### 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 non-negative number.

The quartic root

The **quartic root** or **fourth root** of a number, \(a\ge 0\) is by definition the non-negative 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 \(9375\).

In this case we can write: \[\begin{aligned}9375&=625\times 15\\ \\ &={5}^4\times 15\end{aligned}\] This example follows from the prime factorisation of \(9375\) : \[9375=3\times 5^5\] Instead of prime factorisation of a number you can also proceed in small steps and already extract a recognised forth power from the root. In this case you see perhaps that \(9375\) can be divided by thefourth power \(625\) and you can write down: \[375=5^4\times 15\] Next you can focus on finding a fourth power that divides the newly obtained smaller number under the root sign, namely \(15\). In this way you may get to the greatest square number that divides \(9375\) or you reduce the problem in each step to a less difficult problem of similar nature.

Once the largest fourth power has been found that divides \(9375\) 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: \[\begin{aligned} \sqrt[4]{9375} &= \sqrt[4]{{5}^4\times 15}\\ \\ &=\sqrt[4]{{5}^4}\times \sqrt[4]{15}\\ \\ &=5\sqrt[4]{15}\end{aligned}\]

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: \[720={2}^{4}\times {3}^{2}\times {5}\]

\[\begin{aligned} \sqrt[3]{720} &=\sqrt[3]{{2}^{4}\times {3}^{2}\times {5}} & \blue{\text{prime factorisation}}\\ \\ &= 2\times \sqrt[3]{ {2\times } {{3}^{2}\times } 5 } & \blue{\text{third power extracted from the root}} \\ \\ &= 2 \sqrt[3]{90} & \blue{\text{standard form}}\end{aligned}\]