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{{short description|Arithmetic operation}}
{{about|nth-roots of real and complex numbers|other uses|Root (disambiguation)}}
{{DISPLAYTITLE:{{math|''n''}}th root}}
In [[mathematics]], an '''''n''th root''' of a [[number]] ''x'', where ''n'' is usually assumed to be a positive integer, is a number ''r'' which, when raised to the power ''n'' yields&nbsp;''x'':
:<math>r^n = x,</math>
where ''n'' is the ''degree'' of the root. 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 a {{math|''n''}}th root is a '''root extraction'''.

For example:
* 3 is a square root of 9, since 3<sup>2</sup> = 9.
* −3 is also a square root of 9, since (−3)<sup>2</sup> = 9.

Any non-zero number considered as a [[complex number]] has {{math|''n''}} different complex {{math|''n''}}th roots, including the [[real number|real]] ones (at most two). The {{math|''n''}}th root of 0 is zero for all [[positive integer]]s {{math|''n''}}, since {{math|0<sup>''n''</sup> {{=}} 0}}. In particular, if {{math|''n''}} is even and {{math|''x''}} is a positive real number, one of its {{math|''n''}}th roots is real and positive, one is negative, and the others (when {{math|''n'' > 2}}) are non-real [[complex number]]s; if {{math|''n''}} is even and {{math|''x''}} is a negative real number, none of the {{math|''n''}}th roots is real. If {{math|''n''}} is odd and {{math|''x''}} is real, one {{math|''n''}}th root is real and has the same sign as {{math|''x''}}, while the other ({{math|''n'' – 1}}) roots are not real. Finally, if {{math|''x''}} is not real, then none of its {{math|''n''}}th roots is real.

Roots of real numbers are usually written using the [[radical symbol]] or ''radix'' with <math>\sqrt{x}</math> denoting the positive square root of {{mvar|x}} if {{mvar|x}} is positive, and <math>\sqrt[n]{x}</math> denoting the real {{math|''n''}}th root, if {{math|''n''}} is odd, and the positive square root if {{math|''n''}} is even and {{mvar|x}} is nonnegative. In the other cases, the symbol is not commonly used as being ambiguous. In the expression <math>\sqrt[n]{x}</math>, the integer ''n'' is called the ''index'', <math>\sqrt{{~^~}^~\!\!}</math> is the ''radical sign'' or ''radix'', and {{mvar|x}} is called the ''radicand''. 

When complex {{math|n}}th roots are considered, it is often useful to choose one of the roots as a [[principal value]]. The common choice is the one that makes the {{mvar|n}}th root a [[continuous function]] that is real and positive for {{mvar|x}} real and positive. That is, the principal {{mvar|n}}th root is the {{mvar|n}}th root, with the greatest real part, and, when there are two (for {{math|''x'' < 0}}), the one with positive [[imaginary part]]. 

A difficulty with this choice is that, for a negative real numbers and an odd index, the principal {{mvar|n}}th root is not the real one. For example, <math>-8</math> has three cube roots, <math>-2</math>, <math>1 + i\sqrt{3}</math> and <math>1 - i\sqrt{3}.</math> The real cube root is <math>-2</math> and the principal cube root is <math>1 + i\sqrt{3}.</math>

An unresolved root, especially one using the radical symbol, is sometimes referred to as a ''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#v=onepage&q&f=false}}</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, NJ|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]].

In [[calculus]], '''roots''' are treated as special cases of [[exponentiation]], where the [[exponent]] is a [[Fraction (mathematics)|fraction]]:
:<math>\sqrt[n]{x} = x^{1/n}.</math>

Roots are used for determining the [[radius of convergence]] of a [[power series]] with the [[root test]]. 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]].

==History==

{{Main article|Square root#History|Cube root#History}}
An archaic term 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}}</ref><ref>{{cite web|url=https://en.oxforddictionaries.com/definition/radication|title=radication – Definition of radication in English by Oxford Dictionaries|website=Oxford Dictionaries }}</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</sup>.

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''.<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|accessdate=2008-11-30}}</ref>

===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 a positive real number, 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.

:<math>\sqrt[n]{a^m} \equiv (a^m)^{1/n} \equiv a^{m/n}.</math>

Every [[positive number|positive real number]] has exactly one positive real ''n''th root, and so the rules for operations with surds involving positive radicands <math>a,\; b</math> are straightforward within the real numbers:

:<math>\begin{align}
           \sqrt[n]{ab} &\equiv \sqrt[n]{a} \sqrt[n]{b} \\
  \sqrt[n]{\frac{a}{b}} &\equiv \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/?id=etTbP0rItQ4C&printsec=frontcover&dq=editions:q0hGn6PkOxsC#v=onepage&q&f=false|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 \cdot 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.&nbsp;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>

==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==
The ''n''th root of an [[integer]] ''k'' is only an integer if ''k'' is the product of ''n''th powers of integers. In all other cases the ''n''th root of an integer is an [[irrational number]]. For instance, the fifth root of
:<math>\sqrt[5]{248 832} = \sqrt[5]{3^5\cdot 2^5\cdot 2^5} = 12</math>

and the fifth root of 34 is
:<math> \sqrt[5]{34} = \sqrt[5]{2\cdot 17} = 2.024397458 \ldots , </math>

where here the dots signify not only that the decimal expression does not end after a finite number of digits, but also that the digits never enter a repeating pattern, because the number is irrational.

Since for positive real numbers {{mvar|a}} and {{mvar|b}} the equality <math>\;\sqrt[n]{a/b} = \sqrt[n]{a}/\sqrt[n]{b}\;</math> holds, the above property can be extended to positive rational numbers. Let <math>r=p/q</math>, with {{mvar|p}} and {{mvar|q}} coprime and positive integers, be a rational number, {{nowrap|then {{mvar|r}}}} has a rational ''n''th root, if both positive {{nowrap|integers {{mvar|p}}}} {{nowrap|and {{mvar|q}}}} have an integer ''n''th root, i.e., <math>\;r= p\cdot\tfrac{1}{q}\;</math> is the product of ''n''th powers of rational numbers. If one or both ''n''th roots of {{mvar|p}} or {{mvar|q}} are irrational, the quotient is irrational, too.

===''n''th root algorithm===

The ''n''th root of a number ''A'' can be computed by the [[nth root algorithm|''n''th root algorithm]], a special case of [[Newton's method]]. Start with an initial guess ''x''<sub>0</sub> and then iterate using the [[recurrence relation]]
:<math>x_{k+1} = \frac{1}{n} \left({(n-1)x_k +\frac{A}{x_k^{n-1}}}\right) </math>
until the desired precision is reached.

Depending on the application, it may be enough to use only the first Newton approximant:
:<math> \sqrt[n]{x^n+y} \approx x + \frac{y}{n x^{n-1}}. </math>
For example, to find the fifth root of 34, note that 2<sup>5</sup> = 32 and thus take ''x'' = 2, ''n'' = 5 and ''y'' = 2 in the above formula. This yields
:<math> \sqrt[5]{34} = \sqrt[5]{32 + 2} \approx 2 + \frac{2}{5 \cdot 16} = 2.025. </math>
The error in the approximation is only about 0.03%.

Newton's method can be modified to produce a [[generalized continued fraction#Roots of positive numbers|generalized continued fraction]] for the ''n''th root which can be modified in various ways as described in that article. 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>
:<math>
  \sqrt[n]{z} = x+\cfrac{2x\cdot y}{n(2z - y)-y-\cfrac{(1^2n^2-1)y^2}{3n(2z - y)-\cfrac{(2^2n^2-1)y^2}{5n(2z - y)-\cfrac{(3^2n^2-1)y^2}{7n(2z - y)-\ddots}}}}.
</math>

In the case of the fifth root of 34 above (after dividing out selected common factors):
:<math>
  \sqrt[5]{34} = 2+\cfrac{1} {40+\cfrac{4} {4+\cfrac{6} {120+\cfrac{9} {4+\cfrac{11} {200+\cfrac{14} {4+\ddots}}}}}}
               = 2+\cfrac{4\cdot 1}{165-1-\cfrac{4\cdot 6}{495-\cfrac{9\cdot 11}{825-\cfrac{14\cdot 16}{1155-\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</sup>·1·0<sup>0</sup>·1<sup>2</sup> + 10<sup>1</sup>·2·0<sup>1</sup>·1<sup>1</sup>     ≤      1   <   10<sup>0</sup>·1·0<sup>0</sup>·2<sup>2</sup>   + 10<sup>1</sup>·2·0<sup>1</sup>·2<sup>1</sup>         x = 1
         <u> 01 </u>                     y = 10<sup>0</sup>·1·0<sup>0</sup>·1<sup>2</sup>   + 10<sup>1</sup>·2·0<sup>1</sup>·1<sup>2</sup>   =  1 +    0   =     1
          00 52                10<sup>0</sup>·1·1<sup>0</sup>·2<sup>2</sup> + 10<sup>1</sup>·2·1<sup>1</sup>·2<sup>1</sup>     ≤     52   <   10<sup>0</sup>·1·1<sup>0</sup>·3<sup>2</sup>   + 10<sup>1</sup>·2·1<sup>1</sup>·3<sup>1</sup>         x = 2
         <u> 00 44 </u>                  y = 10<sup>0</sup>·1·1<sup>0</sup>·2<sup>2</sup>   + 10<sup>1</sup>·2·1<sup>1</sup>·2<sup>1</sup>   =  4 +   40   =    44
             08 27             10<sup>0</sup>·1·12<sup>0</sup>·3<sup>2</sup> + 10<sup>1</sup>·2·12<sup>1</sup>·3<sup>1</sup>   ≤    827   <   10<sup>0</sup>·1·12<sup>0</sup>·4<sup>2</sup>  + 10<sup>1</sup>·2·12<sup>1</sup>·4<sup>1</sup>        x = 3
            <u> 07 29 </u>               y = 10<sup>0</sup>·1·12<sup>0</sup>·3<sup>2</sup>  + 10<sup>1</sup>·2·12<sup>1</sup>·3<sup>1</sup>  =  9 +  720   =   729
                98 56          10<sup>0</sup>·1·123<sup>0</sup>·4<sup>2</sup> + 10<sup>1</sup>·2·123<sup>1</sup>·4<sup>1</sup> ≤   9856   <   10<sup>0</sup>·1·123<sup>0</sup>·5<sup>2</sup> + 10<sup>1</sup>·2·123<sup>1</sup>·5<sup>1</sup>       x = 4
               <u> 98 56 </u>            y = 10<sup>0</sup>·1·123<sup>0</sup>·4<sup>2</sup> + 10<sup>1</sup>·2·123<sup>1</sup>·4<sup>1</sup> = 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</sup>·1·0<sup>0</sup>·1<sup>3</sup>    +  10<sup>1</sup>·3·0<sup>1</sup>·1<sup>2</sup>   + 10<sup>2</sup>·3·0<sup>2</sup>·1<sup>1</sup>    ≤          4  <  10<sup>0</sup>·1·0<sup>0</sup>·2<sup>3</sup>     + 10<sup>1</sup>·3·0<sup>1</sup>·2<sup>2</sup>    + 10<sup>2</sup>·3·0<sup>2</sup>·2<sup>1</sup>     x = 1
      <u> 001 </u>                        y = 10<sup>0</sup>·1·0<sup>0</sup>·1<sup>3</sup>   + 10<sup>1</sup>·3·0<sup>1</sup>·1<sup>2</sup>   + 10<sup>2</sup>·3·0<sup>2</sup>·1<sup>1</sup>   =   1 +      0 +          0   =          1
       003 192                  10<sup>0</sup>·1·1<sup>0</sup>·6<sup>3</sup>    +  10<sup>1</sup>·3·1<sup>1</sup>·6<sup>2</sup>   + 10<sup>2</sup>·3·1<sup>2</sup>·6<sup>1</sup>    ≤       3192  <  10<sup>0</sup>·1·1<sup>0</sup>·7<sup>3</sup>     + 10<sup>1</sup>·3·1<sup>1</sup>·7<sup>2</sup>    + 10<sup>2</sup>·3·1<sup>2</sup>·7<sup>1</sup>     x = 6
      <u> 003 096 </u>                    y = 10<sup>0</sup>·1·1<sup>0</sup>·6<sup>3</sup>   + 10<sup>1</sup>·3·1<sup>1</sup>·6<sup>2</sup>   + 10<sup>2</sup>·3·1<sup>2</sup>·6<sup>1</sup>   = 216 +  1,080 +      1,800   =      3,096
           096 000              10<sup>0</sup>·1·16<sup>0</sup>·1<sup>3</sup>   + 10<sup>1</sup>·3·16<sup>1</sup>·1<sup>2</sup>   + 10<sup>2</sup>·3·16<sup>2</sup>·1<sup>1</sup>   ≤      96000  <  10<sup>0</sup>·1·16<sup>0</sup>·2<sup>3</sup>   + 10<sup>1</sup>·3·16<sup>1</sup>·2<sup>2</sup>   + 10<sup>2</sup>·3·16<sup>2</sup>·2<sup>1</sup>    x = 1
          <u> 077 281 </u>                y = 10<sup>0</sup>·1·16<sup>0</sup>·1<sup>3</sup>  + 10<sup>1</sup>·3·16<sup>1</sup>·1<sup>2</sup>  + 10<sup>2</sup>·3·16<sup>2</sup>·1<sup>1</sup>  =   1 +    480 +     76,800   =     77,281
           018 719 000          10<sup>0</sup>·1·161<sup>0</sup>·2<sup>3</sup>  + 10<sup>1</sup>·3·161<sup>1</sup>·2<sup>2</sup>  + 10<sup>2</sup>·3·161<sup>2</sup>·2<sup>1</sup>  ≤   18719000  <  10<sup>0</sup>·1·161<sup>0</sup>·3<sup>3</sup>  + 10<sup>1</sup>·3·161<sup>1</sup>·3<sup>2</sup>  + 10<sup>2</sup>·3·161<sup>2</sup>·3<sup>1</sup>   x = 2
              <u> 015 571 928 </u>        y = 10<sup>0</sup>·1·161<sup>0</sup>·2<sup>3</sup> + 10<sup>1</sup>·3·161<sup>1</sup>·2<sup>2</sup> + 10<sup>2</sup>·3·161<sup>2</sup>·2<sup>1</sup> =   8 + 19,320 + 15,552,600   = 15,571,928
               003 147 072 000  10<sup>0</sup>·1·1612<sup>0</sup>·4<sup>3</sup> + 10<sup>1</sup>·3·1612<sup>1</sup>·4<sup>2</sup> + 10<sup>2</sup>·3·1612<sup>2</sup>·4<sup>1</sup> ≤ 3147072000  <  10<sup>0</sup>·1·1612<sup>0</sup>·5<sup>3</sup> + 10<sup>1</sup>·3·1612<sup>1</sup>·5<sup>2</sup> + 10<sup>2</sup>·3·1612<sup>2</sup>·5<sup>1</sup>  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. 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&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===
[[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,&nbsp;''ω'',&nbsp;''ω''<sup>2</sup>,&nbsp;...&nbsp;''ω''<sup>''n''−1</sup> 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''</sup> = (–1)<sup>''n''</sup> × ''r''<sub>1</sub><sup>''n''</sup> = ''r''<sub>1</sub><sup>''n''</sup>.

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.

==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 a and b are integers.

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}}

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

{{Hyperoperations}}
[[Category:Elementary algebra]]
[[Category:Binary operations]]