3.03b - Rationalising denominators

Removing a single surd denominator

Can $1/\sqrt{2}$ be written with a rational denominator without using a decimal approximation?

Higher only. Since $\sqrt{2}\times\sqrt{2}=2$, multiplying the numerator and denominator by $\sqrt{2}$ gives $\sqrt{2}/2$. The number has not changed: we multiplied by $\sqrt{2}/\sqrt{2}=1$.

With a denominator $d\sqrt{n}$, multiply top and bottom by $\sqrt{n}$. Here take $d$ to be a nonzero rational number and $n$ a positive integer. If $n$ is a square, the denominator can simply be evaluated. If $n=0$, the original fraction is undefined; multiplying cannot repair division by zero. Negative radicands do not give real square roots.

Change only the rational coefficient in the denominator. It stays as a factor when the surd is squared.

The opening value is therefore exactly $\sqrt2/2$. A finite decimal approximation is useful for checking its size, but does not replace the requested exact form.

Carry the rational coefficient through this calculation.

Using a conjugate pair

A sum in the denominator needs a multiplier that removes its surd cross-terms.

For example, multiplying $3+\sqrt5$ by $3-\sqrt5$ gives $9-5=4$. The middle terms cancel. Multiplying by just $\sqrt5$ would give $3\sqrt5+5$, which is still irrational.

Use this multiplier only if it is nonzero, and the original denominator must also be nonzero. For positive nonsquare integer $b$ and rational $a$, both conditions hold. The rational product $a^2-b$ can be negative; it is not always positive. If a root evaluates to a rational number, simplify first: $1+\sqrt1=2$, so there is no reason to multiply by its zero conjugate.

Change only the sign in the denominator. Its conjugate changes with it.