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{{short description|Operasi aritmetika}}
{{about|akar ke-n bilangan real dan kompleks|kegunaan lain|Akar (disambiguasi)#Matematika}}
Dalam [[matematika]], sebuah '''akar ke-''n''''' dari [[bilangan]] ''x'' adalah bilangan ''r'' yang jika dipangkatkan ''n'', menghasilkan ''x'':
:<math>r^n = x,</math>
dimana ''n'' adalah [[bilangan bulat positif]], kadang-kadang disebut ''derajat'' dari akar. Akar derajat 2 disebut ''[[akar kuadrat]]'' dan akar derajat 3, sebuah ''[[akar pangkat tiga]]''. Akar tingkat yang lebih tinggi dirujuk dengan menggunakan bilangan urut, seperti pada ''akar keempat'', ''akar kedua puluh'', dll. Perhitungan akar ke-{{math|''n''}} adalah '''ekstraksi akar'''.
Misalnya, 3 adalah akar kuadrat dari 9, karena 3{{sup|2}} = 9, dan 3 juga merupakan akar kuadrat dari 9, karena (−3){{sup|2}} = 9.
Setiap bilangan bukan nol yang dianggap sebagai [[bilangan kompleks]] memiliki {{math|''n''}} akar ke-{{math|''n''}} yang berbeda, termasuk [[bilangan real|real]] (paling banyak dua). Akar ke-{{math|''n''}} dari 0 adalah nol untuk semua [[bilangan bulat positif]] {{math|''n''}}, setelah {{math|0{{sup|''n''}} {{=}} 0}}. Khususnya, jika {{math|''n''}} genap dan {{math|''x''}} adalah bilangan real positif, satunya adalah negatif, dan yang lainnya (ketika {{math|''n'' > 2}}) [[bilangan kompleks]] non-real; jika {{math|''n''}} genap dan {{math|''x''}} adalah bilangan real negatif, tidak ada satupun akar ke-{{math|''n''}} yang merupakan real. Jika {{math|''n''}} ganjil dan {{math|''x''}} real, satu akar {{math|''n''}} adalah real dan bertanda sama sebagai {{math|''x''}}, sedangkan akar lainnya ({{math|''n'' – 1}}) bukanlah real. Akhirnya, jika {{math|''x''}} bukanlah real, maka tidak ada akar ke-{{math|''n''}} yang merupakan real.
Akar bilangan real biasanya ditulis menggunakan [[simbol radikal]] atau ''radix'' <math>\sqrt{{~^~}^~\!\!}</math>, dengan <math>\sqrt{x}</math> menunjukkan akar kuadrat positif dari {{mvar|x}} jika {{mvar|x}} adalah positif; untuk akar tinggi, <math>\sqrt[n]{x}</math> menunjukkan akar ke-{{math|''n''}} yang sebenarnya jika {{math|''n''}} adalah ganjil, dan akar ke-''n'' positif jika {{math|''n''}} adalah genap dan {{mvar|x}} adalah positif. Dalam kasus lain, simbol tidak umum digunakan sebagai ambigu. Dalam ekspresi <math>\sqrt[n]{x}</math>, bilangan bulat ''n'' disebut ''indeks'' dan {{mvar|x}} disebut ''radikan'' .
Ketika kompleks akar ke-{{mvar|n}} dipertimbangkan, seringkali berguna untuk memilih salah satu akar, yang disebut '''akar utama''', sebagai [[nilai utama]]. Pilihan umum adalah memilih akar ke-{{mvar|n}} utama dari {{mvar|x}} sebagai akar ke-{{mvar|n}}, dengan bagian real terbesar, dan, jika ada dua (untuk {{mvar|x}} real dan negatif), yang memiliki [[bagian imajiner]] positif. Ini membuat akar ke-{{mvar|n}} sebagai [[fungsi (matematika)|fungsi]] real dan positif untuk {{mvar|x}} real dan positif, dan adalah [[fungsi kontinu|kontinu]] diseluruh [[bidang kompleks]], kecuali untuk nilai {{mvar|x}} real dan negatif.
Kesulitan dengan pilihan ini adalah, untuk bilangan real negatif dan indeks ganjil, akar ke-{{mvar|n}} utama yang bukan asli. Misalnya, <math>-8</math> memiliki tiga akar pangkat tiga, <math>-2</math>, <math>1 + i\sqrt{3}</math> dan <math>1 - i\sqrt{3}.</math> Akar pangkat tiga sebenarnya adalah <math>-2</math> dan akar pangkat tiga utama adalah <math>1 + i\sqrt{3}.</math>
Akar yang tidak terselesaikan, terutama yang menggunakan simbol radikal, kadang-kadang disebut sebagai ''surd''<ref>{{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> atau "radikal".<ref name=silver>{{cite book|last=Silver|first=Howard A.|title=Algebra and trigonometry|year=1986|publisher=Prentice-Hall|location=Englewood Cliffs, NJ|isbn=978-0-13-021270-2|url-access=registration|url=https://archive.org/details/algebratrigonome00silv}}</ref> Setiap ekspresi yang mengandung radikal, apakah itu akar kuadrat, akar pangkat tiga, atau akar yang lebih tinggi, disebut ''ekspresi radikal'', dan jika tidak mengandung [[fungsi transendental]] atau [[bilangan transendental]] disebut [[ekspresi aljabar]].
Akar juga didefinisikan sebagai kasus khusus dari [[eksponensial]], dimana [[eksponen]] adalah [[Pecahan (matematika)|pecahan]]:
:<math>\sqrt[n]{x} = x^{1/n}.</math>
<div class="tright">{{Operasi aritmetika}}</div>
Akar digunakan untuk menentukan [[radius konvergensi]] dari [[deret pangkat]] dengan [[uji akar]]. Akar ke-{{mvar|n}} dari 1 disebut [[akar satuan]] dan memainkan peran mendasar dalam berbagai bidang matematika, seperti [[teori bilangan]], [[teori persamaan]], dan [[transformasi Fourier]].
==History==
{{Main article|Square root#History|Cube root#History}}
[[Berkas:Ybc7289-bw.jpg|right|thumb|300px|<center>Вавилонская табличка (около 1800—1600 г. до н. э.) с вычислением <math>\sqrt{2} \approx 1 + 24/60 + 51/60^2 + 10/60^3</math><br> <math>= 1{,}41421296\dots</math></center>]]
Первые задачи, связанные с извлечением квадратного корня, обнаружены в трудах [[Вавилонская математика|вавилонских математиков]] (о достижениях древнего Египта в этом отношении ничего не известно). Среди таких задач{{sfn |История математики|1970—1972|loc=Том I, С. 42—46}}:
* Применение [[Теорема Пифагора|теоремы Пифагора]] для нахождения стороны [[Прямоугольный треугольник|прямоугольного треугольника]] по известным двум другим сторонам.
* Нахождение стороны [[квадрат]]а, площадь которого задана.
* Решение [[Квадратное уравнение|квадратных уравнений]].
Вавилонские математики (II тысячелетие до н. э.) разработали для извлечения квадратного корня особый численный метод. Начальное приближение для <math>\sqrt{a}</math> рассчитывалось исходя из ближайшего к корню (в меньшую сторону) натурального числа <math>n</math>. Представив подкоренное выражение в виде: <math>a=n^2+r</math>, получаем: <math>x_0=n+\frac{r}{2n}</math>, затем применялся итеративный процесс уточнения, соответствующий [[Метод Ньютона|методу Ньютона]]{{sfn |История математики|1970—1972|loc=Том I, С. 47}}:
: <math>x_{n+1}=\frac{1}{2}~\left(x_n + \frac{a}{x_n}\right)\ </math>
Итерации в этом методе очень быстро сходятся. Для <math>\sqrt{5}</math>, например, <math>a=5;\;n=2;\;r=1;\ x_0=\frac{9}{4} = 2{,}25,</math> и мы получаем последовательность приближений:
: <math> x_1=\frac{161}{72} = 2{,}23611;\; x_2=\frac{51841}{23184} = 2{,}2360679779</math>
В заключительном значении верны все цифры, кроме последней.
Аналогичные задачи и методы встречаются в древнекитайской [[Математика в девяти книгах|«''Математике в девяти книгах''»]]{{sfn |История математики|1970—1972|loc=Том I, С. 169—171}}. Древние греки сделали важное открытие: <math>\sqrt{2}</math> — [[иррациональное число]]. Детальное исследование, выполненное [[Теэтет Афинский|Теэтетом Афинским]] (IV век до н. э.), показало, что если корень из натурального числа не извлекается нацело, то его значение иррационально<ref>{{книга|автор=Башмакова И. Г.|заглавие=Становление алгебры (из истории математических идей)|место=М.|издательство=Знание|год=1979|серия=Новое в жизни, науке, технике. Математика, кибернетика, № 9|страницы=23}}</ref>.
Греки сформулировали проблему [[Удвоение куба|удвоения куба]], которая сводилась к построению кубического корня [[Построение с помощью циркуля и линейки|с помощью циркуля и линейки]]. Проблема оказалась неразрешимой. Численные алгоритмы извлечения кубического корня опубликовали [[Герон]] (в трактате «''Метрика''», I век н. э.) и индийский математик [[Ариабхата I]] (V век)<ref>{{статья|автор=Abhishek Parakh.|заглавие=Ariabhata's root extraction methods|ссылка=http://cs.okstate.edu/~parakh/okstate_page/Aryabhatas_Root_Extraction_Methods_IJHS.pdf|издание=Indian Journal of History of Science|год=2007|выпуск=42.2|страницы=149—161|archiveurl=https://web.archive.org/web/20100609142233/http://cs.okstate.edu/%7Eparakh/okstate_page/Aryabhatas_Root_Extraction_Methods_IJHS.pdf|archivedate=2010-06-09}}</ref>.
Алгоритмы извлечения корней любой степени из целого числа, разработанные [[История математики в Индии|индийскими]] и [[Математика исламского средневековья|исламскими]] математиками, были усовершенствованы в средневековой Европе. [[Орем, Николай|Николай Орем]] (XIV век) впервые истолковал{{sfn |История математики|1970—1972|loc=Том I, С. 275—276}} корень <math>n</math>-й степени как возведение в степень <math>\frac{1}{n}</math>.
После появления [[Формула Кардано|формулы Кардано]] (XVI век) началось применение в математике [[Комплексное число|мнимых чисел]], понимаемых как квадратные корни из отрицательных чисел{{sfn |История математики|1970—1972|loc=Том I, С. 296—298}}. Основы техники работы с комплексными числами разработал в XVI веке [[Бомбелли, Рафаэль|Рафаэль Бомбелли]], который также предложил оригинальный метод вычисления корней (с помощью [[Непрерывная дробь|цепных дробей]]). Открытие [[Формула Муавра|формулы Муавра]] (1707) показало, что извлечение корня любой степени из комплексного числа всегда возможно и не приводит к новому типу чисел{{sfn |История математики|1970—1972|loc=Том III, С. 56—59}}.
Комплексные корни произвольной степени в начале XIX века глубоко исследовал [[Гаусс, Карл Фридрих|Гаусс]], хотя первые результаты принадлежат [[Эйлер, Леонард|Эйлеру]]{{sfn |История математики|1970—1972|loc=Том III, С. 62}}. Чрезвычайно важным открытием ([[Галуа, Эварист|Галуа]]) стало доказательство того факта, что не все [[алгебраические числа]] (корни многочленов) могут быть получены из натуральных с помощью четырёх действий арифметики и извлечения корня<ref>{{книга |автор=Колмогоров А. Н., Юшкевич А. П. (ред.). |заглавие=Математика XIX века. Математическая логика, алгебра, теория чисел, теория вероятностей |место=М. |издательство=Наука |том=I |год=1978 |страницы=58—66}}</ref>.
==Definition and notation==
[[File:NegativeOne4Root.svg|thumb|The four 4th roots of −1,<br /> none of which are real]]
[[File:NegativeOne3Root.svg|thumb|The three 3rd roots of −1,<br /> one of which is a negative real]]
An '''''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>r^n = x.</math>
Every positive [[real number]] ''x'' has a single positive ''n''th root, called the [[principal value|principal ''n''th root]], which is written <math>\sqrt[n]{x}</math>. 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}}.
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. 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 number|complex]], has ''n'' different complex number ''n''th roots. (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 number|irrational]]. For example,
:<math>\sqrt{2} = 1.414213562\ldots</math>
All ''n''th roots of integers are [[algebraic number]]s.
The term '''surd''' traces back to [[Khwārizmī|al-Khwārizmī]] (c. 825), who referred to rational and irrational numbers as ''audible'' and ''inaudible'', respectively. This later led to the Arabic word "{{rtl-lang|tg-Arab|أصم}}" (''asamm'', meaning "deaf" or "dumb") for ''irrational number'' being translated into Latin as "surdus" (meaning "deaf" or "mute"). [[Gerard of Cremona]] (c. 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]{i},</math> in which <math>n</math> and <math>i</math> are integer numerals and the whole expression denotes an irrational number.<ref>{{cite web |url=http://jeff560.tripod.com/s.html |title=Earliest Known Uses of Some of the Words of Mathematics|publisher=Mathematics Pages by Jeff Miller|access-date=2008-11-30}}</ref> [[Quadratic irrational numbers]], that is, irrational numbers of the form <math>\sqrt{i},</math> are also known as "quadratic surds".
===Square roots===
[[Image:Square-root function.svg|thumb|right|The graph <math>y=\pm \sqrt{x}</math>.]]
{{Main article|Square root}}
A '''square root''' of a number ''x'' is a number ''r'' which, when [[square (algebra)|squared]], becomes ''x'':
:<math>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''', and is denoted with a radical sign:
:<math>\sqrt{25} = 5.</math>
Since the square of every real number is nonnegative, negative numbers do not have real square roots. However, for every negative real number there are two [[imaginary number|imaginary]] square roots. For example, the square roots of −25 are 5''i'' and −5''i'', where ''[[imaginary unit|i]]'' represents a number whose square is {{math|−1}}.
===Cube roots===
[[Image:cube-root function.svg|thumb|right|The graph <math>y=\sqrt[3]{x}</math>.]]
{{Main article|Cube root}}
A '''cube root''' of a number ''x'' is a number ''r'' whose [[cube (algebra)|cube]] is ''x'':
:<math>r^3 = x.</math>
Every real number ''x'' has exactly one real cube root, written <math>\sqrt[3]{x}</math>. For example,
:<math>\sqrt[3]{8} = 2</math> and <math>\sqrt[3]{-8} = -2.</math>
Every real number has two additional [[complex number|complex]] cube roots.
==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 number|non-negative real number]],
:<math>\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:
:<math>\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 number]]s. For instance:
:<math>\sqrt{-1}\times\sqrt{-1} \neq \sqrt{-1 \times -1} = 1,\quad</math> but, rather, <math>\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.
==Simplified form of a radical expression==
A non-nested radical expression is said to be in '''simplified form''' if<ref>{{cite book|last=McKeague|first=Charles P.|title=Elementary algebra|page=470|year=2011|url=https://books.google.com/books?id=etTbP0rItQ4C&q=editions:q0hGn6PkOxsC|isbn=978-0-8400-6421-9}}</ref>
# There is no factor of the radicand that can be written as a power greater than or equal to the index.
# There are no fractions under the radical sign.
# There are no radicals in the denominator.
For example, to write the radical expression <math>\sqrt{\tfrac{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>\sqrt{\tfrac{32}{5}} = \sqrt{\tfrac{16 \times 2}{5}} = 4 \sqrt{\tfrac{2}{5}}</math>
Next, there is a fraction under the radical sign, which we change as follows:
:<math>4 \sqrt{\tfrac{2}{5}} = \frac{4 \sqrt{2}}{\sqrt{5}}</math>
Finally, we remove the radical from the denominator as follows:
:<math>\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>B.F. Caviness, R.J. Fateman, [http://www.eecs.berkeley.edu/~fateman/papers/radcan.pdf "Simplification of Radical Expressions"], ''Proceedings of the 1976 ACM Symposium on Symbolic and Algebraic Computation'', p. 329.</ref><ref>Richard Zippel, "Simplification of Expressions Involving Radicals", ''Journal of Symbolic Computation'' '''1''':189–210 (1985) {{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>
\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. It is not obvious for instance that:
:<math>\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}</math>
The above can be derived through:
:<math>\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 [[infinite series]]:
:<math>(1+x)^\frac{s}{t} = \sum_{n=0}^\infty \frac{\prod_{k=0}^{n-1} (s-kt)}{n!t^n}x^n</math>
with <math>|x|<1</math>. This expression can be derived from the [[binomial series]].
==Computing principal roots==
===Using Newton's method===
The ''n''th root of a number ''A'' can be computed with [[Newton's method]]. Start with an initial guess ''x''<sub>0</sub> and then iterate using the [[recurrence relation]]
:<math>x_{k+1} = \frac{n-1}{n}x_k+\frac{A}{nx_k^{n-1}}</math>
until the desired precision is reached. For example, to find the fifth root of 34, we plug in ''n'' = 5, ''A'' = 34 and ''x''<sub>0</sub> = 2 (initial guess). The first 5 iterations are, approximately:
<br>
''x''<sub>0</sub> = 2
<br>
''x''<sub>1</sub> = 2.025
<br>
''x''<sub>2</sub> = 2.024397817
<br>
''x''<sub>3</sub> = 2.024397458
<br>
''x''<sub>4</sub> = 2.024397458
<br>
The approximation ''x''<sub>4</sub> is accurate to 25 decimal places.
Newton's method can be modified to produce various [[generalized continued fraction#Roots of positive numbers|generalized continued fraction]] for the ''n''th root. For example,
:<math>
\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>
=== Digit-by-digit calculation of principal roots of decimal (base 10) numbers ===
[[Image:PascalForDecimalRoots.png|right|thumb|[[Pascal's triangle|Pascal's Triangle]] showing <math>P(4,1) = 4</math>.]]
Building on the [[Methods of computing square roots#Decimal (base 10)|digit-by-digit calculation of a square root]], it can be seen that the formula used there, <math>x(20p + x) \le c</math>, or <math>x^2 + 20xp \le c</math>, follows a pattern involving Pascal's triangle. For the ''n''th root of a number <math>P(n,i)</math> is defined as the value of element <math>i</math> in row <math>n</math> of Pascal's Triangle such that <math>P(4,1) = 4</math>, we can rewrite the expression as <math>\sum_{i=0}^{n-1}10^i P(n,i)p^i x^{n-i}</math>. For convenience, call the result of this expression <math>y</math>. Using this more general 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>).
#* 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.'''
<u> 1 2. 3 4 </u>
<u> </u> /
\/ 01 52.27 56
01 10{{sup|0}}·1·0{{sup|0}}·1{{sup|2}} + 10{{sup|1}}·2·0{{sup|1}}·1{{sup|1}} ≤ 1 < 10{{sup|0}}·1·0{{sup|0}}·2{{sup|2}} + 10{{sup|1}}·2·0{{sup|1}}·2{{sup|1}} x = 1
<u> 01 </u> y = 10{{sup|0}}·1·0{{sup|0}}·1{{sup|2}} + 10{{sup|1}}·2·0{{sup|1}}·1{{sup|2}} = 1 + 0 = 1
00 52 10{{sup|0}}·1·1{{sup|0}}·2{{sup|2}} + 10{{sup|1}}·2·1{{sup|1}}·2{{sup|1}} ≤ 52 < 10{{sup|0}}·1·1{{sup|0}}·3{{sup|2}} + 10{{sup|1}}·2·1{{sup|1}}·3{{sup|1}} x = 2
<u> 00 44 </u> y = 10{{sup|0}}·1·1{{sup|0}}·2{{sup|2}} + 10{{sup|1}}·2·1{{sup|1}}·2{{sup|1}} = 4 + 40 = 44
08 27 10{{sup|0}}·1·12{{sup|0}}·3{{sup|2}} + 10{{sup|1}}·2·12{{sup|1}}·3{{sup|1}} ≤ 827 < 10{{sup|0}}·1·12{{sup|0}}·4{{sup|2}} + 10{{sup|1}}·2·12{{sup|1}}·4{{sup|1}} x = 3
<u> 07 29 </u> y = 10{{sup|0}}·1·12{{sup|0}}·3{{sup|2}} + 10{{sup|1}}·2·12{{sup|1}}·3{{sup|1}} = 9 + 720 = 729
98 56 10{{sup|0}}·1·123{{sup|0}}·4{{sup|2}} + 10{{sup|1}}·2·123{{sup|1}}·4{{sup|1}} ≤ 9856 < 10{{sup|0}}·1·123{{sup|0}}·5{{sup|2}} + 10{{sup|1}}·2·123{{sup|1}}·5{{sup|1}} x = 4
<u> 98 56 </u> y = 10{{sup|0}}·1·123{{sup|0}}·4{{sup|2}} + 10{{sup|1}}·2·123{{sup|1}}·4{{sup|1}} = 16 + 9840 = 9856
00 00 Algorithm terminates: Answer is 12.34
'''Find the cube root of 4192 to the nearest hundredth.'''
<u> 1 6. 1 2 4</u>
<u>3</u> /
\/ 004 192.000 000 000
004 10{{sup|0}}·1·0{{sup|0}}·1{{sup|3}} + 10{{sup|1}}·3·0{{sup|1}}·1{{sup|2}} + 10{{sup|2}}·3·0{{sup|2}}·1{{sup|1}} ≤ 4 < 10{{sup|0}}·1·0{{sup|0}}·2{{sup|3}} + 10{{sup|1}}·3·0{{sup|1}}·2{{sup|2}} + 10{{sup|2}}·3·0{{sup|2}}·2{{sup|1}} x = 1
<u> 001 </u> y = 10{{sup|0}}·1·0{{sup|0}}·1{{sup|3}} + 10{{sup|1}}·3·0{{sup|1}}·1{{sup|2}} + 10{{sup|2}}·3·0{{sup|2}}·1{{sup|1}} = 1 + 0 + 0 = 1
003 192 10{{sup|0}}·1·1{{sup|0}}·6{{sup|3}} + 10{{sup|1}}·3·1{{sup|1}}·6{{sup|2}} + 10{{sup|2}}·3·1{{sup|2}}·6{{sup|1}} ≤ 3192 < 10{{sup|0}}·1·1{{sup|0}}·7{{sup|3}} + 10{{sup|1}}·3·1{{sup|1}}·7{{sup|2}} + 10{{sup|2}}·3·1{{sup|2}}·7{{sup|1}} x = 6
<u> 003 096 </u> y = 10{{sup|0}}·1·1{{sup|0}}·6{{sup|3}} + 10{{sup|1}}·3·1{{sup|1}}·6{{sup|2}} + 10{{sup|2}}·3·1{{sup|2}}·6{{sup|1}} = 216 + 1,080 + 1,800 = 3,096
096 000 10{{sup|0}}·1·16{{sup|0}}·1{{sup|3}} + 10{{sup|1}}·3·16{{sup|1}}·1{{sup|2}} + 10{{sup|2}}·3·16{{sup|2}}·1{{sup|1}} ≤ 96000 < 10{{sup|0}}·1·16{{sup|0}}·2{{sup|3}} + 10{{sup|1}}·3·16{{sup|1}}·2{{sup|2}} + 10{{sup|2}}·3·16{{sup|2}}·2{{sup|1}} x = 1
<u> 077 281 </u> y = 10{{sup|0}}·1·16{{sup|0}}·1{{sup|3}} + 10{{sup|1}}·3·16{{sup|1}}·1{{sup|2}} + 10{{sup|2}}·3·16{{sup|2}}·1{{sup|1}} = 1 + 480 + 76,800 = 77,281
018 719 000 10{{sup|0}}·1·161{{sup|0}}·2{{sup|3}} + 10{{sup|1}}·3·161{{sup|1}}·2{{sup|2}} + 10{{sup|2}}·3·161{{sup|2}}·2{{sup|1}} ≤ 18719000 < 10{{sup|0}}·1·161{{sup|0}}·3{{sup|3}} + 10{{sup|1}}·3·161{{sup|1}}·3{{sup|2}} + 10{{sup|2}}·3·161{{sup|2}}·3{{sup|1}} x = 2
<u> 015 571 928 </u> y = 10{{sup|0}}·1·161{{sup|0}}·2{{sup|3}} + 10{{sup|1}}·3·161{{sup|1}}·2{{sup|2}} + 10{{sup|2}}·3·161{{sup|2}}·2{{sup|1}} = 8 + 19,320 + 15,552,600 = 15,571,928
003 147 072 000 10{{sup|0}}·1·1612{{sup|0}}·4{{sup|3}} + 10{{sup|1}}·3·1612{{sup|1}}·4{{sup|2}} + 10{{sup|2}}·3·1612{{sup|2}}·4{{sup|1}} ≤ 3147072000 < 10{{sup|0}}·1·1612{{sup|0}}·5{{sup|3}} + 10{{sup|1}}·3·1612{{sup|1}}·5{{sup|2}} + 10{{sup|2}}·3·1612{{sup|2}}·5{{sup|1}} x = 4
The desired precision is achieved:
The cube root of 4192 is about 16.12
===Logarithmic calculation===
The principal ''n''th root of a positive number can be computed using [[logarithm]]s. 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 [[logarithm#Particular bases|base of the logarithm]] will do) to obtain
:<math>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]]:
:<math>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''{{!}}}}.
==Geometric constructibility==
The [[ancient Greek mathematicians]] knew how to [[compass-and-straightedge construction|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 = [[Monsieur|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}}.</ref>
==Complex roots==
Every [[complex number]] other than 0 has ''n'' different ''n''th roots.
===Square roots===
[[Image:Imaginary2Root.svg|thumb|right|The square roots of '''''i''''']]
The two square roots of a complex number are always negatives of each other. 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>\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:
:<math>\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>\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 ≤ ''θ'' < 2{{pi}}}}, or along the negative real axis with {{math|−{{pi}} < ''θ'' ≤ {{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===
[[File:3rd roots of unity.svg|thumb|right|The three 3rd roots of 1]]
{{Main article|Root of unity}}
The number 1 has ''n'' different ''n''th roots in the complex plane, namely
:<math>1,\;\omega,\;\omega^2,\;\ldots,\;\omega^{n-1},</math>
where
:<math>\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>.
===''n''th roots===
{{visualisation_complex_number_roots.svg}}
Every complex number has ''n'' different ''n''th roots in the complex plane. These are
:<math>\eta,\;\eta\omega,\;\eta\omega^2,\;\ldots,\;\eta\omega^{n-1},</math>
where ''η'' is a single ''n''th root, and 1, ''ω'', ''ω''{{sup|2}}, ... ''ω''{{sup|''n''−1}} are the ''n''th roots of unity. For example, the four different fourth roots of 2 are
:<math>\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 by the formula
:<math>\sqrt[n]{re^{i\theta}} = \sqrt[n]{r} \cdot e^{i\theta/n}.</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 ''a+bi'' then <math>r=\sqrt{a^2+b^2}</math>. Also, <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 = a/r,</math> <math> \sin \theta = b/r,</math> and <math> \tan \theta = 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.
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''<sub>1</sub> is one of the ''n''th roots then ''r''<sub>2</sub> = –''r''<sub>1</sub> is another. This is because raising the latter's coefficient –1 to the ''n''th power for even ''n'' yields 1: that is, (–''r''<sub>1</sub>){{sup|''n''}} = (–1){{sup|''n''}} × ''r''<sub>1</sub>{{sup|''n''}} = ''r''<sub>1</sub>{{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'' / ''n'' is discontinuous.
==Solving polynomials==
{{see also|Root-finding algorithm}}
It was once [[conjecture]]d that all [[polynomial equation]]s could be [[Algebraic solution|solved algebraically]] (that is, that all roots of a [[polynomial]] could be expressed in terms of a finite number of radicals and [[elementary arithmetic|elementary operations]]). However, while this is true for third degree polynomials ([[cubic function|cubics]]) and fourth degree polynomials ([[quartic function|quartics]]), the [[Abel–Ruffini theorem]] (1824) shows that this is not true in general when the degree is 5 or greater. For example, the solutions of the equation
:<math>x^5 = x + 1</math>
cannot be expressed in terms of radicals. (''cf.'' [[quintic equation]])
== 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 ''x'' is not a perfect ''n''th power, this is impossible. Thus <math>\sqrt[n]{x}</math> is irrational.
==See also==
* [[Nth root algorithm]]
* [[Shifting nth root algorithm]]
* [[Radical symbol]]
* [[Algebraic number]]
* [[Nested radical]]
* [[Twelfth root of two]]
* [[Super-root]]
==References==
{{Reflist}}
{{notelist}}
== External links ==
{{Wiktionary|surd}}
{{Wiktionary|radical}}
{{Hyperoperations}}
{{DISPLAYTITLE:{{math|''n''}}th root}}
[[Category:Elementary algebra]]
[[Category:Operations on numbers]]
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