Rationalizing Denominators | Brilliant Math & Science Wiki (2024)

Rationalizing the denominator is when we move any fractional power from the bottom of a fraction to the top. \(\frac{1}{\sqrt{2}}\), for example, has an irrational denominator. We will soon see that it equals \(\frac{\sqrt{2}}{2}\). Let us look at fractions with irrational denominators.

Consider the fraction

\[\frac{a}{\sqrt{b}}.\]

It is perfectly valid to write it as

\[\begin{align}\frac{a}{\sqrt{b}}&=\frac{a}{\sqrt{b} } \times 1\\&=\frac{a}{\sqrt{b} } \times \frac{\sqrt{b}}{\sqrt{b}} \\&=\frac{a\sqrt{b}}{b}.\end{align}\]

We have now rationalized the denominator and most problems will depend on this simple technique.

Rationalize the denominator of \(\frac{1}{\sqrt{2}}.\)

Using what we learned, we have

\[\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2} } \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}.\ _\square\]

Rationalize the denominator of \(\frac{2}{\sqrt {\sqrt {2}}}\).

We have

\[\begin{align}\dfrac{2}{\sqrt {\sqrt {2}}} & = \dfrac{2 × \sqrt{\sqrt {2}}}{\sqrt{\sqrt {2}} × \sqrt{\sqrt {2}}}\\& = \dfrac{2 \sqrt{\sqrt {2}}}{\sqrt {2}}\\& = \dfrac{2 \sqrt{\sqrt {2}} × \sqrt{2}}{\sqrt {2} × \sqrt {2}}\\& = \dfrac{2 \sqrt{2\sqrt{2}}}{2}\\& = \sqrt{2\sqrt{2}}.\ _\square\end{align}\]

Rationalize the denominator of \(\frac{\sqrt{y-1}}{\sqrt{y+1}}\).

We have

\[\begin{align}\dfrac{\sqrt{y-1}}{\sqrt{y+1}} & = \dfrac{\big(\sqrt{y-1}\big)\big(\sqrt{y+1}\big)}{\big(\sqrt{y+1}\big)\big(\sqrt{y+1}\big)}\\& =\dfrac {\sqrt{y^2-1}}{y + 1}.\ _\square\end{align}\]

Rationalize the expression \(\frac{y^2 \sqrt{x^2 - 1}}{\sqrt{x+1}}.\)

Rationalizing the denominator, we get

\[\frac{y^2 \sqrt{x^2 - 1}}{\sqrt{x+1}} \times \frac{\sqrt{x+1}}{\sqrt{x+1}}=\frac{y^2 \sqrt{x^2 -1} \times\sqrt{x+1}}{x+1} .\]

Simplifying even further, we have

\[\begin{align}\frac{y^2 \sqrt{(x-1)(x+1)} \times\sqrt{x+1}}{x+1} &= \frac{y^2 \sqrt{(x-1)(x+1)^2} }{x+1} \\&=\frac{y^2(x+1) \sqrt{x-1}}{x+1}.\end{align}\]

Canceling out \(x+1,\) our final result is

\[ y^2\sqrt{x-1}.\ _\square\]

Rationalize the denominator of \(\frac {3}{\sqrt[3]{3}}\)

We have

\[\begin{align}\dfrac {3}{\sqrt[3]{3}} & = \dfrac {3 × \sqrt[3]{9}}{\sqrt[3]{3} × \sqrt[3]{9}}\\& = \dfrac {3 × \sqrt[3]{9}}{\sqrt[3]{27}}\\& = \dfrac {3 × \sqrt[3]{9}}{3}\\& =\sqrt[3]{9}.\ _\square\end{align}\]

Now consider fractions of the form \[\frac{a}{b + \sqrt{c}}, \] where \(a\),\(b\) and \(c\) are integers. If we multiply by \(\frac{b+\sqrt{c}}{b+\sqrt{c}},\) we will obtain \[\frac{ab +a\sqrt{c}}{b^2 + 2b\sqrt{c} + c} ,\] which as you can see is a more complicated form of the fraction. Therefore, whenever we obtain such type of fractions, we utilize the difference of squares formula \[(x-y)(x+y)=x^2 - y^2.\] This is particularly useful because if \(x = \sqrt{m}\) and \(y=\sqrt{n},\) then \[\big(\sqrt{m}-\sqrt{n}\big)\big(\sqrt{m}+\sqrt{n}\big) = m - n .\]We are now ready to rationalize the denominators of our original expression by multiplying by \(\frac{b-\sqrt{c}}{b-\sqrt{c}}:\) \[\frac{a}{b + \sqrt{c}} \times \frac{b-\sqrt{c}}{b - \sqrt{c} }=\frac{ab - a\sqrt{c}}{b^2 - c} .\]

Simplify the following expression:

\[\frac{3}{\sqrt{3}} + \frac{2}{\sqrt{2} + \sqrt{5} }. \]

Let us approach the two components separately. \(\frac{3}{\sqrt{3}}\) can be simplified by rationalizing with \( \frac{\sqrt{3}}{\sqrt{3}} \) and the right part \( \frac{2}{\sqrt{2} + \sqrt{5} }\) can be simplified by multiplying by \(\frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}-\sqrt{5}} :\) \[\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}+ \frac{2}{\sqrt{2} + \sqrt{5} } \times \frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}-\sqrt{5}} =\frac{3\sqrt{3}}{3} + \frac{2(\sqrt{2}-\sqrt{5})}{-3}.\]From this we obtain our answer \[\frac{3\sqrt{3}-2\sqrt{2}+2\sqrt{5}}{3}.\ _\square \]

Simplify the following expression: \[ \frac{7\sqrt{6} + \sqrt{303} }{ \sqrt{3} } \times (\sqrt{101} - \sqrt{98} ).\]

We begin by writing \(\sqrt{98}\) as \(7\sqrt{2}:\)\[ \frac{7\sqrt{6} + \sqrt{303} }{ \sqrt{3} } \times (\sqrt{101} - 7\sqrt{2} ).\]Rationalizing the denominator by \(\sqrt{3}\) gives\[\begin{align}\frac{7\sqrt{6} \times \sqrt{3} + \sqrt{303} \sqrt{3} }{ 3 } \times (\sqrt{101} - 7\sqrt{2} )= &\frac{7\sqrt{3\times 2 \times 3} + \sqrt{3\times101\times 3} }{ 3 } \times (\sqrt{101} - 7\sqrt{2} )\\=& \frac{7\sqrt{3^2 \times 2 } + \sqrt{3^2 \times101 } }{ 3 } \times (\sqrt{101} - 7\sqrt{2} )\\=& \frac{3\times 7\sqrt{2 } + 3\sqrt{101 } }{ 3 } \times (\sqrt{101} - 7\sqrt{2} ).\end{align}\]Cancelling out the \(3\)'s on the left side gives\[\begin{align}(7\sqrt{2 } + \sqrt{101 } ) \times (\sqrt{101} - 7\sqrt{2} )&=7\sqrt{202} - 98 + 101 - 7\sqrt{202}\\&=3.\ _\square\end{align}\]

Simplify

\[\dfrac{\big({\sqrt{5}\big)}^3 - {\big(\sqrt{4}\big)}^3}{{\big(\sqrt{5}\big)}^3 + {\big(\sqrt{4}\big)}^3}.\]

We have

\[\begin{align}\dfrac{{(\sqrt{5})}^3 - {(\sqrt{4})}^3}{{(\sqrt{5})}^3 + {(\sqrt{4})}^3} & = \dfrac{(\sqrt 5 - \sqrt 4)(5 + \sqrt{20} + 4)}{(\sqrt 5 + \sqrt 4)(5 - \sqrt{20} + 4)}\\& = \dfrac{\big(\sqrt{5} - \sqrt{4}\big)\big(9 + 2\sqrt{5}\big)}{\big(\sqrt{5} + \sqrt{4}\big)\big(9 - 2\sqrt{5}\big)}\\& = \dfrac{\big(\sqrt{5} - \sqrt{4}\big)\big(9 + 2\sqrt{5}\big)\big(9 + 2\sqrt{5}\big)}{\big(\sqrt{5} + \sqrt{4}\big)\big(9 - 2\sqrt{5}\big)\big(9+ 2\sqrt{5}\big)}\\& = \dfrac{\big(\sqrt{5} - 2\big)\big(81 + 20 + 36\sqrt{5}\big)}{\big(\sqrt{5} + 2\big)(81-20)}\\& = \dfrac{\big(\sqrt 5 - 2\big)\big(101 + 36\sqrt {5}\big)}{\big(\sqrt 5 + 2\big)(61)}\\& = \dfrac{\big(101 + 36\sqrt {5}\big)\big(\sqrt 5 - 2\big)}{(61)\big(\sqrt 5 + 2\big)}\\& = \dfrac{\big(101 + 36\sqrt {5}\big)\big(\sqrt 5 - 2\big)\big(\sqrt 5 - 2\big)}{(61)\big(\sqrt 5 + 2\big)\big(\sqrt 5 - 2\big)}\\& = \dfrac{\big(101 + 36\sqrt {5}\big)\big(5 + 4 - 4\sqrt 5\big)}{(61)(5-4)}\\& = \dfrac{\big(101 + 36\sqrt {5}\big)\big(9 - 4\sqrt 5\big)}{61}\\& = \dfrac {189 - 80\sqrt 5}{61}.\ _\square\end{align}\]

Rationalize the denominator of \(\dfrac {{\sqrt{\sqrt[3]{2}}}}{{\sqrt[4]{\sqrt{5}}}-{\sqrt[4]{\sqrt{3}}}}.\)

We have

\[\begin{align}\dfrac{{\sqrt{\sqrt[3]{2}}}}{{\sqrt[4]{\sqrt{5}}}-{\sqrt[4]{\sqrt{3}}}}& = \dfrac {\left(2^{\frac{1}{3}}\right)^{\frac{1}{2}}}{{\left(5^{\frac {1}{4}}\right)^{\frac{1}{2}}-\left(3^{\frac {1}{4}}\right)^{\frac{1}{2}}}}\\& = \dfrac{2^{\frac{1}{6}}}{5^{\frac{1}{8}} - 3^{\frac{1}{8}}}\\& = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)}{\left(5^{\frac{1}{8}} - 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)}\\& = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right)}{\left(5^{\frac{1}{4}} - 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right)}\\& = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right)}{\left(5^{\frac{1}{2}} - 3^{\frac{1}{2}}\right)}\\& = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{\left(5^{\frac{1}{2}} - 3^{\frac{1}{2}}\right)\left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}\\& = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{5-3}\\& = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{2}.\ _\square\end{align}\]

Rationalize the denominator of \(\dfrac{1}{\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}}.\)

We have

\[\begin{align}\dfrac{1}{\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}} & = \dfrac{1 × {\left(\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}\right)}^{4}}{\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}} × {\left(\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}\right)}^{4}}\\& = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}}}{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}\\& = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × {\left(\sqrt[4]{\sqrt[3]{\sqrt{2}}}\right)}^{3}}{\sqrt[4]{\sqrt[3]{\sqrt{2}}} × {\left(\sqrt[4]{\sqrt[3]{\sqrt{2}}}\right)}^{3}}\\& = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}}}{\sqrt[3]{\sqrt{2}}}\\& = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}} × {\left(\sqrt[3]{\sqrt{2}}\right)}^{2}}{\sqrt[3]{\sqrt{2}} × {\left(\sqrt[3]{\sqrt{2}}\right)}^{2}}\\& = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}} × \sqrt[3]{2}}{\sqrt{2}}\\& = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}} × \sqrt[3]{2} × \sqrt{2}}{\sqrt{2} × \sqrt{2}}\\& =\dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}} × \sqrt[3]{2} × \sqrt{2}}{2}.\ _\square\end{align}\]

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