Rationalising denominatorsIB MYP Maths Extended: Revision notes

A surd is a root that cannot be written as a whole number or a fraction, such as $\sqrt2$ or $\sqrt3$. Two rules help: $\sqrt a\times\sqrt b=\sqrt{ab},\qquad\frac{\sqrt a}{\sqrt b}=\sqrt{\frac ab}.$ To simplify, take out the largest square factor: $\sqrt{18}=\sqrt9\times\sqrt2=3\sqrt2$ and $\sqrt{12}\times\sqrt6=\sqrt{72}=6\sqrt2$. You can add or subtract only like surds: $\sqrt{18}+\sqrt8=3\sqrt2+2\sqrt2=5\sqrt2$.

To rationalise a denominator means to rewrite a fraction so that the denominator is a whole number, with no surd. The value stays the same, because you multiply the numerator and denominator by the same number. Rationalised answers are easier to compare and to add. An examiner will expect $\frac{\sqrt3}{3}$, not $\frac{1}{\sqrt3}$, when the question says 'exact'.

Multiply the top and bottom by $\sqrt b$: $\frac{a}{\sqrt b}=\frac{a}{\sqrt b}\times\frac{\sqrt b}{\sqrt b}=\frac{a\sqrt b}{b}.$ Example: $\frac{6}{\sqrt3}=\frac{6\sqrt3}{3}=2\sqrt3$. Always cancel at the end. If the surd can be simplified, do it first: $\frac{4}{\sqrt8}=\frac{4}{2\sqrt2}=\frac{2}{\sqrt2}=\sqrt2$.

Use the conjugate: the same two terms with the sign in the middle changed. The conjugate of $3+\sqrt2$ is $3-\sqrt2$. Multiplying gives the difference of two squares, $(b+\sqrt c)(b-\sqrt c)=b^2-c,$ which is rational. Example: $\frac{1}{3+\sqrt2}=\frac{1}{3+\sqrt2}\times\frac{3-\sqrt2}{3-\sqrt2}=\frac{3-\sqrt2}{9-2}=\frac{3-\sqrt2}{7}.$ Expand the numerator too: $\frac{2}{4-\sqrt3}=\frac{2(4+\sqrt3)}{16-3}=\frac{8+2\sqrt3}{13}$.

To expand brackets with surds, multiply every term by every term, then collect like surds: $(2+\sqrt3)(2-\sqrt3)=4-3=1$ and $(1+\sqrt5)^2=1+2\sqrt5+5=6+2\sqrt5$. Keep answers exact and in simplest form, with a rational denominator. A decimal from a calculator is not an exact answer. Check by estimating: $\frac{1}{3+\sqrt2}\approx\frac{1}{4.41}\approx0.227$ and $\frac{3-\sqrt2}{7}\approx0.227$.

Rationalising can reveal a pattern. $\frac{1}{\sqrt2+\sqrt1}=\sqrt2-1,\quad\frac{1}{\sqrt3+\sqrt2}=\sqrt3-\sqrt2,\quad\frac{1}{\sqrt4+\sqrt3}=2-\sqrt3.$ The general rule is $\frac{1}{\sqrt{n+1}+\sqrt n}=\sqrt{n+1}-\sqrt n$, because the denominator becomes $(n+1)-n=1$. Adding such terms makes the middle terms cancel: the sum from $n=1$ to $8$ is $\sqrt9-\sqrt1=2$. To prove a rule, show the algebra for a general $n$, not just examples.