Calculating with numbers: Calculating with powers and roots
Quartic and higher roots in standard form
We will discuss higher roots, but we start with a special case, namely the quartic root. Like a square, you can only take a quartic root of a nonnegative number.
The quartic root
The quartic root or fourth root of a number, \(a\ge 0\) is by definition the nonnegative integer \(w\) such that \(w^4 = a\). Notation: \(w = \sqrt[4]{a}\) and \(w=a^{\frac{1}{4}}\).
Examples
\[\begin{aligned}\sqrt[4]{16}&={2}\quad\text{because }2^4=16\text{ and }\\[0.2cm] \sqrt{81}&={3}\quad\text{because }(3)^4=81\end{aligned}\]
The rules for calculating quartic roots resemble those of square roots.
For any natural number \(n\) we have: \[\sqrt[4]{n^4}=n\] and \[\left(\sqrt[4]{n}\right)^4=n\]
Example \(n=3\)
\[\sqrt[4]{3^4}=3\] and \[\left(\sqrt[4]{3}\right)^4=3\]
For natural number \(m\) and \(n\) applies: \[\sqrt[4]{m}\times \sqrt[4]{n}= \sqrt[4]{m\times n}\]
example
\[\sqrt[4]{2}\times \sqrt[4]{8}=\sqrt[4]{2\times 8}\] because \[\begin{aligned}\left(\sqrt[4]{2}\times \sqrt[4]{8}\right)^4 &= \left(\sqrt[4]{2}\right)^4\times \left(\sqrt[4]{8}\right)^4\\ &= 2\times 8\\ &= 16\\ &= \left(\sqrt[4]{16}\right)^4 \\ &= \left(\sqrt[4]{2\times 8}\right)^4\end{aligned}\]
Quotient rule of quartic roots
The quartic root of a fraction with positive natural numbers in the numerator and the denominator is equal to the quotient of the quartic root of the numerator and the quartic root of the denominator .
In formula language we have for positive natural numbers \(m\) and \(n\) : \[\sqrt[4]{\frac{m}{n}}=\frac{\sqrt[4]{m}}{\sqrt[4]{n}}\]
example
\[\sqrt[4]{\frac{81}{216}}=\frac{\sqrt[4]{81}}{\sqrt[4]{216}}=\frac{3}{4}\] and indeed \[\left(\frac{3}{4}\right)^4=\frac{3^4}{4^4}=\frac{81}{216}\]
But there is a new rule for a quartic root regarding the reduction of quartic root to square root under special circumstances.
For any natural number \(n\) we have: \[\sqrt[4]{n^2}= \sqrt{n}\]
Example
\[\sqrt[4]{9}=\sqrt[4]{3^2}=\sqrt{3}\] because \[\begin{aligned}\left(\sqrt[4]{9}\right)^4 &= 9\\ &= 3^2\\ &= \bigl((\sqrt{3})^2\bigr)^2\\ &= \sqrt{3}^4 \end{aligned}\]
The above rules can be used to simplify quartic roots.
An irreducible quartic root and the standard form of a quartic root The quartic root of a natural number greater than 1, say \(\sqrt[4]{n}\), is called irreducible if \(n\) has no quartic number (i.e., a fourth power of a natural number) greater than 1 as divisor and also cannot be reduced to a square root because the number under root sign is a square number. So \(\sqrt[4]{24}=\sqrt[3]{2^3\times 3}\) is an irreducible quartic root, but \(\sqrt{36}\) and \(\sqrt[4]{1250}\) are not irreducible, because \[\begin{aligned}\sqrt[4]{36}&=\sqrt[4]{6^2}\\ &=\sqrt{6}\\ \\ \sqrt[4]{1250} &=\sqrt[4]{625\times 2}\\ &=\sqrt[4]{5^4\times 2}\\ &=\sqrt[4]{5^4}\times\sqrt[4]{2}\\ &=5\times \sqrt[4]{2}\end{aligned}\] The last expression we usually write shorter as \(5\sqrt[4]{2}\).
Every quartic root of a natural number greater than 1 can be written in standard form, i.e., as a natural number, a square root, or as the product of a natural number, and an irreducible square or quartic root.
The expression \(m\sqrt[4]{n}\) for integers \(m\) and \(n>1\) is in standard form if
- there exists no fourth power of a natural number greater than 1 that divides \(n\), and
- \(n\) is not equal to a square number.
You find the standard form of a quartic root by 'extracting all fourth powers from the quartic root' and by reducing the quartic root of a square number to a square root. The following examples illustrate this.
First we find the largest possible fourth power that divides \(1701\).
In this case we can write: \[\begin{aligned}1701&=81\times 21\\ \\ &={3}^4\times 21\end{aligned}\] This example follows from the prime factorisation of \(1701\) : \[1701=3^5\times 7\] Once the largest fourth power has been found that divides \(1701\) we apply the calculation rules \[\sqrt[4]{m\times n} = \sqrt[4]{m}\times\sqrt[4]{n}\] and \[\sqrt[4]{n^4}=n\] for natural numbers \(m\) and \(n\) to: \[\begin{aligned} \sqrt[4]{1701} &= \sqrt[4]{{3}^4\times 21}\\ \\ &=\sqrt[4]{{3}^4}\times \sqrt[4]{21}\\ \\ &=3\sqrt[4]{21}\end{aligned}\]
Now that we know how square, cube and quartic roots can be treated mathematically, the way for higher power roots lies open.
Higher roots In general, the \(n\)-th root \(\sqrt[n]{a}\) of \(a\) is the number \(w\) such that \(w^n =a\), provided that \(a\ge 0\) in case \(n\) is even. If \(n\) is even, then we have \(w^n=(-w)^n\) and there are two candidates for the root: the conventional choice is the positive number. We denote this root also as \(a^{\frac{1}{n}}\).
Calculation rules of higher roots
\[\begin{aligned} \sqrt[n]{a}\times \sqrt[n]{b}&=\sqrt[n]{a\times b}\\ \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}&=\sqrt[n]{\frac{a}{b}}\\ \\ \left(\sqrt[n]{a}\right)^m &= \sqrt[n]{a^m}\\ \\ \left(\sqrt[n]{a}\right)^n &=a \\ \\ \sqrt[m\times n]{a^{m}} &= \sqrt[n]{a}\end{aligned}\] If \(n\) is even, these rules only hold for positive values of \(a\) and \(b\).
Standard form of higher power roots
The expression \(\frac{a}{b}\sqrt[n]{c}\), where \(a\), \(b\) and \(c\) are positive natural numbers, is called a standard form of a higher root of a positive rational number if
- \(\frac{a}{b}\) is an irreducible fraction,
- \(c\) has no \(n\)-th power other than \(1\) is divisor,
- \(c\) is not equal to a \(d\)-th power of each divisor \(d\) of \(n\)
The third condition is new compared to the cases \(n=2\) and \(n=3\) and is connected with the last of the above calculation rules.
For example, the standard form of \(\sqrt[6]{144}\) is \(\sqrt[3]{12}\) because \(144 = 2^4\times 3^2=(2^2\times 3)^2 = 18^2\) so \(\sqrt[6]{144}=\sqrt[6]{12^2}=\sqrt[3]{12}\).
You can use the following prime factorisation: \[2733750={2}\times {3}^{7}\times {5}^{4}\]
\[\begin{aligned} \sqrt[4]{2733750} &=\sqrt[4]{{2}\times {3}^{7}\times {5}^{4}} & \blue{\text{prime factorisation}}\\ \\ &= 3\times 5\times \sqrt[4]{ {{3}^{3}\times } {} 2 } & \blue{\text{fourth power extracted from the root}} \\ \\ &= 15 \sqrt[4]{54} & \blue{\text{standard form}}\end{aligned}\]
