Partial Rationalisation Of Irrational Fractions

Let us say you have the irrational number $\dfrac{2\sqrt{5}}{3\sqrt{2}}$.

Do you think it is possible to rationalise the complete fraction. The answer is $\unicode{x201C}No\unicode{x201D}$.

The reason is quite easy to see. we saw that an irrational number cannot be expressed as a ratio of two integers, while we can express a any rational number as a ratio of two integers. Now, if we were able to convert an irrational number completely into a rational number, without changing the value of the irrational number, that would mean that:

$irrational\ number = rational\ number$

which is impossible.

What we can do, however, is convert any one of the numerator or the denominator into rational, while leaving the other as irrational. This is called $rationalising\ the\ denominator\ (or\ the\ numerator)$ as the case may be.

We will take a very simple example to understand this.

Let us take a simple fraction, like $\dfrac{5}{3\sqrt{7}}$

If we multiply the numerator and the denominator by $\sqrt{7}$, the value of the fraction does not change and we get:

$\dfrac{5}{3\sqrt{7}}$

$= \dfrac{5 \times \sqrt{7}}{3\sqrt{7} \times \sqrt{7}}$

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$= \dfrac{5\sqrt{7}}{3 \times 7}$

$= \dfrac{5\sqrt{7}}{21}$

Now, we see that although the denominator has been rationalised, the numerator has become irrational. So our overall fraction is still irrational.

What happens if the denominator contains more than one term? Let's say we have a fraction like the one below:

$\dfrac{2\sqrt{7}-3\sqrt{5}}{3\sqrt{2} - 2\sqrt{3}}$

and we were to rationalise the denominator of this.

We have seen , that $(a+b)(a-b) = a^2-b^2$

In the denominator of our fraction, if we squared the two irrational terms separately, we can get $(3\sqrt{2})^2 = 9 \times 2 = 18$ and $(2\sqrt{3})^2 = 4 \times 3 = 12$.

Therefore, what we need to do is multiply the numerator and the denominator of our fraction by $(3\sqrt{2} + 2\sqrt{3})$, and we get:

$\dfrac{(2\sqrt{7}-3\sqrt{5})\times(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})\times(3\sqrt{2} + 2\sqrt{3})}$

$=\dfrac{2\sqrt{7}\times3\sqrt{2}-3\sqrt{5}\times3\sqrt{2}+2\sqrt{7}\times2\sqrt{3}-3\sqrt{5}\times2\sqrt{3}}{(3\sqrt{2})^2 - (2\sqrt{3})^2}$

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$=\dfrac{6\sqrt{14}-9\sqrt{10}+4\sqrt{21}-6\sqrt{15}}{18 - 12}$

$=\dfrac{6\sqrt{14}-9\sqrt{10}+4\sqrt{21}-6\sqrt{15}}{6}$

Now, we have a rational denominator as required by the problem.

Denominators or numerators containing more than two terms can also be rationalised, we will see how to do that in a higher grade.