Building blocks

## Solution

Fill in the blanks in the equivalent fractions and write the multiplier that gets you from one numerator to the next.

1. $\dfrac{1}{\sqrt{2}} \, \bigg(\times \dfrac{\quad}{\quad\quad}\bigg) = \dfrac{\sqrt{2}}{\quad\quad\quad} \, \bigg(\times \dfrac{\quad}{\quad\quad}\bigg) = \dfrac{\sqrt{6}}{\quad\quad\quad} \, \bigg(\times \dfrac{\quad}{\quad\quad}\bigg) = \dfrac{\quad\quad\quad}{6}$

$\dfrac{1}{\sqrt{2}} \, \bigg(\times \dfrac{\sqrt{2}}{\sqrt{2}}\bigg) = \dfrac{\sqrt{2}}{2} \, \bigg(\times \dfrac{\sqrt{3}}{\sqrt{3}}\bigg) = \dfrac{\sqrt{6}}{2\sqrt{3}} \, \bigg(\times \dfrac{\sqrt{3}}{\sqrt{3}}\bigg) = \dfrac{\sqrt{18}}{6}$

What do you notice about the multipliers when going from a fraction with a surd in the denominator to one without?

1. $\dfrac{2}{5\sqrt{3}} \, \bigg(\times \dfrac{\quad}{\quad\quad}\bigg) = \dfrac{\quad\quad\quad}{15} \, \bigg(\times \dfrac{\quad}{\quad\quad}\bigg) = \dfrac{2\sqrt{6}}{\quad\quad\quad} \, \bigg(\times \dfrac{\quad}{\quad\quad}\bigg) = \dfrac{\quad\quad\quad}{60}$

$\dfrac{2}{5\sqrt{3}} \, \bigg(\times \dfrac{\sqrt{3}}{\sqrt{3}}\bigg) = \dfrac{2\sqrt{3}}{15} \, \bigg(\times \dfrac{\sqrt{2}}{\sqrt{2}}\bigg) = \dfrac{2\sqrt{6}}{15\sqrt{2}} \, \bigg(\times \dfrac{2\sqrt{2}}{2\sqrt{2}}\bigg) = \dfrac{8\sqrt{3}}{60}$

The third to the fourth fraction might have been more challenging that the others. Did you find the multiplier first, or did you work out the numerator from one of the other fractions provided?

1. $\dfrac{5}{2+\sqrt{2}} \, \bigg(\times \dfrac{\quad}{\quad\quad\quad}\bigg) = \dfrac{10-5\sqrt{2}}{\quad\quad\quad} \, \bigg(\times \dfrac{\quad}{\quad\quad\quad}\bigg) = \dfrac{\quad\quad\quad}{20+10\sqrt{2}}$

$\dfrac{5}{2+\sqrt{2}} \, \bigg(\times \dfrac{2-\sqrt{2}}{2-\sqrt{2}}\bigg) = \dfrac{10-5\sqrt{2}}{2} \, \bigg(\times \dfrac{10+5\sqrt{2}}{10+5\sqrt{2}}\bigg) = \dfrac{50}{20+10\sqrt{2}}$

1. $\dfrac{2-\sqrt{3}}{4} \, \bigg(\times \dfrac{\quad}{\quad\quad\quad}\bigg) = \dfrac{\quad\quad\quad\quad}{8+4\sqrt{3}} \, \bigg(\times \dfrac{\quad}{\quad\quad\quad}\bigg) = \dfrac{\quad\quad\quad\quad}{16}$

$\dfrac{2-\sqrt{3}}{4} \, \bigg(\times \dfrac{2+\sqrt{3}}{2+\sqrt{3}}\bigg) = \dfrac{1}{8+4\sqrt{3}} \, \bigg(\times \dfrac{8 - 4\sqrt{3}}{8 - 4\sqrt{3}}\bigg) = \dfrac{8 - 4\sqrt{3}}{16}$

Identify the rationalised fractions in the above lines. What do you notice about the multipliers when moving from a fraction with a surd (square root) in the denominator to a rationalised fraction?

Some examples from above include:

$\dfrac{2}{5\sqrt{3}} \, \bigg(\times \dfrac{\sqrt{3}}{\sqrt{3}}\bigg) = \dfrac{2\sqrt{3}}{15}, \quad$ and $\quad \dfrac{2\sqrt{6}}{15\sqrt{2}} \, \bigg(\times \dfrac{2\sqrt{2}}{2\sqrt{2}}\bigg) = \dfrac{8\sqrt{3}}{60}$

Can you see how to rationalise fractions with a single term in the denominator?

$\dfrac{5}{2+\sqrt{2}} \, \bigg(\times \dfrac{2-\sqrt{2}}{2-\sqrt{2}}\bigg) = \dfrac{10-5\sqrt{2}}{2}, \quad$ and $\quad \dfrac{1}{8+4\sqrt{3}} \, \bigg(\times \dfrac{8 - 4\sqrt{3}}{8 - 4\sqrt{3}}\bigg) = \dfrac{8 - 4\sqrt{3}}{16}$

What is special about the denominators here?

How would you rationalise fractions in the following form: $\dfrac{a}{\sqrt{b}}$, $\dfrac{a}{b\sqrt{c}}$ and $\dfrac{a}{b+\sqrt{c}}$?

• The simplest way to rationalise $\dfrac{a}{\sqrt{b}}$ would be to multiply by $\dfrac{\sqrt{b}}{\sqrt{b}}$, giving $\dfrac{a\sqrt{b}}{b}.$

• For $\dfrac{a}{b\sqrt{c}}$ we could write: $\dfrac{a}{b\sqrt{c}} \times \dfrac{\sqrt{c}}{\sqrt{c}} = \dfrac{a\sqrt{c}}{bc}.$

• For denominators that contain 2 terms such as $\dfrac{a}{b+\sqrt{c}},$ the examples above showed that multiplying by the difference of two squares would rationalise the denominator, as of the terms containing surds will cancel with each other: $\dfrac{a}{b+\sqrt{c}} \times \dfrac{b-\sqrt{c}}{b-\sqrt{c}} = \dfrac{ab-a\sqrt{c}}{b^2 - c}.$

Is there more than one way to rationalise a fraction?

We have seen we can rationalise $\dfrac{2}{5\sqrt{3}}$ by multiplying by $\dfrac{\sqrt{3}}{\sqrt{3}}$. What happens if you try and rationalise by multiplying by $\dfrac{2\sqrt{3}}{2\sqrt{3}}$?

Does this idea work for fractions with two terms in the denominator? What different multipliers could you use to rationalise $\dfrac{5}{2+\sqrt{2}}?$