Rationalizing the Denominator

Rationalizing the denominator allows a fraction that contains a radical in the denominator to be rewritten as an equivalent fraction without a radical in the denominator. This is done by multiplying both the numerator and the denominator by the same nonzero quantity.

This method is based on the invariant property of fractions.

If I multiply both the numerator and the denominator by the same factor k, with k different from zero, I obtain an equivalent fraction whose numerator and denominator are simply multiples of the original ones.

$$ \frac{a}{b} = \frac{a \cdot k}{b \cdot k} $$

This technique is especially useful when studying limits.

A practical example

Consider the following fraction

$$ \frac{3}{\sqrt{2}} $$

The denominator contains a radical.

To eliminate the radical from the denominator, I apply the invariant property and multiply both the numerator and the denominator by the square root of two.

$$ \frac{3}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} $$

This new fraction is equivalent to the previous one because multiplying by \( \frac{\sqrt{2}}{\sqrt{2}} \) is the same as multiplying by 1.

I now multiply the two square roots in the denominator.

$$ \frac{3 \cdot \sqrt{2} }{\sqrt{2} \cdot \sqrt{2}} $$

The two radicals have the same index, so the radicands can be multiplied under the radical.

$$ \frac{3 \cdot \sqrt{2} }{\sqrt{2 \cdot 2}} $$

This produces a radicand whose exponent is equal to the index of the root.

$$ \frac{3 \cdot \sqrt{2} }{\sqrt{4}} $$

$$ \frac{3 \cdot \sqrt{2} }{\sqrt{2^2}} $$

At this point the root in the denominator can be eliminated.

$$ \frac{3 \cdot \sqrt{2} }{2} $$

Example 2

Consider the following fraction

$$ \frac{3}{\sqrt[3]{2}} $$

In this case, to rationalize the denominator I multiply both the numerator and the denominator by the cube root of \(2^2\).

$$ \frac{3}{\sqrt[3]{2}} \cdot \frac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}} $$

Note. In general, if the radical is not a square root, I must multiply by a radical of the same index (n) whose radicand has an exponent that allows the radical in the denominator to be eliminated. In this case the appropriate radicand is \(2^2\), because multiplying it by the other radicand gives \(2 \cdot 2^2 = 2^3\), that is an exponent \(m=3\) equal to the index of the root \(n=3\).

$$ \frac{3 \cdot \sqrt[3]{2^2} }{\sqrt[3]{2} \cdot \sqrt[3]{2^2}} $$

The radicals in the denominator have the same index, so the radicands can be multiplied under the radical.

$$ \frac{3 \cdot \sqrt[3]{2^2} }{\sqrt[3]{2 \cdot 2^2}} $$

This produces a radicand whose exponent (3) is equal to the index of the root (3), allowing the radical to be removed from the denominator of the fraction.

$$ \frac{3 \cdot \sqrt[3]{2^2} }{\sqrt[3]{2^3}} $$

$$ \frac{3 \cdot \sqrt[3]{2^2} }{2} $$

In conclusion, choosing the appropriate multiplying factor is essential for the rationalization to work correctly. It does not always coincide with the radical originally present in the denominator.

Example 3

Consider the following fraction

$$ \frac{3}{\sqrt{3}+\sqrt{2}} $$

To rationalize the denominator, I multiply both the numerator and the denominator by the conjugate of the denominator.

$$ \frac{3}{\sqrt{3}+\sqrt{2}} \cdot \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} $$

Note. If the denominator had contained a difference of radicals, I would have multiplied by the corresponding sum of radicals.

$$ \frac{3 \cdot (\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2}) \cdot (\sqrt{3}-\sqrt{2})} $$

I now perform the algebraic calculations in the denominator.

$$ \frac{3 \cdot (\sqrt{3}-\sqrt{2})}{\sqrt{3}\sqrt{3} - \sqrt{3}\sqrt{2} + \sqrt{2}\sqrt{3} - \sqrt{2}\sqrt{2}} $$

Two terms in the denominator cancel each other out.

$$ \frac{3 \cdot (\sqrt{3}-\sqrt{2})}{\sqrt{3}\sqrt{3} - \sqrt{2}\sqrt{2}} $$

I multiply the radicals in the denominator.

$$ \frac{3 \cdot (\sqrt{3}-\sqrt{2})}{\sqrt{3^2} - \sqrt{2^2}} $$

In the denominator the exponents of the radicands are equal to the indices of the roots. Therefore, the radicals can be eliminated.

$$ \frac{3 \cdot (\sqrt{3}-\sqrt{2})}{3 - 2} $$

$$ \frac{3 \cdot (\sqrt{3}-\sqrt{2})}{1} $$

$$ 3 \cdot (\sqrt{3}-\sqrt{2}) $$

This is another standard technique used to rationalize denominators.

And so on.