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.

(S) Each set over which an induction runs exists, as follows. The rings below are $R$ and the fields $F$, which are commutative rings by Fields §field and Ordered Fields §ordered-field. For such a ring and an element $z$ of it, Maps Given by Iterated Operations: Partial Iterates, Images of Natural Numbers, Powers and Operations over a Finite Set §ring-powers gives a map $\pi_{z}$ from $\mathbb{N}_{0}$ to the ring with $\pi_{z}(k)=z^{k}$ for every $k\in\mathbb{N}_{0}$, and Maps Given by Iterated Operations: Partial Iterates, Images of Natural Numbers, Powers and Operations over a Finite Set §images gives a map $\nu$ from $\mathbb{N}_{0}$ to the ring whose value at each $k\in\mathbb{N}_{0}$ is the image of $k$ in the ring. In the claim of an induction, replace each power $z^{k}$ by $\pi_{z}(k)$ and each image of a natural number $k$ by $\nu(k)$; since these are equal, the resulting property of $n$ holds for $n\in\mathbb{N}$ exactly when the claim does. That property is written with values of the maps $\pi_{z}$ and $\nu$ and of the operations of the ring and of $\mathbb{N}_{0}$, and with $=$ and, in an ordered field, its order $\le$ and strict order $<$. Each value is taken at an argument in its domain, since the property is formed for $n\in\mathbb{N}$ and the arguments ($n$, $mn$, $n+1$) then lie in $\mathbb{N}_{0}$. The maps (for instance $\pi_{z}$ for the element $z=x^{m}$ in the exponents clause), operations and relations are fixed before the induction and are its parameters, and no quantifier is applied to a class variable. So the property is predicative by Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §closure, its values being eliminated as in Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §value and Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §operation, and its relation notation as in Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §pair-membership: for instance, for $k\in\mathbb{N}_{0}$ and every set $w$, $w=\pi_{z}(k)$ holds if and only if $(k,w)\in\pi_{z}$, and this formula is predicative by Predicativity under Renaming of Set Variables by Variables, of Formulas with Ordered Pairs, and of Formulas for Relations, Functions, Maps, Injectivity, Surjectivity and Bijectivity §pair-membership. Hence the set of those $n\in\mathbb{N}$ having the property is a set by Sets and Maps: Ordinary Notation §set-builder, and the induction runs over it. Each induction below names this property.

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 $n$, with the property $\pi_{x^{m}}(n)=\pi_{x}(mn)$ of (S): $(x^{m})^{1}=x^{m}=x^{m\cdot1}$ 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 $1^{1}=1$, with the property $\pi_{1}(n)=1$ of (S). 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, with the property $\pi_{x}(n)\ge0$ of (S). 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 $n$, with the property $\pi_{x}(n)<\pi_{y}(n)$ of (S); 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, with the property $\pi_{x}(n)\le1$ of (S): $x^{1}=x\le1$, 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, with the property $1\le\pi_{x}(n)$ of (S): $x^{1}=x\ge1$, 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, with the property $1+\nu(n)\,x\le\pi_{1+x}(n)$ of (S), $\nu$ and $\pi_{1+x}$ being the maps of (S) for $F$: $(1+x)^{1}=1+x=1+1\cdot x=1+1_{F}\,x$. 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.