Nth root

An archaic term for the operation of taking nth roots is radication.^[4][5]

An nth root of a number x, where n is a positive integer, is any of the n real or complex numbers r whose nth power is x:

$${\displaystyle r^{n}=x.}$$

Every positive real number x has a single positive nth root, called the principal nth root, which is written ${\displaystyle {\sqrt[{n}]{x}}}$. For n equal to 2 this is called the principal square root and the n is omitted. The nth root can also be represented using exponentiation as x^1/n.

For even values of n, positive numbers also have a negative nth root, while negative numbers do not have a real nth root. For odd values of n, every negative number x has a real negative nth root. For example, −2 has a real 5th root, ${\displaystyle {\sqrt[{5}]{-2}}=-1.148698354\ldots }$ but −2 does not have any real 6th roots.

Every non-zero number x, real or complex, has n different complex number nth roots. (In the case x is real, this count includes any real nth roots.) The only complex root of 0 is 0.

The nth roots of almost all numbers (all integers except the nth powers, and all rationals except the quotients of two nth powers) are irrational. For example,

$${\displaystyle {\sqrt {2}}=1.414213562\ldots }$$

All nth roots of rational numbers are algebraic numbers, and all nth roots of integers are algebraic integers.

The term "surd" traces back to Al-Khwarizmi (c. 825), who referred to rational and irrational numbers as "audible" and "inaudible", respectively. This later led to the Arabic word 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 ${\displaystyle {\sqrt[{n}]{r}}}$, in which ${\displaystyle n}$ and ${\displaystyle r}$ are integer numerals and the whole expression denotes an irrational number.^[6] Irrational numbers of the form ${\displaystyle \pm {\sqrt {a}},}$ where ${\displaystyle a}$ is rational, are called "pure quadratic surds"; irrational numbers of the form ${\displaystyle a\pm {\sqrt {b}}}$, where ${\displaystyle a}$ and ${\displaystyle b}$ are rational, are called mixed quadratic surds.^[7]

A square root of a number x is a number r which, when squared, becomes x:

$${\displaystyle r^{2}=x.}$$

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:

$${\displaystyle {\sqrt {25}}=5.}$$

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 square roots. For example, the square roots of −25 are 5i and −5i, where i represents a number whose square is −1.

A cube root of a number x is a number r whose cube is x:

$${\displaystyle r^{3}=x.}$$

Every real number x has exactly one real cube root, written ${\displaystyle {\sqrt[{3}]{x}}}$. For example,

$${\displaystyle {\begin{aligned}{\sqrt[{3}]{8}}&=2\\{\sqrt[{3}]{-8}}&=-2.\end{aligned}}}$$

Every real number has two additional complex cube roots.

Expressing the degree of an nth root in its exponent form, as in ${\displaystyle x^{1/n}}$, makes it easier to manipulate powers and roots. If ${\displaystyle a}$ is a non-negative real number,

$${\displaystyle {\sqrt[{n}]{a^{m}}}=(a^{m})^{1/n}=a^{m/n}=(a^{1/n})^{m}=({\sqrt[{n}]{a}})^{m}.}$$

Every non-negative number has exactly one non-negative real nth root, and so the rules for operations with surds involving non-negative radicands ${\displaystyle a}$ and ${\displaystyle b}$ are straightforward within the real numbers:

$${\displaystyle {\begin{aligned}{\sqrt[{n}]{ab}}&={\sqrt[{n}]{a}}{\sqrt[{n}]{b}}\\{\sqrt[{n}]{\frac {a}{b}}}&={\frac {\sqrt[{n}]{a}}{\sqrt[{n}]{b}}}\end{aligned}}}$$

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

$${\displaystyle {\sqrt {-1}}\times {\sqrt {-1}}\neq {\sqrt {-1\times -1}}=1,\quad }$$

but, rather,

$${\displaystyle \quad {\sqrt {-1}}\times {\sqrt {-1}}=i\times i=i^{2}=-1.}$$

Since the rule ${\displaystyle {\sqrt[{n}]{a}}\times {\sqrt[{n}]{b}}={\sqrt[{n}]{ab}}}$ strictly holds for non-negative real radicands only, its application leads to the inequality in the first step above.

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.^[8]

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

$${\displaystyle {\sqrt {\frac {32}{5}}}={\sqrt {\frac {16\cdot 2}{5}}}={\sqrt {16}}\cdot {\sqrt {\frac {2}{5}}}=4{\sqrt {\frac {2}{5}}}}$$

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

$${\displaystyle 4{\sqrt {\frac {2}{5}}}={\frac {4{\sqrt {2}}}{\sqrt {5}}}}$$

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

$${\displaystyle {\frac {4{\sqrt {2}}}{\sqrt {5}}}={\frac {4{\sqrt {2}}}{\sqrt {5}}}\cdot {\frac {\sqrt {5}}{\sqrt {5}}}={\frac {4{\sqrt {10}}}{5}}={\frac {4}{5}}{\sqrt {10}}}$$

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.^[9][10] For instance using the factorization of the sum of two cubes:

$${\displaystyle {\frac {1}{{\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}}}={\frac {{\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}}{\left({\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}\right)\left({\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}\right)}}={\frac {{\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}}{a+b}}.}$$

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

$${\displaystyle {\sqrt {3+2{\sqrt {2}}}}=1+{\sqrt {2}}.}$$

The above can be derived through:

$${\displaystyle {\sqrt {3+2{\sqrt {2}}}}={\sqrt {1+2{\sqrt {2}}+2}}={\sqrt {1^{2}+2{\sqrt {2}}+{\sqrt {2}}^{2}}}={\sqrt {\left(1+{\sqrt {2}}\right)^{2}}}=1+{\sqrt {2}}}$$

Let ${\displaystyle r=p/q}$, with p and q coprime and positive integers. Then ${\displaystyle {\sqrt[{n}]{r}}={\sqrt[{n}]{p}}/{\sqrt[{n}]{q}}}$ is rational if and only if both ${\displaystyle {\sqrt[{n}]{p}}}$ and ${\displaystyle {\sqrt[{n}]{q}}}$ are integers, which means that both p and q are nth powers of some integer.

The radical or root may be represented by the infinite series:

$${\displaystyle (1+x)^{s/t}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\prod _{k=0}^{n-1}\left({\frac {s}{t}}-k\right)}$$

with ${\displaystyle |x|<1}$. This expression can be derived from the binomial series.^{[citation needed]}

The nth root of a number A can be computed with Newton's method, which starts with an initial guess x₀ and then iterates using the recurrence relation

$${\displaystyle x_{k+1}=x_{k}-{\frac {x_{k}^{n}-A}{nx_{k}^{n-1}}}}$$

until the desired precision is reached. For computational efficiency, the recurrence relation is commonly rewritten

$${\displaystyle x_{k+1}={\frac {n-1}{n}}\,x_{k}+{\frac {A}{n}}\,{\frac {1}{x_{k}^{n-1}}}.}$$

This allows to have only one exponentiation, and to compute once for all the first factor of each term.

For example, to find the fifth root of 34, we plug in n = 5, A = 34 and x₀ = 2 (initial guess). The first 5 iterations are, approximately:

(All correct digits shown.)

The approximation x₄ is accurate to 25 decimal places and x₅ is good for 51.

Newton's method can be modified to produce various generalized continued fractions for the nth root. For example,

$${\displaystyle {\sqrt[{n}]{z}}={\sqrt[{n}]{x^{n}+y}}=x+{\cfrac {y}{nx^{n-1}+{\cfrac {(n-1)y}{2x+{\cfrac {(n+1)y}{3nx^{n-1}+{\cfrac {(2n-1)y}{2x+{\cfrac {(2n+1)y}{5nx^{n-1}+{\cfrac {(3n-1)y}{2x+\ddots }}}}}}}}}}}}.}$$

Building on the digit-by-digit calculation of a square root, it can be seen that the formula used there, ${\displaystyle x(20p+x)\leq c}$, or ${\displaystyle x^{2}+20xp\leq c}$, follows a pattern involving Pascal's triangle. For the nth root of a number ${\displaystyle P(n,i)}$ is defined as the value of element ${\displaystyle i}$ in row ${\displaystyle n}$ of Pascal's Triangle such that ${\displaystyle P(4,1)=4}$, we can rewrite the expression as ${\displaystyle \sum _{i=0}^{n-1}10^{i}P(n,i)p^{i}x^{n-i}}$. For convenience, call the result of this expression ${\displaystyle y}$. 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:

1. 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 ${\displaystyle 10^{n}}$ and add the digits from the next group. This will be the current value c.
2. Find p and x, as follows:
   • Let ${\displaystyle p}$ be the part of the root found so far, ignoring any decimal point. (For the first step, ${\displaystyle p=0}$ and ${\displaystyle 0^{0}=1}$).
   • Determine the greatest digit ${\displaystyle x}$ such that ${\displaystyle y\leq c}$.
   • Place the digit ${\displaystyle x}$ 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.
3. Subtract ${\displaystyle y}$ from ${\displaystyle c}$ to form a new remainder.
4. 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.

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Find the square root of 152.2756.

          1  2. 3  4 
       /
     \/  01 52.27 56          (Results)    (Explanations)
    
         01                   x = 1        100·1·00·12 + 101·2·01·11     ≤      1   <   100·1·00·22   + 101·2·01·21
         01                   y = 1        y = 100·1·00·12   + 101·2·01·11   =  1 +    0   =     1
         00 52                x = 2        100·1·10·22 + 101·2·11·21     ≤     52   <   100·1·10·32   + 101·2·11·31
         00 44                y = 44       y = 100·1·10·22   + 101·2·11·21   =  4 +   40   =    44
            08 27             x = 3        100·1·120·32 + 101·2·121·31   ≤    827   <   100·1·120·42  + 101·2·121·41
            07 29             y = 729      y = 100·1·120·32  + 101·2·121·31  =  9 +  720   =   729
               98 56          x = 4        100·1·1230·42 + 101·2·1231·41 ≤   9856   <   100·1·1230·52 + 101·2·1231·51
               98 56          y = 9856     y = 100·1·1230·42 + 101·2·1231·41 = 16 + 9840   =  9856
               00 00

Algorithm terminates: Answer is 12.34

Find the cube root of 4192 truncated to the nearest thousandth.

        1   6.  1   2   4
 3  /
  \/  004 192.000 000 000          (Results)    (Explanations)
    
      004                          x = 1        100·1·00·13    +  101·3·01·12   + 102·3·02·11    ≤          4  <  100·1·00·23     + 101·3·01·22    + 102·3·02·21
      001                          y = 1        y = 100·1·00·13   + 101·3·01·12   + 102·3·02·11   =   1 +      0 +          0   =          1
      003 192                      x = 6        100·1·10·63    +  101·3·11·62   + 102·3·12·61    ≤       3192  <  100·1·10·73     + 101·3·11·72    + 102·3·12·71
      003 096                      y = 3096     y = 100·1·10·63   + 101·3·11·62   + 102·3·12·61   = 216 +  1,080 +      1,800   =      3,096
          096 000                  x = 1        100·1·160·13   + 101·3·161·12   + 102·3·162·11   ≤      96000  <  100·1·160·23   + 101·3·161·22   + 102·3·162·21
          077 281                  y = 77281    y = 100·1·160·13  + 101·3·161·12  + 102·3·162·11  =   1 +    480 +     76,800   =     77,281
          018 719 000              x = 2        100·1·1610·23  + 101·3·1611·22  + 102·3·1612·21  ≤   18719000  <  100·1·1610·33  + 101·3·1611·32  + 102·3·1612·31
              015 571 928          y = 15571928 y = 100·1·1610·23 + 101·3·1611·22 + 102·3·1612·21 =   8 + 19,320 + 15,552,600   = 15,571,928
              003 147 072 000      x = 4        100·1·16120·43 + 101·3·16121·42 + 102·3·16122·41 ≤ 3147072000  <  100·1·16120·53 + 101·3·16121·52 + 102·3·16122·51

The desired precision is achieved. The cube root of 4192 is 16.124...

The principal nth root of a positive number can be computed using logarithms. Starting from the equation that defines r as an nth root of x, namely ${\displaystyle r^{n}=x,}$ 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

$${\displaystyle n\log _{b}r=\log _{b}x\quad \quad {\text{hence}}\quad \quad \log _{b}r={\frac {\log _{b}x}{n}}.}$$

The root r is recovered from this by taking the antilog:

$${\displaystyle r=b^{{\frac {1}{n}}\log _{b}x}.}$$

(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 ${\displaystyle |r|^{n}=|x|,}$ then proceeding as before to find |r|, and using r = −|r|.

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 nth root of a given length cannot be constructed if n is not a power of 2.^[11]

Every complex number other than 0 has n different nth roots.

The two square roots of a complex number are always negatives of each other. For example, the square roots of −4 are 2i and −2i, and the square roots of i are

$${\displaystyle {\tfrac {1}{\sqrt {2}}}(1+i)\quad {\text{and}}\quad -{\tfrac {1}{\sqrt {2}}}(1+i).}$$

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:

$${\displaystyle {\sqrt {re^{i\theta }}}=\pm {\sqrt {r}}\cdot e^{i\theta /2}.}$$

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

$${\displaystyle {\sqrt {re^{i\theta }}}={\sqrt {r}}\cdot e^{i\theta /2}}$$

which introduces a branch cut in the complex plane along the positive real axis with the condition 0 ≤ θ < 2π, or along the negative real axis with −π < θ ≤ π.

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

The number 1 has n different nth roots in the complex plane, namely

$${\displaystyle 1,\;\omega ,\;\omega ^{2},\;\ldots ,\;\omega ^{n-1},}$$

where

$${\displaystyle \omega =e^{\frac {2\pi i}{n}}=\cos \left({\frac {2\pi }{n}}\right)+i\sin \left({\frac {2\pi }{n}}\right).}$$

These roots are evenly spaced around the unit circle in the complex plane, at angles which are multiples of ${\displaystyle 2\pi /n}$. For example, the square roots of unity are 1 and −1, and the fourth roots of unity are 1, ${\displaystyle i}$, −1, and ${\displaystyle -i}$.

Every complex number has n different nth roots in the complex plane. These are

$${\displaystyle \eta ,\;\eta \omega ,\;\eta \omega ^{2},\;\ldots ,\;\eta \omega ^{n-1},}$$

where η is a single nth root, and 1, ω, ω², ... ωⁿ⁻¹ are the nth roots of unity. For example, the four different fourth roots of 2 are

$${\displaystyle {\sqrt[{4}]{2}},\quad i{\sqrt[{4}]{2}},\quad -{\sqrt[{4}]{2}},\quad {\text{and}}\quad -i{\sqrt[{4}]{2}}.}$$

In polar form, a single nth root may be found by the formula

$${\displaystyle {\sqrt[{n}]{re^{i\theta }}}={\sqrt[{n}]{r}}\cdot e^{i\theta /n}.}$$

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 ${\displaystyle r={\sqrt {a^{2}+b^{2}}}}$. Also, ${\displaystyle \theta }$ 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 ${\displaystyle \cos \theta =a/r,}$ ${\displaystyle \sin \theta =b/r,}$ and ${\displaystyle \tan \theta =b/a.}$

Thus finding nth roots in the complex plane can be segmented into two steps. First, the magnitude of all the nth roots is the nth 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 nth roots is ${\displaystyle \theta /n}$, where ${\displaystyle \theta }$ is the angle defined in the same way for the number whose root is being taken. Furthermore, all n of the nth roots are at equally spaced angles from each other.

If n is even, a complex number's nth roots, of which there are an even number, come in additive inverse pairs, so that if a number r₁ is one of the nth roots then r₂ = −r₁ is another. This is because raising the latter's coefficient −1 to the nth power for even n yields 1: that is, (−r₁)ⁿ = (−1)ⁿ × r₁ⁿ = r₁ⁿ.

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.

It was once conjectured that all polynomial equations could be solved algebraically (that is, that all roots of a polynomial could be expressed in terms of a finite number of radicals and elementary operations). However, while this is true for third degree polynomials (cubics) and fourth degree polynomials (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

$${\displaystyle x^{5}=x+1}$$

cannot be expressed in terms of radicals. (cf. quintic equation)

Assume that ${\displaystyle {\sqrt[{n}]{x}}}$ is rational. That is, it can be reduced to a fraction ${\displaystyle {\frac {a}{b}}}$, where a and b are integers without a common factor.

This means that ${\displaystyle x={\frac {a^{n}}{b^{n}}}}$.

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

Since ${\displaystyle 1^{n}=1}$ and ${\displaystyle {\frac {n}{1}}=n}$, ${\displaystyle {\frac {a^{n}}{b^{n}}}=a^{n}}$.

This means that ${\displaystyle x=a^{n}}$ and thus, ${\displaystyle {\sqrt[{n}]{x}}=a}$. This implies that ${\displaystyle {\sqrt[{n}]{x}}}$ is an integer. Since x is not a perfect nth power, this is impossible. Thus ${\displaystyle {\sqrt[{n}]{x}}}$ is irrational.

• Geometric mean
• Twelfth root of two