Roots, Rational Powers, and Classical Inequalities

Throughout this page $\mathbb{R}$ is the complete ordered field (Complete ordered field (least-upper-bound property)), in particular an ordered field (Ordered field) and a field (Field), and $\mathbb{N}$ is the set of natural numbers (The natural numbers $\mathbb{N}$ (von Neumann)) with successor $\sigma (n)=n+1$ (Addition of natural numbers).

Let $a:\mathbb{N}\to \mathbb{R}$ be a sequence of reals, written $a_k$ for $a(k)$. Finite sums and finite products of $a$ are defined by recursion on the upper index, which is legitimate because of the recursion theorem (The recursion theorem). That theorem produces a function of one variable, so the running index has to be carried along inside the value: applying it to the set $A=\mathbb{N}\times \mathbb{R}$, the starting element $(0,0)$ and the function $f(n,s)=(\sigma (n),s+a_n)$ gives a unique $g:\mathbb{N}\to \mathbb{N}\times \mathbb{R}$ with

$g(0)=(0,0),g(\sigma (n))=f(g(n))(n\in \mathbb{N})\mathrm{.}$

Write $g(n)=({\pi }_1(g(n)),{\mathrm{\Sigma }}_n)$ for its two coordinates.

The first coordinate is the index itself, and that is a small induction, not an observation (The principle of mathematical induction). Indeed ${\pi }_1(g(0))=0$; and if ${\pi }_1(g(n))=n$, then $g(\sigma (n))=f({\pi }_1(g(n)),{\mathrm{\Sigma }}_n)=(\sigma ({\pi }_1(g(n))),{\mathrm{\Sigma }}_n+a_{{\pi }_1(g(n))})=(\sigma (n),{\mathrm{\Sigma }}_n+a_n)$, so ${\pi }_1(g(\sigma (n)))=\sigma (n)$. By induction ${\pi }_1(g(n))=n$ for every $n\in \mathbb{N}$. Only now may the second coordinate of the two displayed clauses be read off, and doing so gives

${\mathrm{\Sigma }}_0=0,{\mathrm{\Sigma }}_{\sigma (n)}={\mathrm{\Sigma }}_n+a_n\mathrm{.}$

$\mathrm{\Sigma }$ is moreover the unique function $\mathbb{N}\to \mathbb{R}$ with those two properties: if ${\mathrm{\Sigma }}^{\mathrm{\prime }}$ also has them then $n\mapsto (n,{\mathrm{\Sigma }}_n^{\mathrm{\prime }})$ satisfies the two clauses defining $g$, hence equals $g$ by the uniqueness clause of The recursion theorem, so ${\mathrm{\Sigma }}^{\mathrm{\prime }}=\mathrm{\Sigma }$.

We write ${\sum }_{k<n}{a}_k:={\mathrm{\Sigma }}_n$. The same construction with starting element $(0,1)$ and $f(n,p)=(\sigma (n),p\cdot a_n)$, with the same induction on the first coordinate and the same uniqueness argument, gives the unique $\mathrm{\Pi }:\mathbb{N}\to \mathbb{R}$ with

${\mathrm{\Pi }}_0=1,{\mathrm{\Pi }}_{\sigma (n)}={\mathrm{\Pi }}_n\cdot a_n,$

and we write ${\prod }_{k<n}{a}_k:={\mathrm{\Pi }}_n$.

Notation. For $m,n\in \mathbb{N}$ we abbreviate

${\sum }_{k=0}^n{a}_k:={\sum }_{k<n+1}{a}_k,{\prod }_{k=0}^n{a}_k:={\prod }_{k<n+1}{a}_k,$

and, for a general lower index $m$ with $m\le n+1$, writing $d=n+1-m$ for the number of terms,

${\sum }_{k=m}^n{a}_k:={\sum }_{j<d}{a}_{m+j},{\prod }_{k=m}^n{a}_k:={\prod }_{j<d}{a}_{m+j}\mathrm{.}$

When $m=n+1$ we have $d=0$ and the sum is empty, with value $0$, while the empty product has value $1$. In the same spirit ${\sum }_{k=0}^{-1}{a}_k$ is notation for the empty sum ${\mathrm{\Sigma }}_0=0$ and ${\prod }_{k=0}^{-1}{a}_k$ for the empty product ${\mathrm{\Pi }}_0=1$; the index $-1$ never occurs as an element of $\mathbb{N}$ and is only a way of writing "no terms".

Only finitely many values of $a$ enter ${\sum }_{k<n}{a}_k$, so the notation ${\sum }_{k<n}{a}_k$ and ${\prod }_{k<n}{a}_k$ is also used for a list $a_0,\dots ,a_{n-1}$ of reals given without reference to any extension of the list to all of $\mathbb{N}$: extend the list by $a_k=0$ (respectively $a_k=1$) for $k\ge n$ and apply the definition above.

• Why recursion and not "$a_0+a_1+\cdots +a_{n-1}$". The dots are not a definition: they presuppose that the displayed pattern determines a value for every $n$, which is exactly what the recursion theorem (The recursion theorem) supplies, and its uniqueness clause is what makes ${\sum }_{k<n}{a}_k$ a single well-determined real rather than a family of choices. Associativity and commutativity of addition are not used in the definition; they are used in the laws proved from it (Laws of finite sums and finite products).
• Naturals and rationals inside $\mathbb{R}$ (a convention used on the whole page). A natural number $n$ and a rational number $r$ are not literally elements of $\mathbb{R}$: they enter $\mathbb{R}$ through the canonical embedding $\iota :\mathbb{Q}\to \mathbb{R}$, which is an injective, order-preserving field homomorphism (The unique embedding of ℚ into an ordered field), restricting on positive naturals to $n\mapsto n\cdot 1_{\mathbb{R}}=1_{\mathbb{R}}+\cdots +1_{\mathbb{R}}$ (Canonical naturals are positive and strictly increasing). Following ordinary practice, and only where no confusion is possible, we write $n$ for $\iota (n)$ and $r$ for $\iota (r)$; so, for instance, $\frac{1}{n}{\sum }_{k<n}{a}_k$ means $\iota (n{)}^{-1}\cdot {\sum }_{k<n}{a}_k$, which makes sense because $\iota (n)>0$ for $n\ge 1$. Exponents are the one place where the identification is deliberately NOT made: in $a^n$ and $a^r$ the exponent stays a natural, an integer or a rational (Integer powers $a^m$, Rational powers $a^r$ of a positive base), never a real.
• The two indexings are related by ${\sum }_{k=0}^n{a}_k={\sum }_{k<n+1}{a}_k$, so a statement proved for one is available for the other. Sums over $k<n$ are the primitive form here because ${\mathrm{\Sigma }}_0$, the empty sum, is then the base case of every induction, and no index outside $\mathbb{N}$ is ever needed.

Let $a,b,c:\mathbb{N}\to \mathbb{R}$ be sequences of reals, let $\lambda \in \mathbb{R}$, and let $m,n\in \mathbb{N}$, with finite sums and finite products as in Finite sums and finite products, by recursion. Then:

1. Additivity. ${\displaystyle \sum _{k<n}(a_k+b_k)=\sum _{k<n}{a}_k+\sum _{k<n}{b}_k}$.
2. Scaling. ${\displaystyle \sum _{k<n}\lambda a_k=\lambda \sum _{k<n}{a}_k}$; in particular ${\displaystyle \sum _{k<n}\lambda =n\lambda }$, where $n$ denotes the canonical natural $\iota (n)\in \mathbb{R}$ (The unique embedding of ℚ into an ordered field, Canonical naturals are positive and strictly increasing).
3. Splitting. If $m\le n$ then ${\displaystyle \sum _{k<n}{a}_k=\sum _{k<m}{a}_k+\sum _{k=m}^{n-1}{a}_k}$, and ${\displaystyle \prod _{k<n}{a}_k=\left(\prod _{k<m}{a}_k\right)\left(\prod _{k=m}^{n-1}{a}_k\right)}$.
4. Monotonicity. If $a_k\le b_k$ for all $k<n$ then ${\displaystyle \sum _{k<n}{a}_k\le \sum _{k<n}{b}_k}$. In particular, if $a_k\ge 0$ for all $k<n$ then ${\sum }_{k<n}{a}_k\ge 0$, every single term satisfies $a_j\le {\sum }_{k<n}{a}_k$ for $j<n$, and ${\sum }_{k<n}{a}_k=0$ forces $a_k=0$ for every $k<n$.
5. Telescoping. ${\displaystyle \sum _{k<n}(c_{k+1}-c_k)=c_n-c_0}$.
6. Products. ${\displaystyle \prod _{k<n}(a_k{b}_k)=\left(\prod _{k<n}{a}_k\right)\left(\prod _{k<n}{b}_k\right)}$; if $a_k\ge 0$ for all $k<n$ then ${\prod }_{k<n}{a}_k\ge 0$, and if $a_k>0$ for all $k<n$ then ${\prod }_{k<n}{a}_k>0$.

Let $a\in \mathbb{R}$, where $\mathbb{R}$ is the ambient ordered field (Ordered field, Field).

Natural exponents. By the recursion theorem (The recursion theorem) applied to the set $\mathbb{R}$, the starting element $1$ and the function $f(x)=x\cdot a$, there is a unique function $\mathbb{N}\to \mathbb{R}$, written $n\mapsto a^n$, with

$a^0=1,a^{n+1}=a^n\cdot a(n\in \mathbb{N})\mathrm{.}$

Thus $a^1=a$, $a^2=a\cdot a$, and so on. Note that this is defined for every $a$, including $a=0$.

Negative exponents. If $a\mathrm{\ne }0$ and $n\in \mathbb{N}$ with $n\ge 1$, set

$a^{-n}:=(a^n{)}^{-1}\mathrm{.}$

Why that is legitimate. The right-hand side presupposes that $a^n$ is invertible, that is, that $a^n\mathrm{\ne }0$. This is a proof obligation and not an observation, and it is discharged by claim 2 of Laws of integer exponents ↗: for $a\mathrm{\ne }0$ in a field, $a^n\mathrm{\ne }0$ for every $n\in \mathbb{N}$, proved there by induction on $n$ from the fact that a field has no zero divisors (A field has no zero divisors: $ab=0\Rightarrow a=0$ or $b=0$). That lemma is a statement about the operation introduced here, so it depends on this definition and is recorded in this item's justified_by rather than in its deps (SCHEMA §3). Given $a^n\mathrm{\ne }0$, the value $(a^n{)}^{-1}$ is a single well-determined element, because multiplicative inverses in a field are unique (Identities and inverses in a field are unique).

Integer exponents. Every integer $m$ (The integers as equivalence classes of pairs of naturals) is either $\iota (n)$ or $-\iota (n)$ for a unique natural $n$, where $\iota$ is the embedding $\mathbb{N}\to \mathbb{Z}$ (The naturals embed in the integers, Arithmetic on the integers). This too is a citation and not a slogan: the order on $\mathbb{Z}$ is total (The integers form a totally ordered ring), so $m\ge 0$ or $m<0$; the image of $\iota$ is exactly the set of nonnegative integers, and each of them is $\iota (n)$ for a unique natural $n$ (The naturals embed in the integers); and if $m<0$ then $-m>0$, by compatibility of the order with addition (The integers form a totally ordered ring), so $-m=\iota (n)$ and $m=-\iota (n)$, with $n$ unique because $\iota$ is injective. The two clauses above therefore define $a^m$ for every $m\in \mathbb{Z}$ whenever $a\mathrm{\ne }0$, and for every $m\in \mathbb{N}$ for arbitrary $a$. The clauses are consistent where they overlap: the only overlap is $m=0$, where $-\iota (0)=\iota (0)$ and $(a^0{)}^{-1}=1^{-1}=1=a^0$.

• The convention $0^0=1$ is adopted here, and it is not a matter of taste but of agreement with the recursion above: $a^0=1$ is the starting value for every $a$, exactly as the empty product is $1$ (Finite sums and finite products, by recursion). This is the convention that makes the empty product, the binomial theorem, and polynomial notation ${\sum }_k{c}_k{x}^k$ work at $x=0$ without an exception. The competing convention "$0^0$ undefined" belongs to contexts where $x^y$ is studied as a function of two real variables and one wants continuity, which is unavailable and irrelevant here: the exponent in $a^m$ is an integer, never a real.
• $0^n=0$ for every $n\ge 1$, since $0^n=0^{n-1}\cdot 0=0$, a product with a zero factor (Multiplication by zero: $0\cdot a=0$); and $0^{-n}$ is not defined for $n\ge 1$, since $0$ has no inverse.
• The exponent is an integer and stays an integer. Rational exponents are a separate construction resting on the existence of roots (Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$, Rational powers $a^r$ of a positive base), and real exponents are not available at this point in the reading order; they are introduced later in Real powers for positive bases, with the zero-base positive-exponent convention ↗ (see Why real exponents are deferred on the rational-powers page).
• The laws $a^{m+n}=a^m{a}^n$, $(a^m{)}^n=a^{mn}$ and $(ab{)}^n=a^n{b}^n$ are proved, not assumed, in Laws of integer exponents ↗; the order behaviour of $a\mapsto a^n$ is Monotonicity of $x\mapsto x^n$ and of $n\mapsto a^n$.

Let $a,b$ be elements of a field (Field) and let integer powers be as in Integer powers $a^m$.

1. For all $m,n\in \mathbb{N}$: $a^{m+n}=a^m{a}^n$, $(a^m{)}^n=a^{mn}$ and $(ab{)}^n=a^n{b}^n$.
2. If $a\mathrm{\ne }0$ then $a^n\mathrm{\ne }0$ for every $n\in \mathbb{N}$, and $a^{-m}=(a^m{)}^{-1}$ for every $m\in \mathbb{Z}$ (Arithmetic on the integers).
3. If $a\mathrm{\ne }0$ and $b\mathrm{\ne }0$ then all three identities of claim 1 hold for all $m,n\in \mathbb{Z}$.

Let $\mathbb{R}$ be an ordered field (Ordered field), let $a,b\in \mathbb{R}$ and let $n\in \mathbb{N}$, with powers as in Integer powers $a^m$.

1. Signs. If $a\ge 0$ then $a^n\ge 0$, and if $a>0$ then $a^n>0$.
2. Strict monotonicity in the base. If $0\le a<b$ and $n\ge 1$ then $a^n<b^n$. Consequently, for $a,b\ge 0$ and $n\ge 1$: $a<b⟺a^n<b^n$ and $a\le b⟺a^n\le b^n$, and $x\mapsto x^n$ is injective on $\{x\in \mathbb{R}:x\ge 0\}$.
3. Comparison with the base. If $a\ge 1$ and $n\ge 1$ then $a^n\ge a\ge 1$; if $a>1$ and $n\ge 1$ then $a^n\ge a>1$; and if $0\le a\le 1$ then $a^n\le 1$.
4. Monotonicity in the exponent. If $a>1$ then $a^m<a^n$ whenever $m<n$ in $\mathbb{N}$; if $a=1$ then $a^n=1$ for all $n$; and if $0<a<1$ then $a^m>a^n$ whenever $m<n$.