# The existence of nth roots

## 1Introduction

We have seen the completeness axiom for the reals, the idea of a monotonic sequence, and we have seen why every bounded monotonic sequence converges to a limit. These are important ideas and essential for much mathematics, including a proper understanding of why square roots (and also cube roots, fourth roots, and solutions of other algebraic equations) exist and how and why we can define and use numbers such as $π$ and $e$. Therefore I consider it essential for this course that you have seen and appreciate at least some of this sort of material, and that's what this web page starts to do.

Some of this material is more difficult than much of the rest of the course. You should at a minimum read and appreciate the main theorem here, in particular noting that it follows in an essential way from the Completeness Axiom for the reals.

## 2Roots

We aim to prove that $s$th roots exist of all positive $x ∈ ℝ$, for all $s ∈ ℕ$. First we need a lemma related to Bernoulli's Inequality.

Lemma 2.1

Suppose $0 < ε < 1 2 s$ where $s ≥ 1$ is a natural number. Then $( 1 + ε ) s < 1 + 2 ⁢ s ⁢ ε$.

Proof

This is proved by induction on $s$. For $s = 1$ we have $( 1 + ε ) 1 = 1 + ε < 1 + 2 ⁢ ε$ for all $ε > 0$, so this case is clear.

Now suppose $s ≥ 2$ and that the statement of the lemma holds for $s - 1$. Let $ε ∈ ℝ$ satisfy $0 < ε < 1 2 s$. Then certainly $ε < 1 2 ⁢ s - 2$ so we have

$( 1 + ε ) s - 1 < 1 + 2 ⁢ ( s - 1 ) ⁢ ε$

by the induction hypothesis. Thus

$( 1 + ε ) s < ( 1 + 2 ⁢ ( s - 1 ) ⁢ ε ) ( 1 + ε ) = 1 + ( 2 s - 1 ) ⁢ ε + 2 ⁢ ( s - 1 ) ⁢ ε 2$

and as $ε < 1 2 s < 1 2 ⁢ ( s - 1 )$ we have $ε > 2 ⁢ ( s - 1 ) ε 2$ and

$( 1 + ε ) s < 1 + 2 ⁢ s ⁢ ε$

as required.

We now prove

Theorem 2.1 (roots)

Let $s ∈ 2 3 4 …$ and $x > 0$. Then there is some $l ∈ ℝ$ such that $l s = x$.

Proof

The idea is to mimic the proof that any real number has a rational sequence converging to it but instead of devising our sequence $a n$ to converge to $x$ we try to arrange that $a n s$ converges to $x$. Then $a n$ will converge to some $l$ with $l s = x$.

Define $( a n )$ by induction as follows. We let $a 0 = a 1 = 0$. Now, supposing $a n$ is defined with both $0 ≤ a n$ and $a n s < x$, let $k ∈ 0 1 2 3 …$ be the greatest integer such that $a n + k n s < x$ and let $a n + 1$ be $a n + k n$.

Such a $k$ exists by the Archimedean Property and the least number principle: in the case when $x > 1$ we clearly have $a n + n ⁢ x n s ≥ x$ as $a n ≥ 0$, so by the Archimedean Property there is some integer $K ∈ ℤ$ with $a n + K n s ≥ x$. If $K 0$ is the least such $K$, then our $k$ is $K 0 - 1$. If $x ≤ 1$ it is easier: again, since $a n ≥ 0$ the number $K = n$ has $a n + K n s ≥ x$ so we take the least such $K 0$ and $k = K 0 - 1$ again.

This construction of the values $a n$ is performed for all $n$, and the result is clearly a monotonic nondecreasing sequence (since we only ever added nonnegative numbers, never subtracted them) which is bounded above by $1$ (if $x ≤ 1$) or by $x$ (if $x > 1$) and hence converges to some limit $l ∈ ℝ$ by the Monotone Convergence Theorem.

We claim that $l s = x$, and we shall prove this by showing that $l s > x$ and $l s < x$ are both impossible.

First:

Subproof

Assume $l s > x$.

Then let $ε = l s - x ⋅ 1 s ⁢ l s - 1$. As $( a n )$ converges to $l$ there is some $a n$ with $a n - l < ε$. Then

$l s - a n s = l - a n ⋅ a n s - 1 + a n s - 2 l + … + a n l s - 2 + l s - 1 < ε ⋅ s l s - 1 < l s - x ,$

so $l s - a n s < l s - x$. This is impossible as it would mean that $a n s > x$; but by construction $a n s < x$ for all $n$.

Also:

Subproof

Assume $l s < x$.

Then choose $n ∈ ℕ$ such that

$1 n < min l 2 ⁢ s , x - l s 2 ⁢ s ⁢ l s - 1 .$

Then

$l + 1 n s = l s 1 + 1 l ⁢ n s ≤ l s 1 + 2 ⁢ s l ⁢ n$

as $1 l ⁢ n < 1 2 ⁢ s$, using the lemma. So

$l s ≤ l + 1 n s ≤ l s + 2 s l s - 1 n < x .$

This last inequality is by choice of $n$ such that $1 n < x - l s 2 s l s - 1$.

This gives a contradiction, for $a n s < x$ and $a n ≤ l$, and so $a n + 1 n s < x$. By construction $a n + 1 = a n + k n$ where $k$ is largest possible giving $a n + 1 s < x$. But $a n + k n ≤ l$ and $a n + k + 1 n ≤ l + 1 n$ so

$a n + k + 1 n s ≤ l + 1 n s < x$

contradicting our choice of $k$.

Therefore the limit $l$ of $( a n )$ satisfies $l s = x$, as required.

This was nice, but quite hard work. As a reward, we get a definition of $x r s$ for positive $x$ and integer $r , s ∈ ℤ$. (Define $x r s$ to be the $s$th root of $x r$.) This idea of rational powers has a couple of nice properties that will be useful later.

Proposition 2.1

If $0 < x < y$ and $s ∈ ℕ$ then $0 < x 1 s < y 1 s$.

Proof

If not, $x 1 s ≥ y 1 s$, so by a result on ordered fields, raising to the $s$th power gives $( x 1 s ) s = x ≥ ( y 1 s ) s = y$.

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