📂Lemmas

Rationalizing Fractions Containing Roots Quickly

Formulas

$${{ x } \over { \sqrt{a} \pm \sqrt{b} }} = {{ x \left( \sqrt{a} \mp \sqrt{b} \right) } \over { a - b }}$$

Explanation

Rationalizing fractions may conceptually be simple, but it becomes difficult due to the extensive calculations involved in multiplying complex terms to both the numerator and denominator and then simplifying them.

However, by using the formula above, one can rationalize quickly and easily, minimizing calculation errors. The key idea is to create a form like $( \alpha^2 - \beta^2 )$ to eliminate the square root from the denominator.

For instance, if we are solving $\displaystyle {{2} \over { \sqrt{5} - 1 }}$, we know that we need to multiply $( \sqrt{5} +1 )$ by both the numerator and the denominator. Knowing that the denominator will ultimately become $(5-1) = 4$, it is a waste of effort and time to write it down explicitly, and since we know $( \sqrt{5} +1 )$ will end up in the numerator, we can just write it directly, resulting in $\displaystyle {{2 ( \sqrt{5} + 1 )} \over { 4 } }$. Then, by simplifying, we get $\displaystyle {{ \sqrt{5} + 1 } \over { 2 } }$.

This technique allows us to do calculations like $\displaystyle {{ \sqrt{2} - 1 } \over { \sqrt{2} + 1 }} = {{ \left( \sqrt{2} - 1 \right) \left( \sqrt{2} - 1 \right) } \over { 2 - 1 }} = 3 - 2 \sqrt{2}$ almost intuitively.

Try solving the examples below to get comfortable with the formula and ready to use it anytime.

Examples

(1)

Rationalize $\displaystyle {{ \sqrt{2} } \over { \sqrt{16} + \sqrt{8} }}$. (Try to do it mentally if possible)

Solution

By simplifying $\sqrt{2}$ in both the numerator and the denominator, $$\displaystyle {{ 1 } \over { \sqrt{8} + \sqrt{4} }} = {{ \sqrt{8} - \sqrt{4} } \over {4} } = {{ \sqrt{2} - 1 } \over {2} }$$

(2)

Rationalize $\displaystyle {{4} \over { \sqrt{5} + \sqrt{7} }}$.

The formula introduced in the post works regardless of the order of the terms in the denominator, but it is better to organize the terms to avoid negative numbers, which can be troublesome.

Solution

$\displaystyle {{4} \over { \sqrt{5} + \sqrt{7} }} = {{4} \over { \sqrt{7} + \sqrt{5} }} = {{4 ( \sqrt{7} - \sqrt{5} ) } \over {2} } = 2 ( \sqrt{7} - \sqrt{5} )$

(3)

Let $$\displaystyle s := { { 1 } \over { 1 + \sqrt{2} }} + { { 1 } \over { \sqrt{2} + \sqrt{3} }} + \cdots + { { 1 } \over { \sqrt{8} + 3 }} + { { 1 } \over { 3 + \sqrt{10} }}$$. Find $s$.

Since the terms in the denominator differ by only $1$ inside the root, rationalizing will definitely result in $1$, and there is no need to worry about the fraction form.

Solution

For example, $${ { 1 } \over { \sqrt{2} + \sqrt{3} }} = { { 1 } \over { \sqrt{3} + \sqrt{2} }} = \sqrt{3} - \sqrt{2}$$ By bringing all terms to the numerator like this, $$( \sqrt{2} - 1 ) + ( \sqrt{3} - \sqrt{2} ) + \cdots + (\sqrt{10} - 3 )$$ And by simplifying the terms, we get $s = \sqrt{10} - 1$.