Prove That Every Nonnegative Real Number Has an nth Root

Let $a\ge 0$ be a real number and let $n$ be a positive integer. We will prove that there is a nonnegative real number $b$ with $b^n=a$. The proof uses the least-upper-bound property of the real numbers and continuity of the function $x\mapsto x^n$.

This argument also explains why the result depends on completeness: it constructs the root as the supremum of a suitable set, rather than assuming in advance that an nth-root function exists.

Construct a set whose supremum will be the root

Define

$$S=\{x\in\mathbb{R}:x\ge 0\text{ and }x^n\le a\}.$$

The set is nonempty because $0\in S$. It is bounded above. For example, $a+1$ is an upper bound: if $x>a+1$, then $x>1$, so $x^n\ge x>a$, and consequently $x\notin S$.

By completeness of $\mathbb{R}$, the nonempty set $S$, which is bounded above, has a supremum. Put

$$b=\sup S.$$

Since every element of $S$ is nonnegative, $b\ge 0$. We now show that its $n$th power is exactly $a$.

First show that $b^n\le a$

Because $b$ is the least upper bound, for every positive integer $k$ there is an element $s_k\in S$ such that

$$b-\frac{1}{k}<s_k\le b.$$

Indeed, if there were no such element, then $b-1/k$ would be an upper bound for $S$, contradicting that $b$ is its least upper bound. These inequalities imply $s_k\to b$. Since the function $f(x)=x^n$ is continuous,

$$s_k^n\longrightarrow b^n.$$

Every $s_k$ belongs to $S$, so $s_k^n\le a$. Taking the limit gives $b^n\le a$.

Rule out the possibility that $b^n<a$

Suppose instead that $b^n<a$. The function $f(x)=x^n$ is continuous at $b$, so values of $f(x)$ for $x$ sufficiently close to $b$ remain below $a$. In particular, there is some $x>b$ close enough to $b$ that $x^n<a$.

This $x$ is nonnegative and satisfies $x^n\le a$, so $x\in S$. But that contradicts the fact that $b$ is an upper bound for $S$, since $x>b$. Therefore $b^n<a$ is impossible. Together with $b^n\le a$, this proves

$$b^n=a.$$

Thus the required nonnegative real number exists.

A shorter proof using the intermediate value theorem

If you already know the intermediate value theorem, you can get existence directly. When $a=0$, take $b=0$. When $a>0$, the polynomial $f(x)=x^n$ is continuous on $[0,a+1]$, and

$$f(0)=0<a<(a+1)^n=f(a+1).$$

The last inequality holds because $a+1>1$ and $n\ge 1$, so $(a+1)^n\ge a+1>a$. By the intermediate value theorem, $f$ takes the value $a$ somewhere between $0$ and $a+1$. That is, there is a $b\ge 0$ such that $b^n=a$.

The supremum proof makes explicit how existence follows from completeness; the intermediate-value proof is convenient once continuity and the intermediate value theorem are available.

Example

Take $a=32$ and $n=5$. The set in the supremum proof is

$$S=\{x\ge 0:x^5\le 32\}.$$

Since $2^5=32$, the boundary between numbers whose fifth powers are at most $32$ and numbers whose fifth powers exceed $32$ is $2$. The proof identifies this boundary as $b=\sup S$ and guarantees that $b^5=32$. For a number without an obvious integer root, such as $a=10$ and $n=3$, the same argument still guarantees a nonnegative real cube root even though it does not express that root as a fraction or decimal.

Is the root unique?

Yes. If $0\le c<d$, then

$$d^n-c^n=(d-c)(d^{n-1}+d^{n-2}c+\cdots+c^{n-1})>0.$$

Thus $x^n$ is strictly increasing on the nonnegative reals. Two distinct nonnegative numbers therefore cannot both have $n$th power equal to $a$. Existence and uniqueness together justify speaking of the nonnegative nth root of $a$.

Common mistakes

• Assuming the root exists because it is written as $a^{1/n}$. That notation presupposes the existence result. The proof must obtain a real number with the required power, for example by the supremum argument or the intermediate value theorem.
• Forgetting the condition $b\ge 0$. For even $n$, a positive $a$ also has a negative real solution. The theorem asks for a nonnegative one; uniqueness above refers only to nonnegative solutions.
• Using an unbounded set in the supremum argument. The least-upper-bound property applies here because $S$ is bounded above. Establishing that bound is part of the proof.
• Treating $n=0$ as an allowed case. The statement requires a positive integer $n$. For $n=0$, the expression $b^0$ does not provide this theorem, and the case must not be folded into the argument.