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# nth root

The phrase “*the $n$-th root of a number*” is a somewhat misleading concept that requires a fair amount of thought to make rigorous.

For $n$ a positive integer, we define *an* $n$-th root of a number $x$ to be a number $y$ such that $y^{n}=x$. The number $n$ is said to be the *index* of the root. Note that the term “number” here is ambiguous, as the discussion can apply in a variety of contexts (groups, rings, monoids, etc.) The purpose of this entry is specifically to deal with $n$-th roots of real and complex numbers.

In an effort to give meaning to the term *the* $n$-th root of a real number $x$, we define it to be the unique real number that $y$ is *an* $n$th root of $x$ and such that $\operatorname{sign}(x)=\operatorname{sign}(y)$, if such a number exists. We denote this number by $\sqrt[n]{x}$, or by $x^{{\frac{1}{n}}}$ if $x$ is positive. This specific $n$th root is also called the *principal $n$th root*.

Example: $\sqrt[4]{81}=3$ because $3^{4}=3\times 3\times 3\times 3=81$, and $3$ is the unique positive real number with this property.

Example: If $x+1$ is a positive real number, then we can write $\sqrt[5]{x^{5}+5x^{4}+10x^{3}+10x^{2}+5x+1}=x+1$ because $(x+1)^{5}=(x^{2}+2x+1)^{2}(x+1)=x^{5}+5x^{4}+10x^{3}+10x^{2}+5x+1$. (See the Binomial Theorem and Pascal’s Triangle.)

The nth root operation is distributive for multiplication and division, but not for addition and subtraction. That is, $\sqrt[n]{x\times y}=\sqrt[n]{x}\times\sqrt[n]{y}$, and $\sqrt[n]{\frac{x}{y}}=\frac{\sqrt[n]{x}}{\sqrt[n]{y}}$. However, except in special cases, $\sqrt[n]{x+y}\not=\sqrt[n]{x}+\sqrt[n]{y}$ and $\sqrt[n]{x-y}\not=\sqrt[n]{x}-\sqrt[n]{y}$.

Example: $\sqrt[4]{\frac{81}{625}}=\frac{3}{5}$ because $\left(\frac{3}{5}\right)^{4}=\frac{3^{4}}{5^{4}}=\frac{81}{625}$.

Note that when we restrict our attention to real numbers, expressions like $\sqrt{-3}$ are undefined. Thus, for a more full definition of $n$th roots, we will have to incorporate the notion of complex numbers: *The nth roots of a complex number* $t=x+yi$ are all the complex numbers $z_{1},z_{2},\ldots,z_{n}\in\mathbb{C}$ that satisfy the condition $z_{k}^{n}=t$. Applying the fundamental theorem of algebra (complex version) to the function $x^{n}-t$ tells us that $n$ such complex numbers always exist (counting multiplicity).

One of the more popular methods of finding these roots is through trigonometry and the geometry of complex numbers. For a complex number $z=x+iy$, recall that we can put $z$ in polar form: $z=(r,\theta)$, where $r=\sqrt[2]{x^{2}+y^{2}}$, and $\theta=\frac{\pi}{2}$ if $x=0$, and $\theta=\arctan{\frac{y}{x}}$ if $x\not=0$. (See the Pythagorean Theorem.) For the specific procedures involved, see calculating the nth roots of a complex number.

## Mathematics Subject Classification

30-00*no label found*12D99

*no label found*

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## Attached Articles

## Corrections

n = ? by pahio ✓

"an" pro "a"? by pahio ✓

fundamental theorem by pahio ✓

index by Wkbj79 ✓

## Comments

## the last example is wrong

I should not have taken the square root of (sqrt[4](2), pi/4). I should have taken the cube root.

As a result, the entire rest of that example is wrong. I'll fix it I promise, but I don't have time right now..

## last example is now fixed

Phew! I thought I remembered trigonometry. Then I did the last example!

## NthRoot.html

Here, in PlanetMath, the real n'th root $\sqrt[n]{x}$ is defined only for non-negative real $x$ and for natural number $n$ (what does mean $\sqrt[0]{x}$?). I have always teached in school so that

the n'th root of a real number x is such a real number y that

(1) $y^n = x$

and

(2) $sign(y) = sign(x)$

(n = 1, 2, 3, ...).

Then the root is _uniquely_ _determined_ in the cases it exists.

Is this right?

The definition in PlanetMath allows two values in the example case

$\sqrt[4]{81}$, although here is given only the value 3.

## odd roots

The odd n'th root (cube root etc.) of a real number b can not be identified with the "fractional power" a^{1/n}, although so has been done in the entries "n'th root" and "cube root". Viz., the fractional power with a negative base is not uniquely determined -- it depends not only on the value of the exponent but also on the form of the exponent; e.g.,

(-1)^{1/3} = the 3'rd root of -1, i.e. = -1

(-1)^(2/6) = the 6'th root of (-1)^2, i.e. = 1

What should we do concerning the fractional powers?

## Re: odd roots

Old discussion, I know, but this entry turned up on the sidebar,

and I think the content of the discussion merits some emphasis

(and an entry on PlanetMath)

The problem is, of course, that there is no algebraic way

to distinguish between the different nth roots of a number.

(Note: positivity does not count because positivity is not part

of the axioms of a field.)

This is already well-known in the case of complex numbers

(the nth root is not even a continuous function then)

Of course it is useful to have a notion of a principal square root

for real numbers. But we must be aware that some of the "distributive properties"

of exponents will then fail to hold, unless we restrict the bases

to be positive too --- naturally, since we imposed a positivity condition

to define real roots to begin with.