Proof of Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits

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 $k\in\mathbb{N}$, $k\ge1$ by Arithmetic and Order of the Natural Numbers §least and $k\in\mathbb{N}_{0}$ 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 §sign and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §monotone-iff, for the ordered field $\mathbb{R}$, apply with exponent $k$ (and with exponent $2$ in the clause abs). The clauses below are proved for arbitrary nonnegative $x,y$, and later clauses apply earlier ones to other nonnegative reals.

Principle (U). Let $j\in\mathbb{N}$ and $w\in\mathbb{R}$ with $w\ge0$. The root $\sqrt[j]{w}$ of The k-th Root and the Square Root of a Nonnegative Real Number §root, taken with $j$ in place of $k$, is the unique $z\ge0$ with $z^{j}=w$, by Existence and Uniqueness of Nonnegative k-th Roots of Nonnegative Real Numbers §root with $j$ in place of $k$. Hence: if $z\in\mathbb{R}$, $z\ge0$ and $z^{j}=w$, then $\sqrt[j]{w}=z$.

Clause power. $(\sqrt[k]{x})^{k}=x$ holds by The k-th Root and the Square Root of a Nonnegative Real Number §root. Next, $x^{k}\ge0$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §sign, so $\sqrt[k]{x^{k}}$ is defined, and (U) with $j=k$, $w=x^{k}$ and $z=x$ gives $\sqrt[k]{x^{k}}=x$. By Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product, $0^{k}=0$ and $1^{k}=1$, and $0\le0$ by reflexivity of $\le$, and $0\le1$ since $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares; so (U) with $j=k$ gives $\sqrt[k]{0}=0$ and $\sqrt[k]{1}=1$.

Clause abs. Let $z\in\mathbb{R}$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §square-abs, $z^{2}=|z|^{2}$, and $|z|\ge0$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value, so $z^{2}\ge0$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §sign and $\sqrt{z^{2}}=\sqrt[2]{z^{2}}$ is defined by The k-th Root and the Square Root of a Nonnegative Real Number §square-root. Principle (U) with $j=2$, which is a natural number by Arithmetic and Order of the Natural Numbers §digits, with $w=z^{2}$ and with the nonnegative element $|z|$, whose square is $z^{2}$, gives $\sqrt{z^{2}}=|z|$.

Clause product. Let $a=\sqrt[k]{x}$ and $b=\sqrt[k]{y}$, so $a,b\ge0$, $a^{k}=x$ and $b^{k}=y$. Then $xy\ge0$ and $ab\ge0$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product, and $(ab)^{k}=a^{k}b^{k}=xy$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product. By (U) with $j=k$, $\sqrt[k]{xy}=ab$.

Clause monotone. With $a,b$ as above, Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §monotone-iff gives $a\le b$ if and only if $a^{k}\le b^{k}$, that is, $x\le y$; and $a=b$ if and only if $x=y$. Since $<$ is the strict relation of $\le$, $a<b$ means $a\le b$ and $a\neq b$, which is therefore equivalent to $x\le y$ and $x\neq y$, that is, to $x<y$.

Clause subadditive. We first show, by induction on $n$ (Arithmetic and Order of the Natural Numbers §induction, applied to the class of $n\in\mathbb{N}$ for which the inequality holds for all $s,t\ge0$), that $s^{n}+t^{n}\le(s+t)^{n}$ for all $s,t\in\mathbb{R}$ with $s,t\ge0$. For $n=1$ both sides equal $s+t$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product. If it holds for $n$, then, since $s+t\ge0$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents give

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

where the last step uses $s^{n}t\ge0$ and $t^{n}s\ge0$, from Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §sign and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product, and Rules of Arithmetic and Order in an Ordered Field §order-sum. Now let $a,b$ be as above. Then $x+y\ge0$, and with $c=\sqrt[k]{x+y}\ge0$,

$$c^{k}=x+y=a^{k}+b^{k}\le(a+b)^{k}.$$

As $c\ge0$ and $a+b\ge0$, Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §monotone-iff gives $c\le a+b$.

Clause difference. Since $y-x=-(x-y)$ by Rules of Arithmetic and Order in an Ordered Field §signs (as $-(x+(-y))=(-x)+(-(-y))=(-x)+y$), we have $|x-y|=|-(x-y)|=|y-x|$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value, and likewise $|\sqrt[k]{x}-\sqrt[k]{y}|=|\sqrt[k]{y}-\sqrt[k]{x}|$, both sides are symmetric in $x$ and $y$, and by totality we may assume $x\le y$. Then $y-x\ge0$ and $|x-y|=|y-x|=y-x$ by Absolute Value in an Ordered Field §absolute-value. The clause subadditive, proved above, applied to the nonnegative reals $x$ and $y-x$, gives $\sqrt[k]{y}\le\sqrt[k]{x}+\sqrt[k]{y-x}$, so $\sqrt[k]{y}-\sqrt[k]{x}\le\sqrt[k]{y-x}$. By the clause monotone, $\sqrt[k]{x}\le\sqrt[k]{y}$, so $\sqrt[k]{y}-\sqrt[k]{x}\ge0$ and $|\sqrt[k]{y}-\sqrt[k]{x}|=\sqrt[k]{y}-\sqrt[k]{x}$. Hence $|\sqrt[k]{x}-\sqrt[k]{y}|\le\sqrt[k]{|x-y|}$.

Clause limits. Since every $a_{n}\ge0$, $\sqrt[k]{a_{n}}\in\mathbb{R}$ is defined for every $n\in\mathbb{N}$, so $n\mapsto\sqrt[k]{a_{n}}$ is a map $\mathbb{N}\to\mathbb{R}$, that is, a sequence in $\mathbb{R}$ by Sequences §sequence. Let $\varepsilon\in\mathbb{R}$ be positive. First put $\eta=\varepsilon^{k}$, which is positive by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §sign. Then, by Convergent Sequences of Real Numbers §converges applied to $a_{n}\to x$ with $\eta$, choose $N\in\mathbb{N}$ with $|a_{n}-x|<\eta$ for every $n\ge N$. Let $n\in\mathbb{N}$ with $n\ge N$. The clause difference, applied to $a_{n}$ and $x$, gives $|\sqrt[k]{a_{n}}-\sqrt[k]{x}|\le\sqrt[k]{|a_{n}-x|}$. Since $0\le|a_{n}-x|<\varepsilon^{k}$, the clause monotone gives $\sqrt[k]{|a_{n}-x|}<\sqrt[k]{\varepsilon^{k}}$, and $\sqrt[k]{\varepsilon^{k}}=\varepsilon$ by the clause power, as $\varepsilon\ge0$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, $|\sqrt[k]{a_{n}}-\sqrt[k]{x}|<\varepsilon$. As $\varepsilon$ was arbitrary, $\sqrt[k]{a_{n}}\to\sqrt[k]{x}$ by Convergent Sequences of Real Numbers §converges.