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 \(\sqrt{24}\): \[\begin{aligned}\frac{\sqrt{14}}{\sqrt{24}} &= \frac{\sqrt{14}}{\sqrt{24}}\times \frac{\sqrt{24}}{\sqrt{24}}\\ \\ &=\frac{\sqrt{14\times 24}}{(\sqrt{24})^2} \\ \\ &= \frac{\sqrt{336}}{24} \\ \\ &= \frac{1}{24}\times \sqrt{336}\end{aligned}\] The remaining root must still be written in standard form.
From the factorisation \(336=2^4\cdot 3\cdot 7\) follows that \[\sqrt{336}= 4\times\sqrt{21}\] So: \[\begin{aligned}\frac{\sqrt{14}}{\sqrt{24}}&=\frac{4}{24}\sqrt{21}\\ \\ &={{1}\over{6}}\sqrt{21}\end{aligned}\] Of course it is here more convenient to first simplify the quotient, because then you can calculate with smaller roots: \[\frac{\sqrt{14}}{\sqrt{24}}=\sqrt{\frac{14}{24}}=\sqrt{{{7}\over{12}}}=\frac{\sqrt{7}}{\sqrt{12}}\]
