Proof of Existence and Uniqueness of Nonnegative k-th Roots of Nonnegative Real Numbers

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named at each use. Since $k\in\mathbb{N}$, we have $k\ge1$ by Arithmetic and Order of the Natural Numbers §least, and $k\in\mathbb{N}_{0}$, as every natural number is, by The Natural Numbers with Zero and Their Embedding into the Integers §naturals; so the clauses Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents, Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product, Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §monotone, Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §monotone-iff, Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §base-above-one and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponent-monotone, for the ordered field $\mathbb{R}$, apply with natural exponents, in particular with exponent $k$. Natural numbers standing where a real number is required are read as in The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §identification. The elementary order steps below, namely $M=1+x\ge1$ and $x<M$ (using $0\le x$ and $0<1$ from Rules of Arithmetic and Order in an Ordered Field §squares), $0\le t-s$, $C\ge1$, $0\le y\le y+h\le M$, $0\le s<y$ and $Ch\le\delta$, follow from Rules of Arithmetic and Order in an Ordered Field §order-sum and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling, with Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed where a strict and a non-strict inequality are chained.

Uniqueness. If $y,y'\in\mathbb{R}$ with $y,y'\ge0$ and $y^{k}=x=y'^{k}$, then $y=y'$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §monotone-iff.

A difference bound. Let $M=1+x$; then $M\ge1$ and $x<M$. We show by induction on $n$ (Arithmetic and Order of the Natural Numbers §induction, applied to the class $A$ of those $n\in\mathbb{N}$ for which the following holds for all $s,t\in\mathbb{R}$): if $0\le s\le t\le M$, then

$$t^{n}-s^{n}\le n\,M^{n}\,(t-s).$$

For $n=1$: $t^{1}-s^{1}=t-s$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product, and $t-s\le M(t-s)$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling, since $1\le M$ and $0\le t-s$; so $1\in A$. Let $n\in A$ and $0\le s\le t\le M$. By Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product, $t^{n+1}=t^{n}t$ and $s^{n+1}=s^{n}s$, so

$$t^{n+1}-s^{n+1}=t^{n}(t-s)+s\,(t^{n}-s^{n}).$$

By Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §monotone-iff, $t^{n}\le M^{n}$ and $0\le t^{n}-s^{n}$; and $M^{n}\le M^{n+1}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §base-above-one. Hence, by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling and $n\in A$,

$$t^{n}(t-s)\le M^{n}(t-s)\le M^{n+1}(t-s),\qquad s\,(t^{n}-s^{n})\le M\,(t^{n}-s^{n})\le M\cdot n\,M^{n}(t-s)=n\,M^{n+1}(t-s),$$

using $0\le s\le M$, $0\le M$ and $M\,M^{n}=M^{n+1}$. Adding these by Rules of Arithmetic and Order in an Ordered Field §order-sum gives $t^{n+1}-s^{n+1}\le(n+1)\,M^{n+1}(t-s)$, the sum $n+1$ computed in $\mathbb{N}$ agreeing with its reading in $\mathbb{R}$ by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement. So $n+1\in A$, and $A=\mathbb{N}$. Put $C=k\,M^{k}$. Since $1\le k$ in $\mathbb{N}$, hence $1\le k$ in $\mathbb{R}$, the order agreeing by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement, and $1\le M^{k}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §base-above-one, $C\ge1>0$. Thus

$$t^{k}-s^{k}\le C\,(t-s)\qquad\text{whenever }0\le s\le t\le M.$$

We also note that $u\le u^{k}$ for every $u\ge1$, by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponent-monotone (with exponents $1\le k$) and $u^{1}=u$.

Existence. Let $A_{x}=\{t\in\mathbb{R}:t\ge0,\ t^{k}\le x\}$. Since $0^{k}=0\le x$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product, $0\in A_{x}$. Every $t\in A_{x}$ satisfies $t\le M$: otherwise $t>M\ge1$, so $t\le t^{k}\le x<M<t$, which is impossible. Thus $A_{x}$ is a subset of $\mathbb{R}$ that is nonempty, as $0\in A_{x}$, and bounded above, $M$ being an upper bound of $A_{x}$ in the sense of Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounds and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §supremum, $A_{x}$ has a supremum, which is unique by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum; let $y=\sup A_{x}$. By Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum, $y$ is an upper bound of $A_{x}$ and $y\le M$; so $0\le y\le M$, since $0\in A_{x}$. The order of $\mathbb{R}$ being total, it suffices to exclude $y^{k}<x$ and $y^{k}>x$.

Suppose $y^{k}<x$ and let $\delta=x-y^{k}>0$. Then $y<M$: if $y=M$, then $y^{k}=M^{k}\ge M>x$, a contradiction. Here $0<M-y$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, and $0<\delta/C=\delta\cdot C^{-1}$, since $0<C^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal (as $0<C$) and then Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product applies to $\delta$ and $C^{-1}$. Let $h=\min\{M-y,\ \delta/C\}$, which exists by The Maximum and Minimum of Two Elements of a Total Order §exists; a least element of a set belongs to it by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §least, so $h$ is one of the two positive elements $M-y$ and $\delta/C$, and $h>0$; by The Maximum and Minimum of Two Elements of a Total Order §bounds, $h\le M-y$ and $h\le\delta/C$. Then $0\le y\le y+h\le M$, so $(y+h)^{k}-y^{k}\le Ch\le\delta$, that is, $(y+h)^{k}\le x$. Since $y+h\ge0$, $y+h\in A_{x}$, so $y+h\le y$, contradicting $h>0$.

Suppose $y^{k}>x$ and let $\delta=y^{k}-x>0$. Then $y>0$, since $0^{k}=0\le x$. As before, $0<\delta/C$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product. Let $h=\min\{y,\ \delta/C\}$, which exists by The Maximum and Minimum of Two Elements of a Total Order §exists and, by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §least, is one of the two positive elements $y$ and $\delta/C$; so $h>0$, and $h\le y$ and $h\le\delta/C$ by The Maximum and Minimum of Two Elements of a Total Order §bounds. Let $s=y-h$. Then $0\le s<y\le M$, so $y^{k}-s^{k}\le Ch\le\delta$, that is, $s^{k}\ge y^{k}-\delta=x$. Now $s<y=\sup A_{x}$, and $A_{x}$ is a nonempty subset of $\mathbb{R}$ that is bounded above, as shown at the start of the existence part ($0\in A_{x}$, and $M$ is an upper bound of $A_{x}$). So Arbitrary Positive Slack, and Approximation of Suprema and Infima, in the Real Numbers §strict-above, applied with $S=A_{x}$ and $b=s$, gives $t\in A_{x}$ with $s<t$. As $0\le s<t$, Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §monotone gives $s^{k}<t^{k}$, so $x\le s^{k}<t^{k}\le x$, which is impossible.

Hence $y^{k}=x$, with $y\ge0$. Together with uniqueness, this proves the clause root.