Building blocks

Main problem

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.

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

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

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

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

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?