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242 LearnersLast updated on December 11, 2025

Rationalization is a mathematical process of removing any irrational number, such as a square root, from a fraction’s denominator. Rationalization also helps in simplifying a complex expression to make it easier to calculate.

Rationalization is the process of rewriting a fraction so that its denominator becomes a rational number. Rationalization in math is not meant to change the value of the expression, but to rewrite it in a more straightforward and more standard mathematical form. In many fractions, you may find radicals, such as square roots and cube roots, or complex numbers in the denominator. Since these are not rational numbers, we use rationalization to eliminate the irrational or imaginary part and rewrite the expression with a rational denominator, which is easier to simplify.
Rationalization Definition
Rationalization is the process of eliminating any radicals or imaginary numbers from the denominator of a fraction, so that it becomes rational. This is done by multiplying the numerator and denominator by a suitable value, known as a conjugate, to simplify the expression without changing its overall value.
The important terms in the concept of rationalization in math are:
Radical: An expression that uses a root, such as a square root or cube root, is called a radical. For example, \(√(a + b)\) is a radical expression representing \((a + b)\).
Radicand: The radicand is a number or expression written inside the radical symbol. For example, in \(³√(a + b)\), \(a + b\) is the radicand, and we need to find its cube root.
Radical Symbol: The symbol √ is read as “the root of.” If the radicand is √16, then it is read as “the square root of 16.” The horizontal bar over the radicand is called the vinculum. It represents variables or constants in the root function. If variables or constants are not beneath the vinculum, they are not part of the root.
Degree: The degree denotes the root type: 2 for square roots, 3 for cube roots, and so on. If the degree is not mentioned, then it is understood that the radical symbol is a square root.
Conjugate: A conjugate of a binomial has the same terms but the opposite sign. For example, the conjugate of the expression \(x + y\) is \(x - y\) and vice versa. Multiplying by the conjugate helps eliminate radicals from the denominator.
Surds are irrational numbers in root form. They are left in root form because they cannot be written as a simple fraction. For example, √2 is a surd because its decimal values are non-repeating and non-terminating. Hence, it cannot be expressed as a simple fraction. The method used for rationalizing these expressions depends on whether the radical is part of a monomial or a binomial expression.
Examples for monomial radical: \(√5, 2√7, 5√3, √11, 4√10\).
Examples for binomial radicals: \(√2 + 3, 2 - √5,\) \(1 + 2√6, 5 - √8.\)
How to Rationalize a Monomial Radical?
A monomial has a single term, such as √2, √7x, or 3√7x. For expressions with simple radicals as their denominators, we can simply multiply both the numerator and the denominator by the denominator.
For a monomial like \(a ÷ ⁿ√xⁿ\) (where n < m), multiply by \(ⁿ√x(m−n)\) to make the denominator a perfect root, which can then be simplified. For example, let us rationalize the fraction: 5/√3.
Step 1: Evaluate the fraction - the provided fraction 5/√3 has √3 in the denominator, which is a monomial radical. It is important to note that the numerator can contain a radical, thus, you should not be concerned about it while evaluating or simplifying the fraction.
Step 2: Multiply the numerator and denominator by √3 (the radical in the denominator).
\(\frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \)
Step 3: The final answer is \(\frac{5\sqrt{3}}{3} \).
How to Rationalize a Binomial Radical?
If the denominator has a binomial with a radical like \(a+l\), multiply both the numerator and the denominator by its conjugate \((a-b)\). For example, let us rationalize the fraction \(\frac{5}{3} + \sqrt{5} \).
Step 1: We want to remove the radical from the denominator. Since the denominator is a binomial with a surd, we multiply both the numerator and denominator by its conjugate.
Step 2: Multiply the fraction by the conjugate of the denominator, which is \(3 − \sqrt{5}\):
\(\left(\frac{5}{3} + \sqrt{5}\right) \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}} \)
Step 3: Use the identity: \((a+b)(a-b) = a^2 - b^2 \)
So the denominator becomes: \((3 + \sqrt{5}) (3 − \sqrt{5})\).
\(3^2-\sqrt{5}^2=9-5=4\)
Step 4: Now multiply the numerators: \(5(3−\sqrt{5}) \).
Step 5: Put it all together,
\(\frac{5(3 - \sqrt{5})}{4} \) \(=\frac{15 - 5\sqrt{5}}{4} \)
Let us look at this with an example for better understanding. Let us rationalize \(\frac{1}{\sqrt[3]{4}} \):
Step 1: Examine the fraction. The denominator contains a radical in the form of a cube root.
Step 2: We need to make \(\sqrt[3]{4} \) a whole number because we don’t want a cube root in the denominator. To achieve this, we need to multiply \(\sqrt[3]{4} \) by something so that we get a perfect cube as the product. Now, the number that we are looking for is \(\sqrt[3]{64}\) because \(\sqrt[3]{64} \) = 4. So we need to multiply the denominator by \(\sqrt[3]{16}\) because \(\sqrt[3]{4} \times \sqrt[3]{16} = \sqrt[3]{64} = 4 \).
Step 3: Multiply the numerator and the denominator by \(\sqrt[3]{16}\)
\(\frac{1}{\sqrt[3]{4}}×\frac{\sqrt[3]{16}}{\sqrt[3]{16}} = \frac{\sqrt[3]{16}}{\sqrt[3]{64}} = \frac{\sqrt[3]{16}}{4}\)


We need to rationalize a denominator to make our calculations easy. Let us take a look at the step-by-step process of rationalizing a denominator:
Step 1: Identify the radical in the denominator and proceed accordingly. For e.g., if the numerator is 1 and the denominator is √3, we need to remove the square root in the denominator. To do so, we can multiply both the numerator and denominator by √3.
Step 2: Do the simplification of surds. Make sure any square roots are simplified as much as possible. For example, √8 should be written as 2√2 because that’s its simplest form.
Step 3: At last, simplify the entire fraction. After rationalizing, you may be able to reduce the expression by dividing or combining like terms. The final answer should be in simplified form.
Rationalization is an important skill in algebra that simplifies expressions by eliminating radicals from denominators. Here are some practical tips and tricks to help you master rationalization:
With rationalization, students can easily simplify complex expressions and solve algebraic problems efficiently. However, they often make errors when working with irrational denominators. Here are some common mistakes and helpful tips to avoid them:
Understanding the value of rationalization allows us to simplify phrases and solve issues more efficiently. Here are some real-world examples of rationalization:
Rationalize: 1/14 + 10
\(\frac{\sqrt{14} - \sqrt{10}}{4} \)
Step 1: Identify the denominator: It is a binomial radical \(\sqrt{14} + \sqrt{10} \)
Step 2: Multiply numerator and denominator by the conjugate of the denominator:
\(\frac{1}{\sqrt{14} + \sqrt{10}} \times \frac{\sqrt{14} - \sqrt{10}}{\sqrt{14} - \sqrt{10}} \)
Step 3: Apply the identity \((a+b)(a-b) = a^2 - b^2 \), \((\sqrt{14})^2 - (\sqrt{10})^2 = 14 - 10 = 4 \)
Step 4: The final result is, \(\frac{\sqrt{14} - \sqrt{10}}{4} \).
Rationalize: 5/3
\(\frac{5\sqrt{3}}{3} \)
Identify the radical in the denominator. The denominator is √3, which is irrational.
\(\frac{5}{\sqrt{3}} \times \frac{3}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \)and the final answer is \(\frac{5\sqrt{3}}{3} \).
Rationalize: 72
\(\frac{7\sqrt{2}}{2} \)
Multiply numerator and denominator by √2: \(\frac{7}{\sqrt{2}} \times \frac{2}{\sqrt{2}} = \frac{7\sqrt{2}}{2} \)
Rationalize: 15 + 2
\(\sqrt{5 - 2} \)
Multiply by the conjugate of the denominator (\(\sqrt{5 - 2}\)):
\(\frac{1}{\sqrt{5} + 2} \times \frac{\sqrt{5} - 2}{\sqrt{5} - 2} \) \(\frac{\sqrt{5} - 2}{\sqrt{(\sqrt{5})^2 - 2^2}} \) = \(\frac{\sqrt{5} - 2}{5 - 4} \)= \(\frac{\sqrt{5} - 2}{1} \) = \(\sqrt{5 - 2}\)
Rationalize: 13 +10
\(-3 + \sqrt{10} \)
Multiply by the conjugate \(3 - \sqrt{10} \):
\(\frac{1}{3 + \sqrt{10}} \) \(\times \frac{3 - \sqrt{10}}{3 - \sqrt{10}}\) \(=\frac{3 - \sqrt{10}}{(3 + \sqrt{10})(3 - \sqrt{10})} \) \(=\frac{3 - \sqrt{10}}{9 - 10} \) \(=\frac{3 - \sqrt{10}}{1} \) \(=-3 + \sqrt{10} \).
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.






