Editing Nth root (section)

==Simplified form of a radical expression==

A [[nested radical|non-nested radical expression]] is said to be in '''simplified form''' if no factor of the radicand can be written as a power greater than or equal to the index; there are no fractions inside the radical sign; and there are no radicals in the denominator.<ref>{{cite book|last=McKeague|first=Charles P.|title=Elementary algebra|page=470|year=2011|publisher=Cengage Learning |url=https://books.google.com/books?id=etTbP0rItQ4C&q=editions:q0hGn6PkOxsC|isbn=978-0-8400-6421-9}}</ref>

For example, to write the radical expression <math>\textstyle \sqrt{32/5}</math> in simplified form, we can proceed as follows. First, look for a perfect square under the square root sign and remove it:

<math display="block">
\sqrt{\frac{32}{5}} = \sqrt{\frac{16 \cdot 2}{5}}
= \sqrt{16} \cdot \sqrt{\frac{2}{5}} = 4 \sqrt{\frac{2}{5}}
</math>

Next, there is a fraction under the radical sign, which we change as follows:

<math display="block">4 \sqrt{\frac{2}{5}} = \frac{4 \sqrt{2}}{\sqrt{5}}</math>

Finally, we remove the radical from the denominator as follows:

<math display="block">\frac{4 \sqrt{2}}{\sqrt{5}} = \frac{4 \sqrt{2}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{4 \sqrt{10}}{5} = \frac{4}{5}\sqrt{10}</math>

When there is a denominator involving surds it is always possible to find a factor to multiply both numerator and denominator by to simplify the expression.<ref>{{Cite conference|first1=B. F. |last1=Caviness|first2=R. J. |last2=Fateman|chapter-url=http://www.eecs.berkeley.edu/~fateman/papers/radcan.pdf|chapter=Simplification of Radical Expressions|title=Proceedings of the 1976 ACM Symposium on Symbolic and Algebraic Computation|page=329}}</ref><ref>{{cite journal|last=Richard|first=Zippel|title=Simplification of Expressions Involving Radicals|journal=Journal of Symbolic Computation|volume=1|number=189–210|year=1985|pages=189–210 |doi=10.1016/S0747-7171(85)80014-6}}</ref> For instance using the [[Factorization#Sum/difference of two cubes|factorization of the sum of two cubes]]:

<math display="block">
  \frac{1}{\sqrt[3]{a} + \sqrt[3]{b}} =
  \frac{\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}}{\left(\sqrt[3]{a} + \sqrt[3]{b}\right)\left(\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}\right)} =
  \frac{\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}}{a + b} .
</math>

Simplifying radical expressions involving [[nested radical]]s can be quite difficult. In particular, denesting is not always possible, and when possible, it may involve advanced [[Galois theory]]. Moreover, when complete denesting is impossible, there is no general [[canonical form]] such that the equality of two numbers can be tested by simply looking at their canonical expressions. 

For example, it is not obvious that

<math display="block">\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}.</math>

The above can be derived through:

<math display="block">\sqrt{3 + 2\sqrt{2}} = \sqrt{1 + 2\sqrt{2} + 2} = \sqrt{1^2 + 2\sqrt{2} + \sqrt{2}^2} = \sqrt{\left(1 + \sqrt{2}\right)^2} = 1 + \sqrt{2}</math>

Let <math>r=p/q</math>, with {{mvar|p}} and {{mvar|q}} coprime and positive integers. Then <math>\sqrt[n]r = \sqrt[n]{p}/\sqrt[n]{q}</math> is rational if and only if both <math>\sqrt[n]{p}</math> and <math>\sqrt[n]{q}</math> are integers, which means that both {{mvar|p}} and {{mvar|q}} are ''n''th powers of some integer.