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

Each result cited below is universally quantified over the data in its own statement and is applied to the data indicated where it is cited.

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 induction from $1$ on $\mathbb{N}$, The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §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 $n\ge1$. All differences of exponents below are differences in $\mathbb{N}_{0}$, and each identity between them is checked from that definition, which characterises $q-p$ for $p\le q$ as the unique $j\in\mathbb{N}_{0}$ with $p+j=q$, using the laws of Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §associative, Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §commutative and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero. For $k\in[n+1]$ we have $1\le k\le n+1$ by Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §segment, $1$ being the least natural number, so $k-1$ and $(n+1)-k$ are defined; put $d_{k}=x^{k-1}y^{(n+1)-k}$. Let $k\in[n]$, so $1\le k\le n$, and $k-1$ and $n-k$ are defined. Then $(k-1)+1=k$, since $1+\big((k-1)+1\big)=\big(1+(k-1)\big)+1=k+1$ and the difference is unique, and in the same way $(n-k)+1=(n+1)-k$, $(k+1)-1=k$ and $(n+1)-(k+1)=n-k$, the last because $(k+1)+(n-k)=\big(k+(n-k)\big)+1=n+1$. Hence $d_{k+1}=x^{k}y^{n-k}$, and by (R), $x^{k-1}x=x^{k}$ and $y^{n-k}y=y^{(n+1)-k}$. So (R), commutativity and (N) give

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

Finally $(n+1)-1=n$, $(n+1)-(n+1)=0$ and $1-1=0$, since $1+n=n+1$ and $p+0=p$, so $d_{n+1}=x^{n}y^{0}$ and $d_{1}=x^{0}y^{n}$. 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) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, applied to $x^{n}x\le x^{n}y$ and $x^{n}y<y^{n}y$.

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 $1\le x^{n}\le x^{n+1}$.

Exponent monotone. Let $m\in\mathbb{N}_{0}$ with $m\le n$. By Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §difference there is $d\in\mathbb{N}_{0}$ with $n=m+d$, so $x^{n}=x^{m}x^{d}$ by the exponents clause. If $m<n$, then $d\neq0$, since $m+0=m$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero; so $d\in\mathbb{N}$ by The Natural Numbers with Zero and Their Embedding into the Integers §naturals, and $d\ge1$ by Arithmetic and Order of the Natural Numbers §least.

Let $x\ge1$. By the base-above-one clause, $1\le x^{m}$ and $1\le x^{d}$; as $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares, $x^{m}>0$. Hence Rules of Arithmetic and Order in an Ordered Field §order-product with $z=x^{m}$ gives $x^{m}=1\cdot x^{m}\le x^{d}x^{m}=x^{n}$, using commutativity. If moreover $x>1$ and $m<n$, then $d\ge1$ and $0\le1<x$, so the monotone clause, with $1$ and $x$ in place of $x$ and $y$, and the product clause give $1=1^{d}<x^{d}$; the same rule then gives $x^{m}=1\cdot x^{m}<x^{d}x^{m}=x^{n}$.

Let $0\le x\le1$. Then $x^{m}\ge0$ by the sign clause, and $x^{d}\le1$ by the base-below-one clause, so $1-x^{d}\ge0$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. By (N) and commutativity, $x^{m}-x^{n}=1\cdot x^{m}-x^{d}x^{m}=(1-x^{d})\,x^{m}$, which is $\ge0$ by Ordered Rings §ordered-ring; that is, $x^{n}\le x^{m}$ by Rules of Arithmetic and Order in an Ordered Field §order-sum.

Bernoulli. Let $x\ge-1$, so $1+x\ge0$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. By Commutative Rings, Fields and Ordered Fields: Standard Notation §numerals, the factor $n$ stands for its image $n_{F}$ in $F$. By the same clause, $0_{F}$ and $1_{F}$ are the zero and the unit of $F$ and the image of a sum is the sum of the images, so $(n+1)_{F}=n_{F}+1$ for every $n\in\mathbb{N}_{0}$; and $0\le n$ in $F$, that is $0\le n_{F}$, by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §naturals. For $n=0$, $(1+x)^{0}=1=1+0\cdot x=1+0_{F}\,x$, as $0\cdot x=x\cdot0=0$ by commutativity and (Z). For $n\ge1$ we induct: $(1+x)^{1}=1+x=1+1\cdot x=1+1_{F}\,x$. 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 middle equality expands the product by the ring laws, using $n_{F}\,x+x=(n_{F}+1)\,x=(n+1)_{F}\,x$, and the last step uses $n\,(x\,x)\ge0$, which holds by $0\le n_{F}$, Rules of Arithmetic and Order in an Ordered Field §squares and Ordered Rings §ordered-ring, together with Rules of Arithmetic and Order in an Ordered Field §order-sum.