Nth root

Definition and notation

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, ${\displaystyle \sqrt[5]{-2}=-1.148698354\dots }$ 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,

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

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

Square roots

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.

Cube roots

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,

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

Every real number has two additional complex cube roots.

Identities and properties

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:

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

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

$${\displaystyle \sqrt{-1}\times \sqrt{-1}\ne \sqrt{-1\times -1}=1,}$$

but, rather,

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

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.

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.

For example, to write the radical expression ${\displaystyle \sqrt{32/5}}$ 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. 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.

Infinite series

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

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

with ${\displaystyle |x|<1}$. This expression can be derived from the binomial series. For the nth root, this becomes

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

For numbers ${\displaystyle r\ge 2}$, choose a value ${\displaystyle p}$ such that

$${\displaystyle \frac{r}{p^n}-1=x^{\prime },\text{ where }|x^{\prime }|<1}$$

then per above, solve for

$${\displaystyle r^{\frac{1}{n}}=p(1+x^{\prime }{)}^{\frac{1}{n}}}$$

As an example, for ${\displaystyle r=30}$ and ${\displaystyle n=2}$, choose ${\displaystyle p=5}$

$${\displaystyle \frac{30}{5^2}-1=\frac{5}{25}=.2}$$

$${\displaystyle \sqrt{30}=5\sqrt{1+.2}=5\left[1+\frac{1}{2}(.2{)}^1-\frac{1}{8}(.2{)}^2+\frac{1}{16}(.2{)}^3-\cdots \right]\approx 5.4775}$$

Nth roots are used to check for convergence of a power series with the root test.

Computing principal roots

Using Newton's method

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

$${\displaystyle x_{k+1}=x_k-\frac{x_k^n-A}{n{x}_k^{n-1}}}$$

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

$${\displaystyle x_{k+1}=\frac{1}{n}\left[(n-1)x_k+\frac{A}{x_k^{n-1}}\right].}$$

This allows the relation to only have one exponentiation, which is computed once for each iteration. The nth root of x can then be defined as the limit of ${\displaystyle x_k}$ as k approaches infinity.

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:

x₀ = 2 x₁ = 2.025 x₂ = 2.02439 7... x₃ = 2.02439 7458... x₄ = 2.02439 74584 99885 04251 08172... x₅ = 2.02439 74584 99885 04251 08172 45541 93741 91146 21701 07311 8...

(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,

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

A suitable initial guess for Newton's method may need to be identified using the bisection method or method of false position. For large values of n and higher requirements for precision, a more rapid algorithm than Newton's method for finding the nth root is to use a truncated Taylor series with a Padé approximant.

Using the Viète technique

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. This method is based on the binomial theorem and is essentially an inverse algorithm solving $${\displaystyle (10x+y{)}^n=\sum _{k=0}^{n}P(n,k)(10x{)}^{n-k}{y}^k}$$ where ${\displaystyle P(n,k)}$, the binomial coefficient, is the kth entry on the nth row of Pascal's triangle.

To compute the root of a number C, choose a series of approximations $${\displaystyle x_i^n,i=0,1,\dots \text{ with }{x}_0=0}$$ that satisfy ${\displaystyle x_i^n\le C}$, where the difference between ${\displaystyle x_{i+1}}$ and ${\displaystyle x_i}$ is the next digit in the approximation. The decimal fraction ${\displaystyle y_i}$ is chosen to be the largest number with a single significant digit that satisfies $${\displaystyle 10x_i+y_i=10x_{i+1},\text{ where }{x}_{i+1}^n\le C}$$ then per the binomial theorem $${\displaystyle (10x_i+y{)}^n-(10x_i{)}^n=\sum _{k=0}^{n-1}P(n,k)(10x_i{)}^{n-k}{y}_i^k\le {10}^n(C-x_i^n)}$$ The term ${\displaystyle {10}^n(C-x_i^n)}$ is just a ${\displaystyle {10}^n}$ multiple of the ith remainder, ${\displaystyle C-x_i^n}$.

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:

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 ${\displaystyle y\le c}$.
   • 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.

Examples

Find the square root of 152.2756

For clarity, the value of the chosen digit x is in red while the current digital tally is in blue.

Algorithm terminates: Answer is 12.34

Find the cube root of 4192 truncated to the nearest thousandth

The desired precision is achieved. The cube root of 4,192 is 16.124...

Logarithmic calculation

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

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

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|.

Complex roots

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

Square 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

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

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:

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

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

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

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 ${\displaystyle \sqrt{z}}$ maps ${\displaystyle z}$ 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

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

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

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 ${\displaystyle -i}$. As a result of this symmetry, the sum of the nth roots of unity equals zero. $${\displaystyle \sum _{k=0}^{n-1}{e}^{\frac{2\pi i}{n}k}=0}$$

nth roots

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

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

where η is a single nth root, and 1, ω, ω², ... ωⁿ⁻¹ are the nth roots of unity. Thus, since they are all just multiplied by the same scalar η, the sum of the nth roots equals zero. For example, the four different fourth roots of 2 are

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

In polar form, a single nth root may be found from Demoivre's theorem:

$${\displaystyle z^{\frac{1}{n}}=\sqrt[n]{r{e}^{i\theta }}=\sqrt[n]{r}\cdot e^{i\theta /n}=r^{\frac{1}{n}}\cdot \left(\cos \left(\frac{\theta }{n}\right)+i\sin \left(\frac{\theta }{n}\right)\right)}$$

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 ${\displaystyle a+ib}$ then ${\displaystyle r=\sqrt{a^2+b^2}}$. The ${\displaystyle \theta }$ 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

$${\displaystyle \cos \theta =\frac{a}{r},\sin \theta =\frac{b}{r},\text{ and }\tan \theta =\frac{b}{a}.}$$

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, as proven by the nth root theorem

$${\displaystyle \sqrt[n]{r}\cdot \left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\text{ for }k=0,1,2,\dots ,n-1.}$$

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.

Polynomial roots

A root of a polynomial ⁠${\displaystyle p(x)}$⁠ is a number ⁠${\displaystyle a}$⁠ such that ⁠${\displaystyle p(a)=0}$⁠. An nth root of a number ⁠${\displaystyle a}$⁠ is by definition a root of the polynomial ⁠${\displaystyle x^n-a}$⁠. 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 nth roots (see Cubic equation and Quartic equation).

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. 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 $${\displaystyle p(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)=0}$$ has radical roots ${\displaystyle a_1,a_2,...,a_5.}$ 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 ⁠${\displaystyle x^5-x-1}$⁠. See Quintic function § Solvable quintics and 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 nth root of a number can always be extracted, even for a quintic function ${\displaystyle x^5-a}$. 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. There are numbers, called transcendental numbers that are not polynomial roots. The number π is an example of such a transcendental number.

It may be unclear why any number ${\displaystyle a^n}$ has n roots rather than just a primary root. To demonstrate this, for the principal root a of the variable x taken to the nth power, the following polynomial relation holds: $${\displaystyle p(x)=x^n-a^n=0}$$ This polynomial can be factored as follows: $${\displaystyle \begin{array}{cc}p(x)& =x^n-a^n\\ & =(x-a)(x^{n-1}+a{x}^{n-2}+a^2{x}^{n-3}+\cdots +a^{n-1})\\ & =(x-a)\left(\sum _{k=0}^{n-1}{x}^{n-k-1}{a}^k\right)\end{array}}$$ Thus, the polynomial ${\displaystyle p(x)}$ is zero for x equal to a, or for any x that solves the equation: $${\displaystyle \sum _{k=0}^{n-1}{x}^{n-k-1}{a}^k=0}$$ By the fundamental theorem of algebra, this series has ${\displaystyle n-1}$ roots, for a combined total of ${\displaystyle n}$.

As an example, let ${\displaystyle n=3}$ and ${\displaystyle a=1}$, then find the cube roots of 1

$${\displaystyle p(x)=x^3-1^3=(x-1)(x^2+x^1+1)=0}$$

Thus the first root is ${\displaystyle x=1}$, and the other two roots can be derived using the quadratic equation with ${\displaystyle a=b=c=1}$

$${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm i\sqrt{3}}{2}}$$

Proof of irrationality for non-perfect nth power x

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 ${\displaystyle b\ne 1}$. This means that if ${\displaystyle b\ne 1}$, ${\displaystyle \frac{a^n}{b^n}}$ 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.