathbb{Q} = \left\{\dfrac{p}{q} \,\,:\,\, p,q \in \mathbb{Z}; \,\, q\neq 0\right\}\)

\(\dfrac{x}{y} \pm \dfrac{m}{n}=\dfrac{x n \pm y m}{y n}\)

\(\dfrac{x}{y} \times \dfrac{m}{n}=\dfrac{x m}{y n}\)

\(\dfrac{x}{y} \div \dfrac{m}{n}=\dfrac{x n}{y m}\)
Note: The set of rational numbers is closed, associative, and commutative under addition and multiplication. The additive identity, 0 and the multiplicative identity, 1 are present in the set of rational numbers. All rational numbers have their additive inverses in the set of rational numbers. All rational numbers other than 0 have their multiplicative inverses in the set of rational numbers.
Let us see the usage of the rational numbers formulas in the following solved examples.
Solved Examples Using Rational Numbers Formulas

Example 1: Identify which of the following are rational numbers using the rational numbers formula: 2, 0, \(\sqrt{2}\), \(\dfrac 1 2\), \(\dfrac 1 3\), and \(\dfrac {1}{\sqrt 2}\).
Solution:
The given numbers can be written as:
2 = \(\dfrac {2} 1\), here both 2 and 1 are integers where 1 \(\neq \) 0.
\(\sqrt 2 \) = \(\dfrac {\sqrt 2} 1\), but here \(\sqrt 2\) is NOT an integer.
\(\dfrac 1 2\), here both 1 and 2 are integers where 2 \(\neq \) 0.
\(\dfrac 1 3\) = \(\dfrac {1} 3\), here both 1 and 3 are integers where 3 \(\neq 0\).
\(\dfrac {1}{\sqrt 2}\), though it is a fraction, \(\sqrt 2\) is NOT an integer.
Thus, only 2, 0, \(\dfrac 1 2\), and \(\dfrac 1 3\) are rational numbers among the given numbers.
Answer: 2, 0, \(\dfrac 1 2\), and \(\dfrac 1 3\) are rational numbers.

Example 2: Find the sum, difference (in the given order), product, and the quotient (in the given order) of the following rational numbers: \(\dfrac 1 3\) and \(\dfrac 2 5\).
Solution:
We will find the sum, difference, product, and the quotient using the rational numbers formulas.
\( \dfrac{1}{3}+ \dfrac{2}{5}= \dfrac{5}{15}+ \dfrac{6}{15} = \dfrac{11}{15}\)
\( \dfrac{1}{3} \dfrac{2}{5}= \dfrac{5}{15} \dfrac{6}{15} = \dfrac{1}{15}\)
\(\dfrac{1}{3}\times \dfrac{2}{5}= \dfrac{1 \times 2}{3 \times 5} = \dfrac{2}{15}\)
\(\dfrac{1}{3}\div \dfrac{2}{5}= \dfrac{1}{3} \times \dfrac{5}{2} = \dfrac{5}{6}\)
Answer: Sum = \( \dfrac{11}{15}\), Difference = \( \dfrac{1}{15}\), Product = \( \dfrac{2}{15}\), and Quotient = \( \dfrac{5}{6}\).