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 \(\frac{\sqrt{10}}{\sqrt{31}}\) 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.

\(\frac{\sqrt{10}}{\sqrt{31}}\,=\,\) \({{1}\over{31}} \sqrt{310}\)

First we remove the root from the denominator by multiplying numerator and denominator by #\sqrt{31}#: \(\frac{\sqrt{10}}{\sqrt{31}}=\frac{\sqrt{10}}{\sqrt{31}}\cdot \frac{\sqrt{31}}{\sqrt{31}}=\frac{\sqrt{10\cdot 31 }}{31}\). The integer under the root sign is now positive. We need to find the quadratic factors. Since #10\cdot 31 = (1)^2\cdot 310 #, we have #\sqrt{10\cdot 31}= 1 \cdot \sqrt{310}#, so \(\frac{\sqrt{10\cdot 31 }}{31}= \frac{1}{31}\sqrt{310}\). Moreover, #310# contains no squares, so the required standard form for \(\frac{\sqrt{10}}{\sqrt{31}}\) is \(\frac{1}{31}\sqrt{310} \).

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