The square root of a natural number

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

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

$\sqrt{9} = 3$ because $3^2 = 9$ Also, $(-3)^2 = 9$ and thus $-3$ would also be a good candidate for the 'square root of 9'. However, as stated in the definition, the term $\sqrt{a}$ stands only for the positive number, the square of which is equal to $a$ , i.e., $\sqrt{9} = +3$ . The square root of a positive integer that is not a square of an integer is always irrational. So $\sqrt{3}$ is not rational, i.e., cannot be written as a fraction.

Square root of a negative number The root of a negative number does not exist within the set of real numbers because squares of such number can never be negative. But you can't catch out mathematicians: they can imagine such a root and extend the set of real number with imaginary numbers. They come in this way with the set of complex numbers, that are commonly used in mathematics, science and engineering. The root of $-1$ is denoted by $\ii = \sqrt{-1}$ .

Root of a square number

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

$\left(\sqrt{4}\right)^2=4$

Product rule of square roots

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

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

How temping it may be, there is no sum rule for roots. For nonnegative numbers $m$ and $n$ we have:

$\sqrt{m}+\sqrt{n}\;\red{\neq}\;\sqrt{m+n}$ A concrete example with numbers illustrates this:

$5=2+3=\sqrt{4}+\sqrt{9}\;\red{\neq}\;\sqrt{4+9}=\sqrt{13}$

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

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 $n$ .

That there exist a standard form of a square root of a positive integer is rooted in the fact that any positive integer $c$ can be uniquely written as $a^2\times b$ , where $a$ and $b$ are positive integers and $b$ cannot be divided by a square of an integer greater than 1. We have: $\sqrt{c}=a\sqrt{b}$ .

If $p$ is a prime divisor of $c$ , but $p^2$ does not divide $c$ , then $p$ is a prime divisor of $b$ and contributes to this number.
If $p$ is a prime divisor of $c$ and $p^2$ divides $c$ , but $p^2$ does not divide $c$ , then $p$ is a divisor of $a$ and contributes to this number.
If $p$ is a prime divisor of $c$ , and $p^2$ and $p^3$ divide $c$ , but $p^4$ does not divide $c$ , then $p$ is a divisor of $a$ an $b$ and contributes to both numbers.
And we can go on like this.

Concrete examples:

$\begin{aligned} 12&=2^2\times 3\quad\text{and thus }\sqrt{12}=2\sqrt{3}\\ 24&=2^3\times 3\quad\text{and thus }\sqrt{24}=2\sqrt{2\times 3}=2\sqrt{6}\\ 48&=2^4\times 3\quad\text{and thus }\sqrt{48}=2^2\sqrt{3}=4\sqrt{3}\end{aligned}$

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

Write $\sqrt{396}$ in standard form.

First we find the largest perfect square that is a divisor of $396$ .
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 $n$ :

$\begin{aligned} \sqrt{396} &= \sqrt{{6}^2\times 11}\\ \\ &=\sqrt{{6}^2}\times \sqrt{11}\\ \\ &=6\sqrt{11}\end{aligned}$

New example

Mathcentre video

Surd and Other Roots (33:54)