Proof of Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality

In the commutative ring $R$, multiplication is associative and commutative by the ring laws, and $1$ is a neutral element of it, since $x\cdot1=x=1\cdot x$ (unique by A Binary Operation Has at Most One Neutral Element §unique, so it is the $1$ of Powers with Exponents in the Natural Numbers with Zero §zero). Inductions below run over the set of those $n\in\mathbb{N}$ for which the claim holds, and conclude by Arithmetic and Order of the Natural Numbers §induction; an element of $\mathbb{N}_{0}$ is $0$ or lies in $\mathbb{N}$ by The Natural Numbers with Zero and Their Embedding into the Integers §naturals.

Preliminaries. (R) For $x\in R$ and $n\in\mathbb{N}_{0}$, $x^{n+1}=x^{n}x$. For $n\ge1$ this is Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion for the constant map $k\mapsto x$ on $[n+1]$, together with Powers with Exponents in the Natural Numbers with Zero §power. For $n=0$, the same clause gives $x^{1}=\prod_{k=1}^{1}x=x=1\cdot x=x^{0}x$ by Powers with Exponents in the Natural Numbers with Zero §zero.

For $u,v,w\in R$ the ring laws give the following. (Z) $u\cdot0=0$: indeed $u\cdot0=u\cdot(0+0)=u\cdot0+u\cdot0$, and adding $-(u\cdot0)$ gives $0=u\cdot0$. (N) $(-u)v=-(uv)$, by the uniqueness in Additive and Multiplicative Inverses Are Unique §negative, since $uv+(-u)v=(u+(-u))v=0\cdot v=v\cdot0=0$ by commutativity and (Z). Hence $(u-v)w=uw-vw$. (D) $-(u-v)=v-u$, by the same uniqueness, since $(u-v)+(v-u)=0$.

Exponents. If $m=0$ or $n=0$, then $x^{m+n}=x^{m}x^{n}$ because $x^{0}=1$ and $0+n=n$, $m+0=m$ (Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero). If $m,n\ge1$, it is Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §splitting for the constant map $k\mapsto x$ on $[m+n]$. For the second law, if $n=0$, then $(x^{m})^{0}=1=x^{0}=x^{m\cdot0}$ by Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §zero. For $n\ge1$ we induct on $n$: $(x^{m})^{1}=x^{m}=x^{m\cdot1}$ by (R) and Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §one, and if $(x^{m})^{n}=x^{mn}$, then (R), the induction hypothesis, the first law and Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §successor give

$$(x^{m})^{n+1}=(x^{m})^{n}x^{m}=x^{mn}x^{m}=x^{mn+m}=x^{m(n+1)}.$$

Product. For $n=0$, $(xy)^{0}=1=1\cdot1=x^{0}y^{0}$. For $n\ge1$, the identity $(xy)^{n}=x^{n}y^{n}$ is Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §termwise for the constant maps $k\mapsto x$ and $k\mapsto y$ on $[n]$. Next, $x^{1}=x$ by (R). For $1^{n}=1$: $1^{0}=1$, and $1^{n+1}=1^{n}\cdot1=1^{n}$ by (R), so $1^{n}=1$ for all $n\ge1$ by induction from $1^{1}=1$. Finally, if $n\ge1$, write $n=p+1$ with $p\in\mathbb{N}_{0}$ ($p=0$ if $n=1$, and otherwise by Arithmetic and Order of the Natural Numbers §predecessor). Then $0^{n}=0^{p}\cdot0=0$ by (R) and (Z).

Factorisation. Let $n\ge1$. For $k\in[n+1]$ put $d_{k}=x^{k-1}y^{n+1-k}$. For $k\in[n]$, (R), commutativity and (N) give

$$(x-y)\,x^{k-1}y^{n-k}=x^{k}y^{n-k}-x^{k-1}y^{n-k+1}=d_{k+1}-d_{k}.$$

Hence Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §distributive and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §telescoping give

$$(x-y)\sum_{k=1}^{n}x^{k-1}y^{n-k}=\sum_{k=1}^{n}(d_{k+1}-d_{k})=d_{n+1}-d_{1}=x^{n}y^{0}-x^{0}y^{n}=x^{n}-y^{n}.$$

Reciprocal. A field is a commutative ring, so the clauses above apply to $F$. If $x\neq0$, then the product clause gives $x^{n}(x^{-1})^{n}=(x\,x^{-1})^{n}=1^{n}=1$. As $0\cdot z=z\cdot0=0$ by (Z) and $0\neq1$ by Fields §field for every $z\in F$, this forces $x^{n}\neq0$, and then $(x^{n})^{-1}=(x^{-1})^{n}$ by the uniqueness in Additive and Multiplicative Inverses Are Unique §reciprocal.

Geometric. Let $x\neq1$ and $n\ge1$. The factorisation clause with $1$ in place of $y$, together with $1^{n}=1$ and $1^{n-k}=1$ from the product clause, gives $x^{n}-1=(x-1)S$ for $S=\sum_{k=1}^{n}x^{k-1}$. By (D) and (N), $1-x^{n}=-\big((x-1)S\big)=\big(-(x-1)\big)S=(1-x)S$. Moreover $1-x\neq0$, since otherwise $1=(1-x)+x=x$. Hence

$$S=S(1-x)(1-x)^{-1}=(1-x^{n})(1-x)^{-1}=\frac{1-x^{n}}{1-x}.$$

From now on $F$ is an ordered field, so the clauses above and the rules of Rules of Arithmetic and Order in an Ordered Field apply.

Sign. Let $x\ge0$. Then $x^{0}=1>0$ by Rules of Arithmetic and Order in an Ordered Field §squares, and $x^{1}=x\ge0$. If $x^{n}\ge0$, then $x^{n+1}=x^{n}x\ge0$ by (R) and Ordered Rings §ordered-ring, so $x^{n}\ge0$ for all $n$ by induction. If $x>0$, then also $x^{n}\neq0$ by the reciprocal clause, so $x^{n}>0$. For the absolute value, $|1|=1$ by Absolute Value in an Ordered Field §absolute-value, which settles $n=0$. For $n\ge1$, the map $u\mapsto|u|$ satisfies $|uv|=|u|\,|v|$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value, so Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §homomorphism with $\ast=\diamond=\cdot$ gives $|x^{n}|=\prod_{k=1}^{n}|x|=|x|^{n}$.

Monotone. Let $0\le x<y$, so $y>0$. We induct on $n$; the case $n=1$ is the hypothesis. Suppose $x^{n}<y^{n}$. Since $x^{n}\ge0$ by the sign clause and $y-x\ge0$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, Ordered Rings §ordered-ring gives $0\le x^{n}(y-x)=x^{n}y-x^{n}x$, so $x^{n}x\le x^{n}y$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Also $x^{n}y<y^{n}y$ by Rules of Arithmetic and Order in an Ordered Field §order-product. Hence $x^{n+1}=x^{n}x<y^{n}y=y^{n+1}$ by (R).

Monotone-iff. Let $n\ge1$ and $x,y\ge0$. If $x\le y$, then $x<y$ or $x=y$, and $x^{n}\le y^{n}$ by the monotone clause in the first case and trivially in the second. Conversely, suppose $x^{n}\le y^{n}$ but not $x\le y$. Then $y<x$ because the order is total, so $y^{n}<x^{n}$ by the monotone clause, contradicting $x^{n}\le y^{n}$. If $x^{n}=y^{n}$, then $x^{n}\le y^{n}$ and $y^{n}\le x^{n}$, so $x\le y$ and $y\le x$, hence $x=y$; the converse is trivial.

Base below one. Let $0\le x\le1$. For every $n\in\mathbb{N}_{0}$, $x^{n}\ge0$ by the sign clause and $1-x\ge0$, so $x^{n}-x^{n+1}=x^{n}(1-x)\ge0$ by (R) and Ordered Rings §ordered-ring; that is, $x^{n+1}\le x^{n}$, and $x^{n+1}\ge0$ by the sign clause. It remains to show $x^{n}\le1$. This holds for $n=0$, and for $n\ge1$ by induction: $x^{1}=x\le1$, and $x^{n+1}\le x^{n}\le1$.

Base above one. Let $x\ge1$, so $x>0$ by Rules of Arithmetic and Order in an Ordered Field §squares. For every $n\in\mathbb{N}_{0}$, $x^{n}>0$ by the sign clause and $x-1\ge0$, so $x^{n+1}-x^{n}=x^{n}(x-1)\ge0$, that is, $x^{n}\le x^{n+1}$. It remains to show $1\le x^{n}$. This holds for $n=0$, and for $n\ge1$ by induction: $x^{1}=x\ge1$, and $1\le x^{n}\le x^{n+1}$.

Bernoulli. Let $x\ge-1$, so $1+x\ge0$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Integers read in $F$ via $\kappa_{F}$ (Ordinary Mathematical Language for Analysis: Sets, Maps, Numbers and Ordered Fields §fields) satisfy $0\le n$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §order, and $n+1$ read in $F$ is $n+1$ computed in $F$ (Ordinary Mathematical Language for Analysis: Sets, Maps, Numbers and Ordered Fields §numbers and The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique). For $n=0$, $(1+x)^{0}=1=1+0\cdot x$, using The Rational Numbers Embed in Exactly One Way into Every Ordered Field §zero, and $0\cdot x=x\cdot0=0$ by commutativity and (Z). For $n\ge1$ we induct: $(1+x)^{1}=1+x=1+1\cdot x$. Suppose $(1+x)^{n}\ge1+n\,x$. Then $\big((1+x)^{n}-(1+n\,x)\big)(1+x)\ge0$ by Rules of Arithmetic and Order in an Ordered Field §order-sum and Ordered Rings §ordered-ring, so by (R) and Rules of Arithmetic and Order in an Ordered Field §order-sum,

$$(1+x)^{n+1}=(1+x)^{n}(1+x)\ge(1+n\,x)(1+x)=1+(n+1)\,x+n\,(x\,x)\ge1+(n+1)\,x,$$

where the last step uses $n\,(x\,x)\ge0$, which holds by Rules of Arithmetic and Order in an Ordered Field §squares and Ordered Rings §ordered-ring.