### Calculating with numbers: Calculating with powers and roots

### The square root of a fraction

From the general definition of the square root of a number follows more or less follows the following definition of the root of a fraction.

Root of a fraction

The root of a fraction with positive integers for numerator and denominator is equal to the quotient of the root of the numerator and the root of the denominator .

In formula language we have for positive integers \(m\) and \(n\): \[\sqrt{\frac{m}{n}}=\frac{\sqrt{m}}{\sqrt{n}}\]

**example**

\[\sqrt{\frac{9}{16}}=\frac{\sqrt{9}}{\sqrt{16}}=\frac{3}{4}\] and indeed \[\left(\frac{3}{4}\right)^2=\frac{3^2}{4^2}=\frac{9}{16}\]

Actually, the above definition of the root of a fraction is the only meaningful definition that also leads to the following calculation rule (in fact the definition read in the reverse direction).

Quotient rule of roots For positive integers \(m\) and \(n\) we have: \[\frac{\sqrt{m}}{\sqrt{n}}=\sqrt{\frac{m}{n}}\]

The definition and calculation rule can be used to simplify the root of a fraction to a **standard form** of such root. The root of a positive fraction can namely be written as an irreducible fraction, or as the product of a irreducible fraction and an irreducible root.

So, the expression \(\frac{p}{q}\sqrt{r}\) for integers \(p\), \(q\neq0\) and \(r>1\) is in standard form if

- \(\frac{p}{q}\) is an irreducible fraction;
- there exists no square number greater than 1 that divides\(r\).

Conversely, you can write a quotient of roots in standard form by using the calculation rules of roots.

First we get rid of the root in the denominator by multiplying the numerator and the denominator by \(\sqrt{2}\): \[\begin{aligned}\frac{\sqrt{3}}{\sqrt{2}} &= \frac{\sqrt{3}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ \\ &=\frac{\sqrt{3\times 2}}{(\sqrt{2})^2} \\ \\ &= \frac{\sqrt{6}}{2} \\ \\ &= \frac{1}{2}\times \sqrt{6}\end{aligned}\] The remaining root must still be written in standard form.

From the factorisation \(6=2\cdot 3\) follows that \(\sqrt{6}\) al in standaardvorm is. So: \[\frac{\sqrt{3}}{\sqrt{2}}=\frac{1}{2}\sqrt{6}\]