Standard notation of roots

Numbers: Powers and roots

Standard notation of roots

We have seen some calculation rules for roots and roots of fractions. Using these calculation rules, we can simplify roots.

When a root occurs in an expression, the expression is in standard notation if it meets the following requirements.

The number in front of the radical sign is as large as possible.
This means that we eliminate the squares from the number inside the radical sign. Keep in mind that we only want a single square root. Therefore, you should not 'simplify' #\sqrt{14}# to #\sqrt{2}\cdot\sqrt{7}#. We do the following.
\[\sqrt{12}=\sqrt{2^2 \cdot 3}=\sqrt{2^2} \cdot \sqrt{3}=2\sqrt{3}\]
There are no fractions inside the radical sign. We use that the root of a fraction equals the fraction of the roots. \[\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}}\]
There are no roots in the denominator of the fraction.
We achieve this by multiplying the numerator and the denominator of the fraction by the root from the denominator.
\[\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=\frac{\sqrt{3}}{3}\]
The rational numbers in the expression are simplified fractions. \[\frac{3}{\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}\]

Examples

\[\begin{array}{rcl}\sqrt{56}&=&\sqrt{2^2 \cdot 14} \\ &=& \sqrt{2^2} \cdot \sqrt{14} \\ &=& 2 \sqrt{14} \\ \\\sqrt{108}&=&\sqrt{2^2 \cdot 3^2 \cdot 3} \\&=& \sqrt{2^2} \cdot \sqrt{3^2} \cdot \sqrt{3} \\ &=& 2 \cdot 3 \cdot \sqrt{3} \\ &=& 6 \sqrt{3} \\ \\ \dfrac{2}{\sqrt{5}}&=& \dfrac{2 \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}}\\ &=& \dfrac{2 \sqrt{5}}{5} \\ \dfrac{3}{\sqrt{63}}&=&\dfrac{3}{\sqrt{3^2\cdot7}}\\&=& \dfrac{3}{3\cdot\sqrt{7}}\\&=&\dfrac{1}{\sqrt{7}}\\&=& \dfrac{\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}\\&=&\dfrac{\sqrt{7}}{7}\end{array}\]

In practice

In practice, we address the first requirement by checking if the number inside the radical sign is divisible by any squares ( #4#, #9#, #16#, #25,\ldots# ). We first check whether the number is divisible by #4#. If so, we take #4# out of the square root, and repeat the process with the new number inside the radical sign. If not, we check if it is divisible by #9#, and so on.

We can also do this by writing the number in its prime factorisation and seeing if there are prime factors that occur more than once.

Two notations The expressions #\frac{2}{3}\sqrt{5}# and #\frac{2\sqrt{5}}{3}# are standard notations of the same number.

Write \(\sqrt{60}\) in standard notation.

#\sqrt{60}=# #2\sqrt{15}#

#\begin{array}{rcl}
\sqrt{60}&=&\sqrt{4\times 15} \\ &&\phantom{xxx}\blue{60 \text{ written as a product inside the radical symbol}} \\
&=& \sqrt{{2}^2\times 15} \\ &&\phantom{xxx}\blue{4 \text{ written as a square}} \\
&=& \sqrt{{2}^2} \times \sqrt{15} \\ &&\phantom{xxx}\blue{\text{calculation rule: the root of a product is the product of roots}} \\
&=&2\sqrt{15} \\ &&\phantom{xxx}\blue{\text{root of a square is the absolute value of the number}} \end{array}#

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