Each line is a set of equivalent fractions. Fill in the blanks in the fractions to make each line complete, including the multiplier used to get from one fraction to the next.

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

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

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

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

A *rationalised fraction* is one whose denominator is a whole number. These are usually easier to work with than fractions with square roots in their denominators.

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?

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

Is there more than one way to rationalise a fraction?