$n$ -th Root¶

Definition¶

Let $x\geq0$ be a nonnegative real number, and let $n\geq1$ be a positive integer. The principal $n$ -th root of $x$ is

x^{1/n}=\sup\{y\in\mathbb{R}\colon y\geq 0 \text{ and } y^n\leq x\}.

It is the unique nonnegative real number whose $n$ -th power is $x\text{.}$ When $n$ is odd, the root extends to every real $x$ : $\sqrt[n]{x}$ is the unique real number $y$ satisfying $y^n=x\text{,}$ and $\sqrt[n]{-x}=-\sqrt[n]{x}\text{.}$

Interpretation¶

The $n$ -th root undoes the $n$ -th power: $x^{1/n}$ is the nonnegative number whose $n$ -th power is $x$ . It extends exponentiation from integer exponents to fractional ones.

Graphs of x squared and the square root of x, mirrored in the line y = x, and of the cube root of x, which is also defined for negative x; the point (4, 2) shows that the square root of 4 is 2.

The graph of $\sqrt{x}$ is the mirror image in the line $y=x$ of the graph of $x^2$ for $x\ge0$ $0$ , because the root swaps input and output. The marked point $(4,2)$ records $\sqrt{4}=2$ $2$ , that is $2^2=4$ $4$ , and $(1,1)$ lies on all three curves. The cube root continues to the left of $0$ : every real number, negative or not, has exactly one real cube root.

Change $n$ : the $n$ -th root undoes $x^n$ , so its graph is the mirror image in $y=x$ $x$ . For odd $n$ both are defined on all of $\mathbb R$ ; for even $n$ only for $x\ge0$ $0$ .

The definition uses a supremum instead of a formula because the existence of a number such as $\sqrt{2}$ , which is irrational, is exactly what has to be proved. The set $\{y\ge0\colon y^n\le x\}$ contains $0$ and is bounded above by $\max\{1,x\}$ , so its supremum exists by the completeness of the real numbers; one then shows that its $n$ -th power equals $x$ .

On $[0,\infty)$ the map $y\mapsto y^n$ is a bijection, and the principal $n$ -th root is its inverse. For odd $n$ the power is a bijection of all of $\mathbb{R}$ , which is why odd roots of negative numbers exist and why only even roots restrict the domain to $[0,\infty)$ .

A common slip is to write $\sqrt{4}=\pm2$ $\pm 2$ . The equation $y^2=4$ has the two solutions $\pm2$ , but $\sqrt{4}$ denotes only the nonnegative one. For the same reason $\sqrt{x^2}=|x|$ $∣ x ∣$ , not $x$ .

Key Results¶

For nonnegative inputs, the principal $n$ -th root is defined as the supremum of a bounded set.
Even roots are real-valued only for nonnegative inputs; odd roots are real-valued on all of $\mathbb{R}\text{.}$
If $p/q$ is in lowest terms and $q$ is odd, the course uses $x^{p/q}=(\sqrt[q]{x})^p$ for every real $x$ (with $x\neq0$ when $p<0$ $0$ ).
The principal $n$ -th-root function is increasing on $[0,+\infty)\text{.}$

Examples¶

$\sqrt{4}=2\text{.}$
$\sqrt[3]{8}=2\text{.}$
$\sqrt[3]{-8}=-2$ and $(-8)^{2/3}=4\text{.}$

Prerequisites¶

Sources¶

Math I: Calculus · Lectures 3–4

Aliases: nth root Symbols: element of, infinity, real numbers, square root

Last updated: 20 October 2026 03:02

nnn-th Root¶