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

### The square root of a natural number

The square root of a number

The **square root** \(a\ge 0\) is by definition the nonnegative number \(w\) such that \(w^2 = a\). Notation: \(w = \sqrt{a}\) and \(w=a^{\frac{1}{2}}\)

**Examples**

\[\begin{aligned}\sqrt{4}&={2}\quad\text{because }2^2=4\text{ and }2\ge 0\\[0.2cm] \sqrt{16}&={4}\quad\text{because }4^2=16\end{aligned}\]

For any natural number \(n\) we have: \[\sqrt{n^2}=n\] and \[\left(\sqrt{n}\right)^2=n\]

**Example** \(n=4\)

\[\sqrt{4^2}=4\] and \[\left(\sqrt{4}\right)^2=4\]

For any natural numbers \(m\) and \(n\) we have: \[\sqrt{m}\times \sqrt{n}= \sqrt{m\times n}\]

**Example**

\[\sqrt{2}\times \sqrt{3}=\sqrt{2\times 3}\] because \[\begin{aligned}\left(\sqrt{2}\times \sqrt{3}\right)^2 &= \left(\sqrt{2}\right)^2\times \left(\sqrt{3}\right)^2\\ &= 2\times 3\\ &= 6\\ &= \left(\sqrt{6}\right)^2 \\ &= \left(\sqrt{2\times 3}\right)^2\end{aligned}\]

The above properties of roots can be used to simplify roots and expressions containing roots.

An irreducible square root and the standard form of a square root The square root of a natural number greater than 1, say \(\sqrt{n}\) is called **irreducible** if \(n\) cannot be divided by a square number greater than 1. Thus, \(\sqrt{6}=\sqrt{2\times 3}\) and \(\sqrt{30}=\sqrt{2\times 3\times 5}\) are irreducible roots, but \(\sqrt{18}\) is not, because \(\sqrt{18}=\sqrt{9\times 2}=\sqrt{3^2\times 2}=3\sqrt{2}\)

Each square root of a positive integer can be written in the standard form, i.e., as a positive integer or as the product of a positive number and an irreducible square root.

So, the expression \(m\sqrt{n}\) for positive integers \(m\) and \(n>1\) is in standard form if there exist no square number greater than 1 that divides \(n\).

You can find the standard form by 'extracting all squares from the root'. The example below illustrate this.

In this case we can write: \[\begin{aligned}396&=36\times 11\\ \\ &={6}^2\times 11\end{aligned}\] This follows, for example, from the prime factorisation of \(396\): \[396=2^2\times 3^2\times 11\] Instead of prime factorisation of a number you can also proceed in small steps and already extract a recognised square from the root. In this case you see perhaps that \(396\) can be divided by the square number \(9\) and you can write down: \[396=3^2\times 44\] Next you can focus on finding a square number that divides the newly obtained smaller number under the root sign, namely \(44\). In this way you may get to the greatest square number that divides \(396\) or you reduce the problem in each step to a less difficult problem of similar nature.

Once the largest perfect square is found that divides \(396\) we can apply the computational rules \[\sqrt{m\times n} = \sqrt{m}\times\sqrt{n}\] and \[\sqrt{n^2}=n\] for natural numbers \(m\) and \(n\): \[\begin{aligned} \sqrt{396} &= \sqrt{{6}^2\times 11}\\ \\ &=\sqrt{{6}^2}\times \sqrt{11}\\ \\ &=6\sqrt{11}\end{aligned}\]

Mathcentre video

Surd and Other Roots (33:54)