Roots of fractions

Numbers: Roots

Roots of fractions

For roots of positive rational numbers, there is also a standard notation.

Rule of calculation for roots of fractions

If $p$ and $q$ are positive integers, then $\sqrt{\frac{p}{q}}=\frac{\sqrt{p}}{\sqrt{q}} = \frac{1}{q}\sqrt{p\cdot q}$ .

The definition of root says $b=\sqrt{\frac{p}{q}}$ is the unique non-negative number with $b^2=\frac{p}{q}$ . But $\frac{\sqrt{p}}{\sqrt{q}}$ also satisfies both properties: it is clearly positive because numerator and denominator are positive and $\left(\frac{\sqrt{p}}{\sqrt{q}}\right)^2 = \frac{\left(\sqrt{p}\right)^2}{\left(\sqrt{q}\right)^2}=\frac{p}{q}$ . The conclusion is that $\frac{\sqrt{p}}{\sqrt{q}}$ equals $b$ , hence to $\sqrt{\frac{p}{q}}$ . This proves the first equality.

But $\frac{1}{q}\sqrt{p\cdot q}$ is also positive and has $b$ (as above) as a square: $\left(\frac{1}{q}\sqrt{p\cdot q}\right)^2 = \frac{1}{q^2}\cdot\left(\sqrt{p\cdot q}\right)^2=\frac{1}{q^2}\cdot p\cdot q =\frac{p}{q}$ . Hence $\frac{1}{q}\sqrt{p\cdot q}$ also equals $b$ , and hence also equals $\frac{\sqrt{p}}{\sqrt{q}}$ . This proves the second equality.

Standard notation for the root of a rational number

If we take the square root of a positive rational number $\frac{p}{q}$ , we first write the result as $\frac{1}{q}\sqrt{p\cdot q}$ , next we write $\sqrt{p\cdot q}$ in the standard notation for the root of an integer. The result therefore has the form $\frac{a}{b}\sqrt{c}$ , wherein $a$ , $b$ , and $c$ are positive integers and $c$ has no more squares as factors has. This form is called the standard notation of $\sqrt{\frac{p}{q}}$ .

The standard notation of $\sqrt{\frac{2}{27}}$ is $\frac{1}{9}\sqrt{6}$ . This can be calculated as follows: $\sqrt{\frac{2}{27}} = \frac{1}{27}\sqrt{2\cdot27}=\frac{1}{27}\sqrt{2\cdot 3^3}=\frac{3}{27}\sqrt{2\cdot 3}=\frac{1}{9}\sqrt{6}$ It makes sense not to perform the multiplication under the root sign before removing all squares from under the root.

Another reduction of this answer is $\sqrt{\frac{2}{27}} =\sqrt{\frac{2}{3^2\cdot 3}} =\frac{1}{3}\sqrt{\frac{2}{ 3}}=\frac{1}{3}\frac{\sqrt{2}}{ \sqrt{3}} = \frac{1}{3\cdot 3}\sqrt{2\cdot 3} = \frac{1}{9}\sqrt{6}$

Write the expression $\frac{\sqrt{5}}{\sqrt{2}}$ in standard notation. In other words, write it as $\frac{a}{b}\cdot \sqrt{c}$ , where $c$ is a positive integer that does not have a square as a divisor.

$\frac{\sqrt{5}}{\sqrt{2}}\,=\,$ ${{1}\over{2}} \sqrt{10}$

First we remove the root from the denominator by multiplying numerator and denominator by $\sqrt{2}$ : $\frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{5}}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{5\cdot 2 }}{2}$ . The integer under the root sign is now positive. We need to find the quadratic factors. Since $5\cdot 2 = (1)^2\cdot 10$ , we have $\sqrt{5\cdot 2}= 1 \cdot \sqrt{10}$ , so $\frac{\sqrt{5\cdot 2 }}{2}= \frac{1}{2}\sqrt{10}$ . Moreover, $10$ contains no squares, so the required standard form for $\frac{\sqrt{5}}{\sqrt{2}}$ is $\frac{1}{2}\sqrt{10}$ .

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Numbers: Roots

Roots of fractions

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