Nth root

{{short description|Arithmetic operation, inverse of nth power}} {{about|nth-roots of real and complex numbers|other uses|Root (disambiguation)#Mathematics}} {{cs1 config|mode=cs1}} [[File:Root-rendered-by-TeX.svg|right|thumb|upright=0.8|Modern notation for the ''n''th root of the variable ''x'']]

In mathematics, an '''{{mvar|n}}th root''' of a number {{mvar|x}} is the number {{mvar|r}} which, when multiplied by itself {{mvar|n}} times, yields&nbsp;{{mvar|x}}: <math display="block">r^n = \underbrace{r \times r \times \dotsb \times r}_{n\text{ factors}} = x.</math> The positive integer {{mvar|n}} is called the ''index'' or ''degree'', and the number {{mvar|x}} of which the root is taken is the ''radicand.'' A root of degree 2 is called a ''square root'' and a root of degree 3, a ''cube root''. Roots of higher degree are referred by using ordinal numbers, as in ''fourth root'', ''twentieth root'', etc. The computation of an {{mvar|n}}th root is a '''root extraction'''.

The {{mvar|n}}th root of {{mvar|x}} is written as <math>\sqrt[n]{x}</math> using the radical symbol <math>\sqrt{\phantom x}</math>. The square root is usually written as {{tmath|\sqrt x}}, with the degree omitted. Taking the {{mvar|n}}th root of a number, for fixed {{tmath|n}}, is the inverse of raising a number to the {{mvar|n}}th power,<ref>{{cite web |url=https://www.nagwa.com/en/explainers/985195836913 |access-date=22 July 2023 |title=Lesson Explainer: nth Roots: Integers}}</ref> and can be written as a fractional exponent:

<math display="block">\sqrt[n]{x} = x^{1/n}.</math>

For a positive real number {{mvar|x}}, <math>\sqrt{x}</math> denotes the positive square root of {{mvar|x}} and <math>\sqrt[n]{x}</math> denotes the positive real {{mvar|n}}th root. For example, {{math|3}} is a square root of {{math|9}}, since {{math|1=3{{sup|2}} = 9}}, and {{math|−3}} is also a square root of {{math|9}}, since {{math|1=(−3){{sup|2}} = 9}}.<ref name=Zuckerman_1986/> A negative real number {{math|−''x''}} has no real-valued square roots, but when {{mvar|x}} is treated as a complex number it has two imaginary square roots, {{tmath|+i\sqrt x }} and {{tmath|-i\sqrt x }}, where {{mvar|i}} is the imaginary unit.

In general, any non-zero complex number has {{mvar|n}} distinct complex-valued {{mvar|n}}th roots, equally distributed around a complex circle of constant absolute value. (The {{mvar|n}}th root of {{math|0}} is zero with multiplicity {{mvar|n}}, and this circle degenerates to a point.) Extracting the {{mvar|n}}th roots of a complex number {{mvar|x}} can thus be taken to be a multivalued function. By convention the principal value of this function, called the '''principal root''' and denoted {{tmath|\sqrt[n]{x} }}, is taken to be the {{mvar|n}}th root with the greatest real part and in the special case when {{mvar|x}} is a negative real number, the one with a positive imaginary part. The principal root of a positive real number is thus also a positive real number. As a function, the principal root is continuous in the whole complex plane, except along the negative real axis. The {{mvar|n}}th roots of 1 are called roots of unity and play a fundamental role in various areas of mathematics, such as number theory, theory of equations, and Fourier transform.

An unresolved root, especially one using the radical symbol, is sometimes referred to as a '''surd'''<ref name=Bansal_2006>{{cite book | title=New Approach to CBSE Mathematics IX | first=R. K. | last=Bansal | page=25 | year=2006 | isbn=978-81-318-0013-3 | publisher=Laxmi Publications | url=https://books.google.com/books?id=1C4iQNUWLBwC&pg=PA25 }}</ref> or a '''radical'''.<ref name=silver>{{cite book | last=Silver | first=Howard A. | title=Algebra and trigonometry | year=1986 | publisher=Prentice-Hall | location=Englewood Cliffs, New Jersey | isbn=978-0-13-021270-2 | url-access=registration | url=https://archive.org/details/algebratrigonome00silv }}</ref> Any expression containing a radical, whether it is a square root, a cube root, or a higher root, is called a '''''radical expression''''', and if it contains no transcendental functions or transcendental numbers it is called an ''algebraic expression''.

{{Arithmetic operations}}

==History==

{{Main article|Square root#History|Cube root#History}} The Babylonians, as early as 1800 BCE, demonstrated numerical approximations of irrational quantities such as the square root of 2 on clay tablets, with an accuracy analogous to six decimal places, as in the tablet YBC 7289.<ref name=Fowler_Eleanor_1998>{{cite journal | title=Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context | first1=David | last1=Fowler | first2=Eleanor | last2=Robson | journal=Historia Mathematica | volume=25 | issue=4 | date=November 1998 | pages=366–378 | publisher=Elsevier | doi=10.1006/hmat.1998.2209 }}</ref> Cuneiform tablets from Larsa include tables of square and cube roots of integers.<ref>{{cite book | title=The History of Mathematics: A Brief Course | first=Roger L. | last=Cooke | edition=3rd | publisher=John Wiley & Sons | year=2012 | isbn=978-1-118-46029-0 | url=https://books.google.com/books?id=CFDaj0WUvM8C&pg=PT49 }}</ref> The first to prove the irrationality of √2 was most likely the Pythagorean Hippasus.<ref>{{cite journal | title=The Discovery of Incommensurability by Hippasus of Metapontum | first=Kurt | last=Von Fritz | journal=Annals of Mathematics | volume=46 | issue=2 | date=April 1945 | pages=242–264 | doi=10.2307/1969021 | jstor=1969021 }}</ref> Plato in his ''Theaetetus'', then describes how Theodorus of Cyrene (c. 400 BC) proved the irrationality of <math>\sqrt3</math>, <math>\sqrt5</math>, etc. up to <math>\sqrt{17}</math>.<ref>{{cite book | first=T. L. | last=Heath | author-link=Thomas Heath (classicist) | year=1921 | title=A History of Greek Mathematics, Volume&nbsp;1, From Thales to Euclid | page=155 }}</ref> In the first century AD, Heron of Alexandria devised an iterative method to compute the square root, which is actually a special case of the more general Newton's method.<ref>{{cite book | title=History of Continued Fractions and Padé Approximants | volume=12 | series=Springer Series in Computational Mathematics | first=Claude | last=Brezinski | publisher=Springer Science & Business Media | year=2012 | isbn=978-3-642-58169-4 | url=https://books.google.com/books?id=rxzsCAAAQBAJ&pg=PA16 }}</ref>

The term ''surd'' traces back to Al-Khwarizmi ({{circa|825}}), who referred to rational and irrational numbers as "audible" and "inaudible", respectively. This later led to the Arabic word {{lang|ar|أصم}} ({{lang|ar-Latn|asamm}}, meaning "deaf" or "dumb") for "irrational number" being translated into Latin as {{lang|la|surdus}} (meaning "deaf" or "mute"). Gerard of Cremona ({{circa|1150}}), Fibonacci (1202), and then Robert Recorde (1551) all used the term to refer to "unresolved irrational roots", that is, expressions of the form <math>\sqrt[n]{r}</math>, in which <math>n</math> and <math>r</math> are integer numerals and the whole expression denotes an irrational number.<ref>{{cite web | title=Earliest Known Uses of Some of the Words of Mathematics | website=Mathematics Pages | first=Jeff | last=Miller | url=http://jeff560.tripod.com/s.html | access-date=2008-11-30 }}</ref> Irrational numbers of the form <math>\pm\sqrt{a},</math> where <math>a</math> is rational, are called "pure quadratic surds"; irrational numbers of the form <math>a \pm\sqrt{b}</math>, where <math>a</math> and <math>b</math> are rational, are called ''mixed quadratic surds''.<ref>{{cite book | last=Hardy | first=G. H. | author-link=G. H. Hardy | at=§1.13 "Quadratic Surds" – §1.14, {{pgs|19–23}} | title=A Course of Pure Mathematics | year=1921 | edition=3rd | publisher=Cambridge | url=https://archive.org/details/coursepuremath00hardrich/page/n36/mode/2up }}</ref> An archaic term from the late 15th century for the operation of taking ''n''th roots is ''radication'',<ref>{{cite web | url=https://www.merriam-webster.com/dictionary/radication | title=Definition of RADICATION | website=www.merriam-webster.com | access-date=2025-11-10 }}</ref><ref>{{cite web | url=https://en.oxforddictionaries.com/definition/radication | archive-url=https://web.archive.org/web/20180403112348/https://en.oxforddictionaries.com/definition/radication | url-status=dead | archive-date=April 3, 2018 | title=radication – Definition of radication in English by Oxford Dictionaries | website=Oxford Dictionaries }}</ref> and an unresolved root is a ''radical''.<ref name=silver/>

In the fourteenth century, Jamshid al-Kashi used an iterative technique now called the Ruffini-Horner method to extract ''n''th roots for an arbitrary ''n''. This technique has been used since antiquity to determine square roots, then by China and Kushyar ibn Labban during the tenth century to determine cube roots.<ref>{{cite book | chapter=Pure mathematics in Islamic civilization | first=J. P. | last=Hogendijk | title=Companion Encyclopedia of the History and Philosophy of the Mathematical Sciences | volume=1 | editor-first=Ivor | editor-last=Grattan-Guiness | publisher=Routledge | year=2004 | isbn=978-1-134-88748-4 | chapter-url=https://books.google.com/books?id=kGUPEAAAQBAJ&pg=PA77 }}</ref> In 1665, Isaac Newton discovered the general binomial theorem, which can convert an ''n''th root into an infinite series.<ref>{{cite book | title=Mathematics Emerging: A Sourcebook 1540 – 1900 | first=Jacqueline | last=Stedall | publisher=OUP Oxford | year=2008 | isbn=978-0-19-152771-5 | pages=190–191 | url=https://books.google.com/books?id=VwxREAAAQBAJ&pg=PA190 }}</ref> Based on approach developed by François Viète, Newton devised an iterative method for solving a non-linear function of the form <math>f(x) = 0</math>, which can be used to extract an ''n''th root. This technique was further refined by Joseph Raphson and became known as the Newton–Raphson method.<ref>{{cite web | title=Historical Development of the Newton-Raphson Method | first=Tjalling | last=Ypma | year=1995 | publisher=Western Washington University | url=https://wwu.elsevierpure.com/ws/portalfiles/portal/40026944/Historical%20Development%20of%20the%20Newton-Raphson%20Method.pdf | access-date=2025-11-08 }}</ref> In 1690, Michel Rolle introduced the notation <math>\sqrt[n]{a}</math> for the ''n''th root of the value ''a''.<ref>{{cite book | title=Calculus: Early Transcendentals | display-authors=1 | first1=Howard | last1=Anton | first2=Irl C. | last2=Bivens | first3=Stephen | last3=Davis | edition=12th | publisher=John Wiley & Sons | year=2021 | isbn=978-1-119-77818-9 | page=236 | url=https://books.google.com/books?id=sRU-EAAAQBAJ&pg=PA236 }}</ref>

In 1629, Albert Girard proposed the fundamental theorem of algebra, but failed to produce a proof.<ref>{{cite book | title=A Source Book in Mathematics, 1200-1800 | series=Princeton Legacy Library | first=Dirk Jan | last=Struik |author-link=Dirk Jan Struik | publisher=Princeton University Press | year=2014 | isbn=978-1-4008-5800-2 | url=https://books.google.com/books?id=o-3_AwAAQBAJ&pg=PA99 }}</ref> This theorem states that every single-variable polynomial of degree ''n'' has ''n'' roots.<ref name=Clark_2012>{{cite book | title=Elements of Abstract Algebra | series=Dover Books on Mathematics | first=Allan | last=Clark | publisher=Courier Corporation | year=2012 | isbn=978-0-486-14035-3 | page=xi | url=https://books.google.com/books?id=B6LDAgAAQBAJ&pg=PR11 }}</ref> Further, a polynomial with complex coefficients has at least one complex root. Equivalently, the theorem states that the field of complex numbers is algebraically closed. Among the notable mathematicians who worked on a proof during the 18th and 19th centuries were d'Alembert, Gauss, Bolzano, and Weierstrass, with Gauss usually being credited with the first correct proof. A consequence of this proof is that any ''n''th root of a real or complex number will be on the complex plane.<ref>{{cite book | title=Mathematics and Its History | series=Undergraduate Texts in Mathematics | first=John | last=Stillwell | edition=2nd | publisher=Springer Science & Business Media | year=2013 | isbn=978-1-4684-9281-1 | pages=266–267 | url=https://books.google.com/books?id=YfDiBwAAQBAJ&pg=PA267 }}</ref><ref>{{cite book | title=Algebra and Number Theory: A Selection of Highlights | series=De Gruyter Textbook | display-authors=1 | first1=Benjamin | last1=Fine | first2=Anthony | last2=Gaglione | first3=Anja | last3=Moldenhauer | first4=Gerhard | last4=Rosenberger | first5=Dennis | last5=Spellman | publisher=Walter de Gruyter GmbH & Co KG | year=2017 | isbn=978-3-11-051614-2 | pages=203–209 | url=https://books.google.com/books?id=xwA2DwAAQBAJ&pg=PA203 }}</ref>

The ancient Greek mathematicians knew how to use compass and straightedge to construct a length equal to the square root of a given length, when an auxiliary line of unit length is given. In 1837 Pierre Wantzel proved that an ''n''th root of a given length cannot be constructed if ''n'' is not a power of 2.<ref>{{Citation | first=M. L. | last=Wantzel | title=Recherches sur les moyens de reconnaître si un Problème de Géométrie peut se résoudre avec la règle et le compas | journal=Journal de Mathématiques Pures et Appliquées | year = 1837 | volume=1 | issue=2 | pages=366–372 | url=http://visualiseur.bnf.fr/ConsulterElementNum?O=NUMM-16381&Deb=374&Fin=380&E=PDF | archive-url=https://web.archive.org/web/20240209033028/http://visualiseur.bnf.fr/ConsulterElementNum?O=NUMM-16381&Deb=374&Fin=380&E=PDF | access-date=2024-02-09 | archive-date=2024-02-09 }}</ref>

==Definition and notation== {{multiple image <!-- Layout parameters --> | direction = vertical | caption_align = left | width = 240 <!--image 1--> | image1 = NegativeOne4Root.svg | alt1 = A unit circle on the complex plane with roots at 45 degree angles to the axes | caption1 = The four 4th roots of −1, none of which are real <!--image 2--> | image2 = NegativeOne3Root.svg | alt2 = A unit circle on the complex plane with roots at 60 degrees to the positive x axis, and a third root at negative one | caption2 = The three 3rd roots of −1, one of which is a negative real }} An ''{{mvar|n}}th root'' of a number ''x'', where ''n'' is a positive integer, is any of the ''n'' real or complex numbers ''r'' whose ''n''th power is ''x'':

<math display="block">r^n = x.</math>

Every positive real number ''x'' has a single positive ''n''th root, called the principal ''n''th root, which is written <math>\sqrt[n]{x}</math>.<ref name=Brink_1951>{{cite book | title=College Algebra | edition=2nd | first=Raymond W. | last=Brink | year=1951 | page=124 | series=The Appleton-Century Mathematics Series | location=New York | publisher=Appleton-Century-Crofts | url=https://books.google.com/books?id=n-EjEuKqk1YC&pg=PA124 }}</ref> For ''n'' equal to 2 this is called the principal square root and the ''n'' is omitted. The ''n''th root can also be represented using exponentiation as ''x''{{sup|1/n}}.<ref name=Zuckerman_1986/>

For even values of ''n'', positive numbers also have a negative ''n''th root, while negative numbers do not have a real ''n''th root. For odd values of ''n'', every negative number ''x'' has a real negative ''n''th root.<ref name=Brink_1951/> For example, −2 has a real 5th root, <math>\sqrt[5]{-2} = -1.148698354\ldots</math> but −2 does not have any real 6th roots.

Every non-zero number ''x'', real or complex, has ''n'' different complex number ''n''th roots.<ref name=Beilina_et_al_2017>{{cite book | title=Numerical Linear Algebra: Theory and Applications | display-authors=1 | first1=Larisa | last1=Beilina | first2=Evgenii | last2=Karchevskii | first3=Mikhail | last3=Karchevskii | publisher=Springer | year=2017 | page=7 | isbn=978-3-319-57304-5 | url=https://books.google.com/books?id=iaU2DwAAQBAJ&pg=PA7 }}</ref> (In the case ''x'' is real, this count includes any real ''n''th roots.) The only complex root of 0 is 0.

The ''n''th roots of almost all numbers (all integers except the ''n''th powers, and all rationals except the quotients of two ''n''th powers) are irrational.<ref name=Fine_Rosenberger_2007/> For example,

<math display="block">\sqrt{2} = 1.414213562\ldots</math>

All ''n''th roots of rational numbers are algebraic numbers, and all ''n''th roots of integers are algebraic integers.

===Square roots=== {{Main article|Square root}} thumb|right|The graph <math>y=\pm \sqrt{x}</math>. A '''square root''' of a number ''x'' is a number ''r'' which, when squared, becomes ''x'':

<math display="block">r^2 = x.</math>

Every positive real number has two square roots, one positive and one negative. For example, the two square roots of 25 are 5 and −5. The positive square root is also known as the '''principal square root''',<ref name=Zuckerman_1986>{{cite book | title=Intermediate Algebra: A Straightforward Approach | first=Martin M. | last=Zuckerman | publisher=Bloomsbury Publishing PLC | year=1986 | isbn=978-1-4616-3783-7 | pages=249–251 | url=https://books.google.com/books?id=qaj9AQAAQBAJ&pg=PA249 }}</ref> and is denoted with a radical sign:

<math display="block">\sqrt{25} = 5.</math>

Since the square of every real number is nonnegative, negative numbers do not have real square roots.<ref name=Gullberg_1997/> However, for every negative real number there are two imaginary square roots. For example, the square roots of −25 are 5''i'' and −5''i'', where ''i'' represents a number whose square is {{math|−1}}.

===Cube roots=== {{Main article|Cube root}} [[Image:cube-root function.svg|thumb|right|The graph <math>y=\sqrt[3]{x}</math>.]] A '''cube root''' of a number ''x'' is a number ''r'' whose cube is ''x'':

<math display="block">r^3 = x.</math>

Every real number ''x'' has exactly one real cube root,<ref name=Zuckerman_1986/> written <math>\sqrt[3]{x}</math>. For example,

<math display="block">\begin{align} \sqrt[3]{8} &= 2\\ \sqrt[3]{-8} &= -2. \end{align}</math>

Every real number has two additional complex cube roots.<ref>{{cite book | title=The Theory of Equations: With an Introduction to the Theory of Binary Algebraic Forms | series=Dublin University Press series | first1=William Snow | last1=Burnside | first2=Arthur William | last2=Panton | publisher=Hodges, Figgis & Company | year=1881 | page=45 | url=https://books.google.com/books?id=0J4AAAAAMAAJ&pg=PA45 }}</ref><ref>{{cite book | chapter=root of a number | title=Mathematics Dictionary | first=R. C. | last=James | edition=5th | publisher=Springer Science & Business Media | year=1992 | isbn=978-0-412-99041-0 | chapter-url=https://books.google.com/books?id=UyIfgBIwLMQC&pg=PA364 }}</ref>

==Identities and properties== Expressing the degree of an ''n''th root in its exponent form, as in <math>x^{1/n}</math>, makes it easier to manipulate powers and roots. If <math>a</math> is a non-negative real number,<ref name=Gullberg_1997>{{cite book | title=Mathematics From the Birth of Numbers | first=Jan | last=Gullberg | publisher=W. W. Norton & Company | year=1997 | isbn=978-0-393-04002-9 | pages=138–139 | url=https://books.google.com/books?id=E09fBi9StpQC&pg=PA138 }}</ref>

<math display="block">\sqrt[n]{a^m} = (a^m)^{1/n} = a^{m/n} = (a^{1/n})^m = (\sqrt[n]a)^m.</math>

Every non-negative number has exactly one non-negative real ''n''th root, and so the rules for operations with surds involving non-negative radicands <math>a</math> and <math>b</math> are straightforward within the real numbers:<ref name=Gullberg_1997/>

<math display="block">\begin{align} \sqrt[n]{ab} &= \sqrt[n]{a} \sqrt[n]{b} \\ \sqrt[n]{\frac{a}{b}} &= \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \end{align}</math>

Subtleties can occur when taking the ''n''th roots of negative or complex numbers. For instance:

<math display="block">\sqrt{-1}\times\sqrt{-1} \neq \sqrt{-1 \times -1} = 1,\quad</math>

but, rather,

<math display="block">\quad\sqrt{-1}\times\sqrt{-1} = i \times i = i^2 = -1.</math>

Since the rule <math>\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab} </math> strictly holds for non-negative real radicands only, its application leads to the inequality in the first step above.<ref>{{cite book | title=Negative Math: How Mathematical Rules Can be Positively Bent | first=Alberto A. | last=Martínez | publisher=Princeton University Press | year=2006 | pages=122–123 | isbn=978-0-691-12309-7 | url=https://books.google.com/books?id=8HSodlYby9MC&pg=PA122 }}</ref>

==Simplified form of a radical expression==

A 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 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 radicals 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.

==Infinite series== The radical or root may be represented by the generalized binomial theorem:

<math display="block">(1+x)^{s/t} = \sum_{m=0}^\infty \frac{x^m}{m!} \prod_{k=0}^{m-1} \left(\frac st - k\right)</math>

with <math>|x|<1</math>. This expression can be derived from the binomial series.<ref name=Roberts_Tesman_2024>{{cite book | title=Applied Combinatorics | edition=3rd | series=Discrete Mathematics and Its Applications | first1=Fred S. | last1=Roberts | first2=Barry | last2=Tesman | publisher=CRC Press | year=2024 | isbn=978-1-040-12029-3 | page=277 | url=https://books.google.com/books?id=iIIaEQAAQBAJ&pg=PA277 }}</ref> For the ''n''th root, this becomes

<math display="block">(1+x)^\frac{1}{n} = \sum_{m=0}^\infty \frac{x^m}{m!} \prod_{k=0}^{m-1} \left(\frac{1}{n} - k\right)</math>

For numbers <math>r \ge 2</math>, choose a value <math>p</math> such that

<math display="block">\frac{r}{p^n} - 1 = x', \text{ where } |x'| < 1</math>

then per above, solve for

<math display="block">r^\frac{1}{n} = p (1 + x')^\frac{1}{n}</math>

As an example, for <math>r = 30</math> and <math>n = 2</math>, choose <math>p = 5</math><ref name=Roberts_Tesman_2024/>

<math display="block">\frac{30}{5^2} - 1 = \frac{5}{25} = .2</math>

<math display="block">\sqrt{30} = 5 \sqrt{1 + .2} = 5\left[ 1 + \frac{1}{2} (.2)^1 - \frac{1}{8} (.2)^2 + \frac{1}{16}(.2)^3 - \cdots \right] \approx 5.4775</math>

''N''th roots are used to check for convergence of a power series with the root test.<ref>{{cite book | title=An Introduction to Fourier Analysis | first=Russell L. | last=Herman | publisher=CRC Press | year=2016 | isbn=978-1-4987-7371-3 | page=12 | url=https://books.google.com/books?id=ljENDgAAQBAJ&pg=PA12 }}</ref>

==Computing principal roots==

===Using Newton's method===

The {{mvar|n}}th root of a positive real number {{math|''A''}} can be computed with Newton's method, which starts with an initial guess {{math|''x''₀}}, which is also a positive real number, and then iterates using the recurrence relation<ref name=Lange_2013>{{cite book | title=Optimization | series=Springer Texts in Statistics | first=Kenneth | last=Lange | publisher=Springer Science & Business Media | year=2013 | isbn=978-1-4757-4182-7 | page=156 | url=https://books.google.com/books?id=CeUlBQAAQBAJ&pg=PA156 }}</ref>

<math display="block">x_{k+1} = x_k-\frac{x_k^n-A}{nx_k^{n-1}}</math>

until the desired precision is reached. For computational efficiency, the recurrence relation can be rewritten<ref name=Lange_2013/>

<math display="block">x_{k+1} = \frac{1}{n} \left[(n-1)\,x_k+\,\frac{A}{x_k^{n-1}}\right].</math>

This allows the relation to only have one exponentiation, which is computed once for each iteration. The {{mvar|n}}th root of {{mvar|x}} can then be defined as the limit of <math>x_k</math> as {{mvar|k}} approaches infinity.

For example, to find the fifth root of 34, we plug in {{math|1=''n'' = 5, ''A'' = 34}} and {{math|1=''x''₀ = 2}} (initial guess). The first 5 iterations are, approximately:

{{block indent|{{math|1=''x''₀ = 2}}}} {{block indent|{{math|1=''x''₁ = 2.025}}}} {{block indent|{{math|1=''x''₂ = 2.02439 7...}}}} {{block indent|{{math|1=''x''₃ = 2.02439 7458...}}}} {{block indent|{{math|1=''x''₄ = 2.02439 74584 99885 04251 08172...}}}} {{block indent|{{math|1=''x''₅ = 2.02439 74584 99885 04251 08172 45541 93741 91146 21701 07311 8...}}}}

(All correct digits shown.)

The approximation {{math|''x''₄}} is accurate to 25 decimal places and {{math|''x''₅}} is good for 51.

Newton's method can be modified to produce various generalized continued fractions for the ''n''th root. For example,{{cn|date=November 2025}}

<math display="block"> \sqrt[n]{z} = \sqrt[n]{x^n+y} = x+\cfrac{y} {nx^{n-1}+\cfrac{(n-1)y} {2x+\cfrac{(n+1)y} {3nx^{n-1}+\cfrac{(2n-1)y} {2x+\cfrac{(2n+1)y} {5nx^{n-1}+\cfrac{(3n-1)y} {2x+\ddots}}}}}}. </math>

A suitable initial guess for Newton's method may need to be identified using the bisection method or method of false position.<ref>{{cite book | title=Methods in Algorithmic Analysis | series=Chapman & Hall/CRC Computer and Information Science Series | first=Vladimir A. | last=Dobrushkin | publisher=CRC Press | year=2016 | isbn=978-1-4200-6830-6 | page=242 | url=https://books.google.com/books?id=PJWusG2Yp-wC&pg=PA242 }}</ref> For large values of ''n'' and higher requirements for precision, a more rapid algorithm than Newton's method for finding the ''n''th root is to use a truncated Taylor series with a Padé approximant.<ref>{{cite journal | title=Fast computation of the Nth root | first1=S.-G. | last1=Chen | first2=P. Y. | last2=Hsieh | journal=Computers & Mathematics with Applications | volume=17 | issue=10 | year=1989 | pages=1423–1427 | doi=10.1016/0898-1221(89)90024-2 }}</ref>

=== Using the Viète technique === [[Image:PascalForDecimalRoots.svg|right|thumb|Pascal's triangle showing <math>P(4,1) = 4</math>.]] The technique of François Viète, published c. 1600, can be used to perform digit-by-digit calculation of principal roots of decimal (base 10) numbers.<ref>{{cite journal | title=A problem on the approximation of n-roots based on the Viète’s work | last1=Herrero Piñeyro | first1=P. J. | last2=Linero Bas | first2=A. | last3=Massa Esteve | first3=M. R. | last4=Mellado Romero | first4=A. | year=2023 | journal=MATerials MATemàtics | volume=5 | pages=1–27 | url=https://ddd.uab.cat/pub/matmat/matmat_a2023/matmat_a2023a5.pdf | access-date=2025-11-15 }} See p. 8.</ref> This method is based on the binomial theorem and is essentially an inverse algorithm solving <math display="block">(10 x+y)^n = \sum_{k=0}^n P(n, k) (10 x)^{n-k} y^k</math> where <math>P(n, k)</math>, the binomial coefficient, is the ''k''th entry on the ''n''th row of Pascal's triangle.

To compute the root of a number ''C'', choose a series of approximations <math display="block">x_i^n, i = 0, 1, \ldots \text{ with } x_0 = 0</math> that satisfy <math>x_i^n \le C</math>, where the difference between <math>x_{i+1}</math> and <math>x_i</math> is the next digit in the approximation. The decimal fraction <math>y_i</math> is chosen to be the largest number with a single significant digit that satisfies <math display="block">10 x_i + y_i = 10 x_{i+1}, \text{ where } x_{i+1}^n \le C</math> then per the binomial theorem <math display="block">(10 x_i+y)^n - (10 x_i)^n = \sum_{k=0}^{n-1} P(n, k) (10 x_i)^{n-k} y_i^k \le 10^n(C - x_i^n)</math> The term <math>10^n(C - x_i^n)</math> is just a <math>10^n</math> multiple of the ''i''th remainder, <math>C - x_i^n</math>.

Using this expression, any positive principal root can be computed, digit-by-digit, as follows.

Write the original number in decimal form. The numbers are written similar to the long division algorithm, and, as in long division, the root will be written on the line above. Now separate the digits into groups of digits equating to the root being taken, starting from the decimal point and going both left and right. The decimal point of the root will be above the decimal point of the radicand. One digit of the root will appear above each group of digits of the original number.

Beginning with the left-most group of digits, do the following procedure for each group:

# Starting on the left, bring down the most significant (leftmost) group of digits not yet used (if all the digits have been used, write "0" the number of times required to make a group) and write them to the right of the remainder from the previous step (on the first step, there will be no remainder). In other words, multiply the remainder by <math>10^n</math> and add the digits from the next group. This will be the '''current value ''c'''''. # Find ''p'' and ''x'', as follows: #* Let <math>p</math> be the '''part of the root found so far''', ignoring any decimal point. (For the first step, <math>p = 0</math> and <math>0^0 = 1</math>). #* Determine the greatest digit <math>x</math> such that <math>y \le c</math>. #* Place the digit <math>x</math> as the next digit of the root, i.e., above the group of digits you just brought down. Thus the next ''p'' will be the old ''p'' times 10 plus ''x''. # Subtract <math>y</math> from <math>c</math> to form a new remainder. # If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated. Otherwise go back to step 1 for another iteration.

====Examples====

=====Find the square root of 152.2756===== For clarity, the value of the chosen digit ''x'' is in <span style="color:red;">red</span> while the current digital tally is in <span style="color:blue;">blue</span>. {| style="white-space: nowrap; background: #F9F9F9; border: solid lightgrey 1px; border-spacing: 4px;" |- |style="text-align: right;"| 1 |style="text-align: right;"| 2. |style="text-align: right;"| 3 |style="text-align: right;"| 4 |- |colspan=4 style="border-bottom: 1px solid;"| |- | 01 || 52. || 27 || 56 || |style="text-align: right;"| (Results) |colspan="3" style="text-align: center;"| (Explanation) |- |colspan=9 style="border-bottom: 1px solid;"| |- | 01 || || || || |style="text-align: right;"| x = 1: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">0</span>⁰·<span style="color:red;">1</span>² + 2·10¹·<span style="color:blue;">0</span>¹·<span style="color:red;">1</span>¹) ≤ |style="text-align: center;"| 1 | < (1·10⁰·0⁰·2² + 2·10¹·0¹·2¹) |- | 01 || || || || |style="text-align: right;"| y = 1: |colspan=3 style="text-align: center;"| y = 1·10⁰·0⁰·1² + 2·10¹·0¹·1¹ = 1 + 0 = <span style="color:red;">1</span>

|- |colspan=4 style="border-bottom: 1px solid;"| |- | || 52. || || || |style="text-align: right;"| x = 2: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">1</span>⁰·<span style="color:red;">2</span>² + 2·10¹·<span style="color:blue;">1</span>¹·<span style="color:red;">2</span>¹) ≤ |style="text-align: center;"| 52 | < (1·10⁰·1⁰·3² + 2·10¹·1¹·3¹) |- | || 44. || || || |style="text-align: right;"| y = 44: |colspan=3 style="text-align: center;"| y = 1·10⁰·<span style="color:blue;">1</span>⁰·2² + 2·10¹·<span style="color:blue;">1</span>¹·2¹ = 4 + 40 = <span style="color:red;">44</span>

|- |colspan=4 style="border-bottom: 1px solid;"| |- | || 08. || 27 || || |style="text-align: right;"| x = 3: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">12</span>⁰·<span style="color:red;">3</span>² + 2·10¹·<span style="color:blue;">12</span>¹·<span style="color:red;">3</span>¹) ≤ |style="text-align: center;"| 827 | < (1·10⁰·12⁰·4² + 2·10¹·12¹·4¹) |- | || 07. || 29 || || |style="text-align: right;"| y = 729: |colspan=3 style="text-align: center;"| y = 1·10⁰·12⁰·3² + 2·10¹·12¹·3¹ = 9 + 720 = <span style="color:red;">729</span>

|- |colspan=4 style="border-bottom: 1px solid;"| |- | || || 98 || 56 || |style="text-align: right;"| x = 4: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">123</span>⁰·<span style="color:red;">4</span>² + 2·10¹·<span style="color:blue;">123</span>¹·<span style="color:red;">4</span>¹) ≤ |style="text-align: center;"| 9856 | < (1·10⁰·123⁰·5² + 2·10¹·123¹·5¹) |- | || || 98 || 56 || |style="text-align: right;"| y = 9856: |colspan=3 style="text-align: center;"| y = 1·10⁰·123⁰·4² + 2·10¹·123¹·4¹ = 16 + 9840 = <span style="color:red;">9856</span>

|- |colspan=4 style="border-bottom: 1px solid;"| |- | 00 | 00. | 00 | 00 |}

Algorithm terminates: Answer is 12.34

=====Find the cube root of 4192 truncated to the nearest thousandth===== {| style="white-space: nowrap; background: #F9F9F9; border: solid lightgrey 1px; border-spacing: 4px;" |- |style="text-align: right;"| 1 |style="text-align: right;"| 6. |style="text-align: right;"| 1 |style="text-align: right;"| 2 |style="text-align: right;"| 4 |- |colspan=5 style="border-bottom: 1px solid;"| |- | 004 || 192. || 000 || 000 || 000 || |style="text-align: right;"| (Results) |colspan="3" style="text-align: center;"| (Explanation) |- |colspan=10 style="border-bottom: 1px solid;"|

|- | 004 || || || || || |style="text-align: right;"| x = 1: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">0</span>⁰·<span style="color:red;">1</span>³ + 3·10¹·<span style="color:blue;">0</span>¹·<span style="color:red;">1</span>² + 3·10²·<span style="color:blue;">0</span>²·<span style="color:red;">1</span>¹) ≤ |style="text-align: center;"| 4 | < (1·10⁰·0⁰·2³ + 3·10¹·0¹·2² + 3·10²·0²·2¹) |- | 001 || || || || || |style="text-align: right;"| y = 1: |colspan=3 style="text-align: center;"| y = 1·10⁰·0⁰·1³ + 3·10¹·0¹·1² + 3·10²·0²·1¹ = 1 + 0 + 0 = <span style="color:red;">1</span> |- |colspan=5 style="border-bottom: 1px solid;"|

|- | 003 || 192 || || || || |style="text-align: right;"| x = 6: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">1</span>⁰·<span style="color:red;">6</span>³ + 3·10¹·<span style="color:blue;">1</span>¹·<span style="color:red;">6</span>² + 3·10²·<span style="color:blue;">1</span>²·<span style="color:red;">6</span>¹) ≤ |style="text-align: center;"| 3,192 | < (1·10⁰·1⁰·7³ + 3·10¹·1¹·7² + 3·10²·1²·7¹) |- | 003 || 096 || || || || |style="text-align: right;"| y = 3,096: |colspan=3 style="text-align: center;"| y = 1·10⁰·1⁰·6³ + 3·10¹·1¹·6² + 3·10²·1²·6¹ = 4 + 40 = <span style="color:red;">3,096</span> |- |colspan=5 style="border-bottom: 1px solid;"|

|- | || 096 || 000 || || || |style="text-align: right;"| x = 1: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">16</span>⁰·<span style="color:red;">1</span>³ + 3·10¹·<span style="color:blue;">16</span>¹·<span style="color:red;">1</span>² + 3·10²·<span style="color:blue;">16</span>²·<span style="color:red;">1</span>¹) ≤ |style="text-align: center;"| 96,000 | < (1·10⁰·16⁰·2³ + 3·10¹·16¹·2² + 3·10²·16²·2¹) |- | || 077 || 281 || || || |style="text-align: right;"| y = 77,281: |colspan=3 style="text-align: center;"| y = 1·10⁰·16⁰·1³ + 3·10¹·16¹·1² + 3·10²·16²·1¹ = 1 + 480 + 76,800 = <span style="color:red;">77,281</span> |- |colspan=5 style="border-bottom: 1px solid;"|

|- | || 018 || 719 || 000 || || |style="text-align: right;"| x = 2: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">161</span>⁰·<span style="color:red;">2</span>³ + 3·10¹·<span style="color:blue;">161</span>¹·<span style="color:red;">2</span>² + 3·10²·<span style="color:blue;">161</span>²·<span style="color:red;">2</span>¹) ≤ |style="text-align: center;"| 18,719,000 | < (1·10⁰·161⁰·3³ + 3·10¹·161¹·3² + 3·10²·161²·3¹) |- | || 015 || 571 || 928 || || |style="text-align: right;"| y = 15,571,928: |colspan=3 style="text-align: center;"| y = 1·10⁰·161⁰·2³ + 3·10¹·161¹·2² + 3·10²·161²·2¹ = 8 + 19,320 + 15,552,600 = <span style="color:red;">15,571,928</span>

|- |colspan=5 style="border-bottom: 1px solid;"| |- | || 003 || 147 || 072 || 000 || |style="text-align: right;"| x = 4: |style="text-align: right;"| (1·10⁰·<span style="color:blue;">1612</span>⁰·<span style="color:red;">4</span>³ + 3·10¹·<span style="color:blue;">1612</span>¹·<span style="color:red;">4</span>² + 3·10²·<span style="color:blue;">1612</span>²·<span style="color:red;">4</span>¹) ≤ |style="text-align: center;"| 3,147,072,000 | < (1·10⁰·1612⁰·5³ + 3·10¹·1612¹·5² + 3·10²·1612²·5¹) |} The desired precision is achieved. The cube root of 4,192 is 16.124...

===Logarithmic calculation===

The principal ''n''th root of a positive number can be computed using logarithms. Starting from the equation that defines ''r'' as an ''n''th root of ''x'', namely <math>r^n=x,</math> with ''x'' positive and therefore its principal root ''r'' also positive, one takes logarithms of both sides (any base of the logarithm will do) to obtain

<math display="block">n \log_b r = \log_b x \quad \quad \text{hence} \quad \quad \log_b r = \frac{\log_b x}{n}.</math>

The root ''r'' is recovered from this by taking the antilog:<ref>{{cite book | title=Complex Analysis, Determinants and Matrices | volume=1 | series=Mathematical Methods for Engineers and Scientists | first=Kwong-Tin | last=Tang | publisher=Springer Science & Business Media | year=2006 | pages=13–14 | isbn=978-3-540-30274-2 | url=https://books.google.com/books?id=orOTiguKIR4C&pg=PA14 }}</ref>

<math display="block">r = b^{\frac{1}{n}\log_b x}.</math>

(Note: That formula shows ''b'' raised to the power of the result of the division, not ''b'' multiplied by the result of the division.)

For the case in which ''x'' is negative and ''n'' is odd, there is one real root ''r'' which is also negative. This can be found by first multiplying both sides of the defining equation by −1 to obtain <math>|r|^n = |x|,</math> then proceeding as before to find |''r''|, and using {{nowrap|''r'' {{=}} −{{!}}''r''{{!}}}}.

==Complex roots== Every complex number other than 0 has ''n'' different ''n''th roots.<ref name=Beilina_et_al_2017/>

===Square roots=== thumb|right|The square roots of '''''i''''' The two square roots of a complex number are always negatives of each other.<ref>{{cite book | title=Engineering Mathematics | first=John | last=Bird | edition=5th, revised | publisher=Routledge | year=2007 | page=326 | isbn=978-1-136-34697-2 | url=https://books.google.com/books?id=PthCDLSAmu4C&pg=PA326 }}</ref> For example, the square roots of {{math|−4}} are {{math|2''i''}} and {{math|−2''i''}}, and the square roots of {{math|''i''}} are

<math display="block">\tfrac{1}{\sqrt{2}}(1 + i) \quad\text{and}\quad -\tfrac{1}{\sqrt{2}}(1 + i).</math>

If we express a complex number in polar form, then the square root can be obtained by taking the square root of the radius and halving the angle:<ref>{{cite book | title=Mathematical Methods: For Students of Physics and Related Fields | first=Sadri | last=Hassani | edition=2nd | publisher=Springer Science & Business Media | year=2008 | isbn=978-0-387-09504-2 | page=488 | url=https://books.google.com/books?id=JcdGAAAAQBAJ&pg=PA488 }}</ref>

<math display="block">\sqrt{re^{i\theta}} = \pm\sqrt{r} \cdot e^{i\theta/2}.</math>

A ''principal'' root of a complex number may be chosen in various ways, for example

<math display="block">\sqrt{re^{i\theta}} = \sqrt{r} \cdot e^{i\theta/2}</math>

which introduces a branch cut in the complex plane along the positive real axis with the condition {{math|0&nbsp;≤&nbsp;''θ''&nbsp;<&nbsp;2{{pi}}}}, or along the negative real axis with {{math|−{{pi}}&nbsp;<&nbsp;''θ''&nbsp;≤&nbsp;{{pi}}}}.

Using the first(last) branch cut the principal square root <math>\scriptstyle \sqrt z</math> maps <math>\scriptstyle z</math> to the half plane with non-negative imaginary(real) part. The last branch cut is presupposed in mathematical software like Matlab or Scilab.

===Roots of unity=== {{Main article|Root of unity}} thumb|right|The three 3rd roots of 1

The number 1 has ''n'' different ''n''th roots in the complex plane,<ref name=Beilina_et_al_2017/> namely

<math display="block">1,\;\omega,\;\omega^2,\;\ldots,\;\omega^{n-1},</math>

where

<math display="block">\omega = e^\frac{2\pi i}{n} = \cos\left(\frac{2\pi}{n}\right) + i\sin\left(\frac{2\pi}{n}\right).</math>

These roots are evenly spaced around the unit circle in the complex plane, at angles which are multiples of <math>2\pi/n</math>. For example, the square roots of unity are 1 and −1, and the fourth roots of unity are 1, <math>i</math>, −1, and <math>-i</math>. As a result of this symmetry, the sum of the ''n''th roots of unity equals zero.<ref name=Stefanucci_van_Leeuwen_2013>{{cite book | title=Nonequilibrium Many-Body Theory of Quantum Systems: A Modern Introduction | first1=Gianluca | last1=Stefanucci | first2=Robert | last2=van Leeuwen | publisher=Cambridge University Press | year=2013 | isbn=978-1-107-35457-9 | page=503 | url=https://books.google.com/books?id=1tYgAwAAQBAJ&pg=PA503 }}</ref> <math display="block">\sum_{k=0}^{n-1} e^{\frac{2\pi i}{n} k} = 0</math>

===''n''th roots=== {{visualisation_complex_number_roots.svg}} Every complex number has ''n'' different ''n''th roots in the complex plane.<ref name=Beilina_et_al_2017/> These are

<math display="block">\eta,\;\eta\omega,\;\eta\omega^2,\;\ldots,\;\eta\omega^{n-1},</math>

where ''η'' is a single ''n''th root, and 1,&nbsp;''ω'',&nbsp;''ω''{{sup|2}},&nbsp;...&nbsp;''ω''{{sup|''n''−1}} are the ''n''th roots of unity. Thus, since they are all just multiplied by the same scalar ''η'', the sum of the ''n''th roots equals zero.<ref name=Stefanucci_van_Leeuwen_2013/> For example, the four different fourth roots of 2 are

<math display="block">\sqrt[4]{2},\quad i\sqrt[4]{2},\quad -\sqrt[4]{2},\quad\text{and}\quad -i\sqrt[4]{2}.</math>

In polar form, a single ''n''th root may be found from Demoivre's theorem:<ref name=Burd_2019>{{cite book | title=Mathematical Methods in the Earth and Environmental Sciences | first=Adrian | last=Burd | publisher=Cambridge University Press | year=2019 | isbn=978-1-108-63126-6 | page=571 | url=https://books.google.com/books?id=IUmPDwAAQBAJ&pg=PA571 }}</ref>

<math display="block">z^\frac{1}{n} = \sqrt[n]{re^{i\theta}} = \sqrt[n]{r} \cdot e^{i\theta/n} = r^\frac{1}{n} \cdot \left( \cos\left(\frac{\theta}{n}\right) + i \sin\left(\frac{\theta}{n}\right) \right)</math>

Here ''r'' is the magnitude (the modulus, also called the absolute value) of the number whose root is to be taken; if the number can be written as <math>a + i b</math> then <math>r=\sqrt{a^2+b^2}</math>. The <math>\theta</math> is the angle formed as one pivots on the origin counterclockwise from the positive horizontal axis to a ray going from the origin to the number; it has the properties that

: <math>\cos \theta = \frac{a}{r}, \sin \theta = \frac{b}{r}, \text{ and } \tan \theta = \frac{b}{a}.</math>

Thus finding ''n''th roots in the complex plane can be segmented into two steps. First, the magnitude of all the ''n''th roots is the ''n''th root of the magnitude of the original number. Second, the angle between the positive horizontal axis and a ray from the origin to one of the ''n''th roots is <math>\theta / n</math>, where <math>\theta</math> is the angle defined in the same way for the number whose root is being taken. Furthermore, all ''n'' of the ''n''th roots are at equally spaced angles from each other, as proven by the ''n''th root theorem<ref>{{cite book | title=Precalculus: A Functional Approach to Graphing and Problem Solving | series=The Jones & Bartlett learning series in mathematics | first=Karl | last=Smith | publisher=Jones & Bartlett Publishers | year=2013 | isbn=978-0-7637-5177-7 | page=418 | url=https://books.google.com/books?id=ZUJbVQN37bIC&pg=PA418 }}</ref>

:<math>\sqrt[n]{r} \cdot \left( \cos\left(\frac{\theta + 2\pi k}{n}\right) + i \sin\left(\frac{\theta + 2\pi k}{n}\right) \right) \text{ for } k = 0, 1, 2, \ldots, n - 1.</math>

If ''n'' is even, a complex number's ''n''th roots, of which there are an even number, come in additive inverse pairs, so that if a number ''r''₁ is one of the ''n''th roots then ''r''₂ = −''r''₁ is another. This is because raising the latter's coefficient −1 to the ''n''th power for even ''n'' yields 1: that is, (−''r''₁){{sup|''n''}} = (−1){{sup|''n''}} × ''r''₁{{sup|''n''}} = ''r''₁{{sup|''n''}}.

As with square roots, the formula above does not define a continuous function over the entire complex plane, but instead has a branch cut at points where ''θ''&nbsp;/&nbsp;''n'' is discontinuous.

==Polynomial roots==

A ''root'' of a polynomial {{tmath|p(x)}} is a number {{tmath|a}} such that {{tmath|1=p(a)=0}}. An {{mvar|n}}th root of a number {{tmath|a}} is by definition a root of the polynomial {{tmath|x^n-a}}. Algebraic numbers are the numbers that are polynomial roots.

The quadratic formula expresses the roots of quadratic polynomials in terms of square roots. During the 16th century, Gerolamo Cardano and other Italian mathematicians discovered that, similarly, the roots of the polynomials of degree 3 and 4 can always be expressed in terms of {{mvar|n}}th roots (see Cubic equation and Quartic equation).<ref>{{cite book | title=Essentials of Mathematics: Introduction to Theory, Proof, and the Professional Culture | volume=21 | series=Classroom resource materials | first=Margie | last=Hale | publisher=Mathematical Association of America | year=2003 | isbn=978-0-88385-729-8 | page=138 | url=https://books.google.com/books?id=iz6syg59kgsC&pg=PA138 }}</ref>

During the two next centuries, a considerable effort was devoted to the question of whether every algebraic number can be expressed in terms of radicals. In 1824, the proof of the Abel–Ruffini theorem showed that there is no ''general'' formula for the degree 5.<ref>{{cite journal | last=Rosen | first=Michael I. | author-link=Michael Rosen (mathematician) | title=Niels Hendrik Abel and Equations of the Fifth Degree | journal=American Mathematical Monthly | year=1995 | volume=102 | issue=6 | pages=495–505 | doi=10.2307/2974763 | jstor=2974763 | zbl=0836.01015 | mr=1336636}}</ref> This did not completely exclude the possibility of expressing polynomial roots in terms of radicals with formulas depending on each specific polynomial. For example, the quintic polynomial : <math>p(x) = (x - a_1)(x - a_2)(x - a_3)(x - a_4)(x - a_5) = 0</math> has radical roots <math>a_1, a_2, ..., a_5.</math> Galois theory, introduced in 1830 showed that there are polynomials of degree 5 and higher whose roots cannot be expressed in terms of radicals, the simplest example being {{tmath|1= \textstyle x^5 - x - 1}}.<ref>{{cite book | title=Basic Abstract Algebra | display-authors=1 | first1=P. B. | last1=Bhattacharya | first2=S. K. | last2=Jain | first3=S. R. | last3=Nagpaul | publisher=Cambridge University Press | year=1994 | isbn=978-0-521-46081-1 | url=https://books.google.com/books?id=RmgW7kHRYScC&pg=PA348 }}</ref> See {{slink|Quintic function#Solvable quintics}} and {{slink|Galois theory#A non-solvable quintic example}}. In summary, radicals are not always sufficient for expressing polynomial roots.

In spite of this obstacle, Demoivre's theorem demonstrates that an ''n''th root of a number can always be extracted, even for a quintic function <math>x^5 - a</math>.<ref name=Burd_2019/> The two following results, proved in the 19th century resolve fundamental problems on polynomial roots that were set in the 17th century. The fundamental theorem of algebra asserts that every polynomial has complex roots.<ref name=Clark_2012/> There are numbers, called transcendental numbers that are not polynomial roots. The number {{pi}} is an example of such a transcendental number.<ref>{{cite book | title=A History of Pi | series=Griffin Books | first=Petr | last=Beckmann | publisher=Macmillan | year=1971 | isbn=978-0-312-38185-1 | pages=167–169 | url=https://books.google.com/books?id=gDW3BQAAQBAJ&pg=PA167 }}</ref>

It may be unclear why any number <math>a^n</math> has ''n'' roots rather than just a primary root. To demonstrate this, for the principal root ''a'' of the variable ''x'' taken to the ''n''th power, the following polynomial relation holds: <math display="block">p(x) = x^n - a^n = 0</math> This polynomial can be factored as follows:<ref>{{cite book | title=Cambridge 3 Unit Mathematics Year 11 | display-authors=1 | first1=William | last1=Pender | first2=David | last2=Saddler | first3=Julia | last3=Shea | first4=Derek | last4=Ward | publisher=Cambridge University Press | year=2011 | isbn=978-1-107-63332-2 | url=https://books.google.com/books?id=710s7NUinZcC&pg=PA229 }}</ref> <math display="block"> \begin{align} p(x) & = x^n - a^n \\ & = (x-a) (x^{n-1} + a x^{n-2} + a^2 x^{n-3} + \cdots + a^{n-1}) \\ & = (x-a) \left( \sum_{k=0}^{n-1} x^{n-k-1} a^k \right) \\ \end{align} </math> Thus, the polynomial <math>p(x)</math> is zero for ''x'' equal to ''a'', or for any ''x'' that solves the equation:<ref>{{cite book | title=Galois Theory | series=Pure and Applied Mathematics: A Wiley Series of Texts, Monographs and Tracts | first=David A. | last=Cox | edition=2nd | publisher=John Wiley & Sons | year=2012 | isbn=978-1-118-07205-9 | page=526 | url=https://books.google.com/books?id=TshTYrh7MDYC&pg=PA526 }}</ref> <math display="block"> \sum_{k=0}^{n-1} x^{n-k-1} a^k = 0 </math> By the fundamental theorem of algebra, this series has <math>n-1</math> roots, for a combined total of <math>n</math>.

As an example, let <math>n = 3</math> and <math>a = 1</math>, then find the cube roots of 1<ref>{{cite book | title=Polynomials and Equations: A Chinese Merchant Elite in Colonial Hong Kong (with a new preface) | display-authors=1 | first1=K. T. | last1=Leung | first2=S. N. | last2=Suen | first3=Ida A. C. | last3=Mok | publisher=Hong Kong University Press | year=1992 | isbn=978-962-209-271-6 | url=https://books.google.com/books?id=g09qDwAAQBAJ&pg=PA86 }}</ref>

<math display="block"> p(x) = x^3 - 1^3 = (x - 1) (x^2 + x^1 + 1) = 0 </math>

Thus the first root is <math>x = 1</math>, and the other two roots can be derived using the quadratic equation with <math>a = b = c = 1</math><ref>{{cite book | title=Numerical Methods for Roots of Polynomials - Part II | volume=16 | series=Studies in Computational Mathematics | first1=J. M. | last1=McNamee | first2=Victor | last2=Pan | publisher=Newnes | year=2013 | page=533 | isbn=978-0-08-093143-2 | url=https://books.google.com/books?id=j0rY3D9fx-0C&pg=PA533 }}</ref>

<math display="block"> x = \frac{-b \pm\sqrt{b^2 - 4ac} }{2a} = \frac{-1 \pm \sqrt{1^2 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2} </math>

== Proof of irrationality for non-perfect ''n''th power ''x'' == Assume that <math>\sqrt[n]{x}</math> is rational. That is, it can be reduced to a fraction <math>\frac{a}{b}</math>, where {{mvar|a}} and {{mvar|b}} are integers without a common factor.

This means that <math>x = \frac{a^n}{b^n}</math>.

Since ''x'' is an integer, <math>a^n</math>and <math>b^n</math>must share a common factor if <math>b \neq 1</math>. This means that if <math>b \neq 1</math>, <math>\frac{a^n}{b^n}</math> is not in simplest form. Thus ''b'' should equal 1.

Since <math>1^n = 1</math> and <math>\frac{n}{1} = n</math>, <math>\frac{a^n}{b^n} = a^n</math>.

This means that <math>x = a^n</math> and thus, <math>\sqrt[n]{x} = a</math>. This implies that <math>\sqrt[n]{x}</math> is an integer. Since {{mvar|x}} is not a perfect {{mvar|n}}th power, this is impossible. Thus <math>\sqrt[n]{x}</math> is irrational.<ref name=Fine_Rosenberger_2007>{{cite book | title=Number Theory: An Introduction via the Distribution of Primes | first1=Benjamin | last1=Fine | first2=Gerhard | last2=Rosenberger | publisher=Springer Science & Business Media | year=2007 | isbn=978-0-8176-4541-0 | url=https://books.google.com/books?id=RG242zhBD8kC&pg=PA20 }}</ref>

==See also== * Geometric mean * Twelfth root of two

==References== {{Reflist|30em}}

== External links == {{Wiktionary|surd}} {{Wiktionary|radical}}

{{Hyperoperations}} {{DISPLAYTITLE:{{mvar|n}}th root}}

Category:Elementary algebra Category:Operations on numbers