Proof of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities

Throughout, the properties of $\exp$ are those of Basic Properties of the Exponential Function: claim 1 there gives $\exp(0)=1$ and $\exp(u+v)=\exp(u)\exp(v)$, claim 2 gives $\exp(u)>0$ and $\exp(-u)=1/\exp(u)$, claim 4 gives that $\exp$ is strictly increasing (hence also nondecreasing) and that $\exp(u)\ge1+u$ for $u\ge0$. The properties of $\log$ are those recorded in The Natural Logarithm: $\log(\exp(u))=u$ for real $u$, $\exp(\log t)=t$ for $t>0$, and $\log(st)=\log s+\log t$ for $s,t>0$. Uniqueness of the nonnegative square root is Existence and Uniqueness of the Nonnegative Square Root, and the monotonicity of squaring on nonnegative reals is Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field (claim 1 the strict form, claim 2 the weak form). Limits of sequences are as in Limit of a Sequence of Real Numbers; we use Arithmetic of Limits of Real Sequences (claim 1 sums, claim 3 scalar multiples) and Order Properties of Limits of Real Sequences (claim 1 comparison, claim 2 squeeze, claim 4 absolute values), and the fact, immediate from the definition of the limit, that a constant sequence $(c)_{N\in\mathbb{N}}$ converges to $c$ (in particular the zero sequence converges to $0$). The absolute value satisfies the rules of Properties of the Absolute Value in an Ordered Field (claim 4 multiplicativity, claim 9 strict two-sided bound). The canonical map $\iota$ has the properties of Properties of the Canonical Map from the Natural Numbers to an Ordered Field (claim 1 base and step, claim 2 lower bound $1$, claim 6 strict monotonicity). The order facts used (products of positive reals are positive, multiplication by a positive real preserves $\le$ and $<$, passing to inverses reverses the order between positive reals, and so on) are those of the ordered field $\mathbb{R}$. The claims are proved in the order 1, 2, 4, 5, 6, 3: the proof of claim 3 uses claims 1, 2, 4, 5 and 6, claim 2 uses claim 1, and claims 4, 5 and 6 use no other claim.

Proof of claim 1. Since $\log t$ is a real number, $t^{a}=\exp(a\log t)>0$. Next $t^{0}=\exp(0)=1$ and $t^{1}=\exp(\log t)=t$. By additivity of $\exp$, $t^{a+b}=\exp(a\log t+b\log t)=\exp(a\log t)\exp(b\log t)=t^{a}t^{b}$, and $t^{-a}=\exp(-(a\log t))=1/\exp(a\log t)=1/t^{a}$. By $\log(st)=\log s+\log t$ and additivity, $(st)^{a}=\exp(a\log s+a\log t)=s^{a}t^{a}$. Since $\log(\exp(u))=u$, $\log(t^{a})=\log(\exp(a\log t))=a\log t$, hence $(t^{a})^{b}=\exp(b\log(t^{a}))=\exp(ba\log t)=t^{ab}$. For the natural number powers we argue by induction on $n$ (Natural Numbers): the natural power $t^{1}$ is $t$ (claim 1 of Properties of Natural Number Powers in a Field), which equals the real power $t^{1}$; and if the two powers agree for $n$, then the natural power $t^{n+1}$ is, by the recursion of that claim, the product of the natural power $t^{n}$ with $t$, which equals $t^{n}\cdot t^{1}=t^{n+1}$ in the sense of real powers by additivity. Consequently $(t^{a})^{n}=t^{na}$ in either sense, by the formula $(t^{a})^{b}=t^{ab}$ with $b=n$. Finally $t^{a/2}>0$ and $(t^{a/2})^{2}=t^{a/2}t^{a/2}=t^{a/2+a/2}=t^{a}$, so $t^{a/2}$ is a nonnegative real number whose square is $t^{a}$; by uniqueness of the nonnegative square root, $t^{a/2}=\sqrt{t^{a}}$. With $a=1$, $a=\tfrac12$, $a=\tfrac14$, and so on, this gives $t^{1/2}=\sqrt{t}$, $t^{1/4}=\sqrt{t^{1/2}}=\sqrt{\sqrt{t}}$, $t^{1/8}=\sqrt{t^{1/4}}$; the formulas for $t^{-1/2}$ and $t^{-1/4}$ follow from $t^{-a}=1/t^{a}$.

Proof of claim 2. Let $0<s<t$ and suppose $\log s\ge\log t$. Since $\exp$ is nondecreasing, $s=\exp(\log s)\ge\exp(\log t)=t$, contradicting $s<t$; hence $\log s<\log t$, and $\log$ is strictly increasing. Also $\log1=\log(\exp(0))=0$. If $t\ge1$ then $\log t\ge\log1=0$; if $0<t<1$ then $\log t<\log 1=0$; this proves the equivalence. Now let $a>0$. If $0<s\le t$ then $\log s\le\log t$, so $a\log s\le a\log t$ and $s^{a}=\exp(a\log s)\le\exp(a\log t)=t^{a}$, since $\exp$ is nondecreasing; if $s<t$ all three inequalities are strict, $\exp$ being strictly increasing. If $t\ge1$ and $b\le c$, then $\log t\ge0$ gives $b\log t\le c\log t$ and hence $t^{b}\le t^{c}$; with $b=0$ this gives $1=t^{0}\le t^{c}$ for $c\ge0$. If $0<t\le1$ and $b\le c$, then $\log t\le0$ gives $b\log t\ge c\log t$ and $t^{b}\ge t^{c}$. For the equivalences let $t>0$ and $x>0$. Since $\exp$ is strictly increasing, $\exp(u)\ge\exp(v)$ holds if and only if $u\ge v$, and $\exp(u)>\exp(v)$ if and only if $u>v$. Hence $t^{a}\ge x=\exp(\log x)$ if and only if $a\log t\ge\log x$, if and only if $\log t\ge\log x/a$ (as $a>0$), if and only if $t=\exp(\log t)\ge\exp(\log x/a)=x^{1/a}$; the strict version is identical with $>$ in place of $\ge$. Likewise $t^{-a}<x$ if and only if $-a\log t<\log x$, if and only if $\log t>-\log x/a$, if and only if $t>\exp(-\log x/a)=x^{-1/a}$.

Proof of claim 4. By Existence and Uniqueness of the Integer Part of a Real Number, $\lfloor x\rfloor\le x<\lfloor x\rfloor+1$; hence $n=\lfloor x\rfloor+1>x$ and $n\le x+1$. Let $x\ge0$. Then $\lfloor x\rfloor>x-1\ge-1$. By The Integers as a Subset of the Real Numbers, $\lfloor x\rfloor$ is $0$, or $\iota(k)$, or $-\iota(k)$ for some natural number $k$, where $\iota$ is the canonical map; the last case is excluded because $\iota(k)\ge1$ (claim 2 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field) gives $-\iota(k)\le-1$. If $\lfloor x\rfloor=0$ then $n=1=\iota(1)$; if $\lfloor x\rfloor=\iota(k)$ then $n=\iota(k)+1=\iota(k+1)$, by claim 1 of that lemma. In both cases $n$ is the image of a natural number. For the consequence, given a real $x$, put $N_{0}=1$ if $x<0$ and $N_{0}=\lfloor x\rfloor+1$ if $x\ge0$; in both cases $N_{0}\in\mathbb{N}$ and $N_{0}>x$ (for $x<0$ because $1>0>x$). If $N\in\mathbb{N}$ satisfies $N\ge N_{0}$ in $\mathbb{N}$, then $\iota(N)\ge\iota(N_{0})$ by the strict monotonicity of $\iota$ (claim 6 of that lemma), so $N>x$ as real numbers.

Proof of claim 5. Let $a,b\ge0$. The number $\sqrt{a}\sqrt{b}$ is nonnegative with square $(\sqrt{a})^{2}(\sqrt{b})^{2}=ab$, so it is the nonnegative square root of $ab$; likewise $a\ge0$ has square $a^{2}$, so $\sqrt{a^{2}}=a$. If $a\le b$ and $\sqrt{a}>\sqrt{b}$, then the strict form of the monotonicity of squaring gives $a=(\sqrt a)^{2}>(\sqrt b)^{2}=b$, a contradiction; so $\sqrt{a}\le\sqrt{b}$. If $a<b$ and $\sqrt a\ge\sqrt b$, the weak form gives $a\ge b$, a contradiction; so $\sqrt a<\sqrt b$. Next, $(\sqrt a+\sqrt b)^{2}=a+2\sqrt a\sqrt b+b\ge a+b=(\sqrt{a+b})^{2}$, and both $\sqrt a+\sqrt b$ and $\sqrt{a+b}$ are nonnegative, so the weak form gives $\sqrt{a+b}\le\sqrt a+\sqrt b$. For the next inequality we may, by symmetry of both sides in $(a,b)$, assume $a\ge b$; then $|a-b|=a-b\ge0$, $\sqrt a=\sqrt{b+(a-b)}\le\sqrt b+\sqrt{a-b}$ by what was just shown, and $\sqrt a\ge\sqrt b$ by monotonicity, whence $|\sqrt a-\sqrt b|=\sqrt a-\sqrt b\le\sqrt{a-b}=\sqrt{|a-b|}$. Next, $(1+a)^{2}=1+2a+a^{2}\ge a$ and $1+a\ge0$, so $\sqrt a\le\sqrt{(1+a)^{2}}=1+a$ by monotonicity. Finally $2a^{2}+2b^{2}-(a+b)^{2}=a^{2}-2ab+b^{2}=(a-b)^{2}\ge0$.

Proof of claim 6. Throughout this proof we use the convention $x^{0}=1$ of The Real Exponential Function and the convention $0!=1$ (adopted as in Series Formula, Exponential Moments, and Chernoff Tail Bounds for the Poisson Distribution), under which the series of that definition is read; factorials of natural numbers are the finite products of that definition, so that $k!\ge1$ is a natural number for $k\in\mathbb{N}$, and $1!=1=1\cdot0!$, while $(k+1)!=(k+1)\,k!$ for $k\in\mathbb{N}$ by extracting the last factor of the product (claim 3 of Extraction of a Term from a Finite Sum or Product in a Field). All finite sums are those of Finite Sum Notation in a Field, indexed by initial segments of $\mathbb{N}$; we use termwise comparison of finite sums (claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers), the splitting of a sum at an index (Splitting a Finite Sum at an Index), and the independence of a finite sum from the extension of its summands, homogeneity and nonnegativity of finite sums (claims 1, 3 and 5 of Properties of Finite Sums). Fix $x\ge0$ and, for $n\in\mathbb{N}$, put

$$S_{n}=1+\sum_{k=1}^{n}\frac{x^{k}}{k!},$$

which is what the partial sum $\sum_{k=0}^{n}x^{k}/k!$ of the series defining $\exp(x)$ abbreviates; thus $(S_{n})_{n\in\mathbb{N}}$ is the sequence of partial sums of that series from the index $n=1$ onward, and by that definition (the series converging to $\exp(x)$, and omitting the single partial sum with index $0$ not affecting convergence or the limit) the sequence $(S_{n})_{n\in\mathbb{N}}$ converges to $\exp(x)$. Each term $x^{k}/k!$ is nonnegative, $x^{k}\ge0$ by claim 5 of Properties of Natural Number Powers in a Field and $k!>0$. Fix $p\in\mathbb{N}$. For $j\in\mathbb{N}$, splitting the sum defining $S_{p+j}$ at the index $p$ gives $S_{p+j}=S_{p}+\sum_{i=1}^{j}x^{p+i}/(p+i)!\ge S_{p}$, the last sum being nonnegative as a sum of nonnegative summands. The sequence $(S_{p+j})_{j\in\mathbb{N}}$ converges to $\exp(x)$: given $\varepsilon>0$, choose $J$ with $|S_{i}-\exp(x)|<\varepsilon$ for all $i\ge J$; since $p\ge1$, every $j\ge J$ has $p+j\ge J$, hence $|S_{p+j}-\exp(x)|<\varepsilon$. Comparison of this sequence with the constant sequence $S_{p}$ gives

$$S_{p}\le\exp(x)\qquad(p\in\mathbb{N}).$$

Next, for $n\ge2$ we record the reindexing identity

$$\sum_{k=1}^{n}\frac{x^{k-1}}{(k-1)!}=1+\sum_{i=1}^{n-1}\frac{x^{i}}{i!}=S_{n-1},$$

obtained by splitting the left-hand sum at the index $1$: its first term is $x^{0}/0!=1$, and its remaining terms are $x^{(1+i)-1}/((1+i)-1)!=x^{i}/i!$ for $i$ in the initial segment of $n-1$. For $k\in\mathbb{N}$ one has $k!=k\cdot(k-1)!\ge(k-1)!>0$ and $x^{k}=x\cdot x^{k-1}$ (for $k=1$ both by the conventions $0!=1$, $x^{0}=1$ together with $x^{1}=x$; for $k\ge2$ by the factorial recursion and by the power recursion $x^{k}=x^{k-1}x$ of claim 1 of Properties of Natural Number Powers in a Field), hence, passing to inverses of positive reals and multiplying by $x^{k-1}\ge0$ and by $x\ge0$, $x^{k}/k!\le x\cdot x^{k-1}/(k-1)!$. Therefore, for $n\ge2$, by termwise comparison, extraction of the factor $x$, the reindexing identity and the bound $S_{n-1}\le\exp(x)$,

$$S_{n}-1=\sum_{k=1}^{n}\frac{x^{k}}{k!}\le x\sum_{k=1}^{n}\frac{x^{k-1}}{(k-1)!}=xS_{n-1}\le x\exp(x),$$

while for $n=1$, $S_{1}-1=x\le x\exp(x)$ because $\exp(x)\ge1+x\ge1$. Since $(S_{n}-1)$ converges to $\exp(x)-1$, comparison with the constant sequence $x\exp(x)$ gives $\exp(x)-1\le x\exp(x)$. For the second inequality, splitting at the index $1$ gives, for $n\ge2$,

$$S_{n}-1-x=\sum_{i=1}^{n-1}\frac{x^{i+1}}{(i+1)!}\ \ge\ 0,$$

and for $i\in\mathbb{N}$ one has $(i+1)!=(i+1)\,i\,(i-1)!\ge(i-1)!>0$ and $x^{i+1}=x^{2}x^{i-1}$ (for $i=1$ by the conventions $0!=1$, $x^{0}=1$, as $2!=2\cdot1!$ and $x^{2}=x\cdot x^{1}$; for $i\ge2$ by two applications of the factorial recursion and of the power recursion), hence $x^{i+1}/(i+1)!\le x^{2}\,x^{i-1}/(i-1)!$. For $n\ge3$, termwise comparison, extraction of the factor $x^{2}$, the reindexing identity with $n-1\ge2$ in place of $n$, and the bound $S_{n-2}\le\exp(x)$ give

$$S_{n}-1-x\le x^{2}\sum_{i=1}^{n-1}\frac{x^{i-1}}{(i-1)!}=x^{2}S_{n-2}\le x^{2}\exp(x),$$

while $S_{1}-1-x=0$ and $S_{2}-1-x=x^{2}/2$, both in $[0,x^{2}\exp(x)]$ because $\exp(x)\ge1$. Since $(S_{n}-1-x)$ converges to $\exp(x)-1-x$, comparison gives $0\le\exp(x)-1-x\le x^{2}\exp(x)$.

Now let $h$ be real. If $h\ge0$ then $\exp(h)-1\ge h\ge0$, so $|\exp(h)-1|=\exp(h)-1\le h\exp(h)=|h|\exp(|h|)$. If $h<0$ then $\exp(h)<\exp(0)=1$, and, using $\exp(h)\exp(-h)=\exp(0)=1$ and the first inequality with $x=-h>0$,

$$|\exp(h)-1|=1-\exp(h)=\exp(h)\bigl(\exp(-h)-1\bigr)\le\exp(h)\,(-h)\exp(-h)=-h=|h|\le|h|\exp(|h|),$$

because $\exp(|h|)\ge1+|h|\ge1$. Finally, for $x\ge0$, $\exp(x)\ge1+x\ge1>0$ gives $\exp(-x)=1/\exp(x)\le1/(1+x)\le1$; and $\exp$ is nondecreasing because it is strictly increasing, so $u\ge v$ implies $-u\le-v$ and $\exp(-u)\le\exp(-v)$.

Proof of claim 3. (a) Let $a>0$ and $\varepsilon>0$. By claim 4 there is $N_{0}\in\mathbb{N}$ with $N>\varepsilon^{-1/a}$ for every natural number $N\ge N_{0}$; for such $N$, the last equivalence of claim 2 (with $t=N$, $x=\varepsilon$) gives $N^{-a}<\varepsilon$, and $N^{-a}>0$ by claim 1, so $|N^{-a}-0|<\varepsilon$. Hence $N^{-a}\to0$.

(b) Let $a>0$, $c>0$, $k\ge0$, and let $m\in\mathbb{N}$ satisfy $am\ge k$ (such $m$ exists: $m=\lfloor k/a\rfloor+1\in\mathbb{N}$ by claim 4, and $m>k/a$). Fix $N\in\mathbb{N}$ and put $t=cN^{a}>0$. By The Exponential Function Dominates Every Power (with the natural number $m$, whose successor is $m+1$), $t^{m}\exp(-t)\le(m+1)^{m+1}t^{-1}$; multiplying by $t^{-m}>0$ gives $\exp(-t)\le(m+1)^{m+1}t^{-(m+1)}$, where $t^{-(m+1)}=1/t^{m+1}$. By claim 1, $t^{m+1}=(cN^{a})^{m+1}=c^{m+1}(N^{a})^{m+1}=c^{m+1}N^{a(m+1)}$ (natural number powers agreeing with real powers). Hence $\exp(-cN^{a})\le(m+1)^{m+1}c^{-(m+1)}N^{-a(m+1)}$, and multiplying by $N^{k}>0$,

$$0\le N^{k}\exp(-cN^{a})\le(m+1)^{m+1}c^{-(m+1)}N^{k-a(m+1)}\le(m+1)^{m+1}c^{-(m+1)}N^{-a},$$

the last step by claim 2, since $N\ge1$ and $k-a(m+1)=(k-am)-a\le-a$. The right-hand side is a scalar multiple of $N^{-a}$, which converges to $0$ by (a), so it converges to $0$ by the scalar-multiple limit law, and the squeeze (between the zero sequence and this majorant) gives $N^{k}\exp(-cN^{a})\to0$.

(c) Let $\varepsilon>0$. Since $b_{N}\to0$ there is $N_{1}$ with $|b_{N}|<\varepsilon$ for $N\ge N_{1}$. Let $N_{2}$ be the larger of $N_{0}$ and $N_{1}$. For $N\ge N_{2}$ we have $b_{N}\ge|c_{N}-L|\ge0$, so $|c_{N}-L|\le b_{N}=|b_{N}|<\varepsilon$. Hence $c_{N}\to L$.

(d) If $L<x$, apply the definition of the limit with $\varepsilon=x-L>0$: there is $N_{0}$ with $|c_{N}-L|<x-L$ for $N\ge N_{0}$, whence $c_{N}-L<x-L$ (strict two-sided bound), that is, $c_{N}<x$. If $L>x$, use $\varepsilon=L-x$ to get $-(L-x)<c_{N}-L$, that is, $c_{N}>x$.

(e) Comparison of $(a_{N})$ with the zero sequence gives $A\ge0$. By claim 5, $|\sqrt{a_{N}}-\sqrt{A}|\le\sqrt{|a_{N}-A|}$. Let $\varepsilon>0$; there is $N_{0}$ with $|a_{N}-A|<\varepsilon^{2}$ for $N\ge N_{0}$, and then $\sqrt{|a_{N}-A|}<\sqrt{\varepsilon^{2}}=\varepsilon$ by the strict monotonicity and the identity $\sqrt{\varepsilon^{2}}=\varepsilon$ of claim 5; hence $|\sqrt{a_{N}}-\sqrt{A}|<\varepsilon$ for $N\ge N_{0}$, and $\sqrt{a_{N}}\to\sqrt{A}$. Applying this to the nonnegative sequence $(\sqrt{a_{N}})$ gives $\sqrt{\sqrt{a_{N}}}\to\sqrt{\sqrt{A}}$.

(f) By additivity, $\exp(a_{N})-\exp(A)=\exp(A)(\exp(a_{N}-A)-1)$, so by multiplicativity of the absolute value and claim 6, $|\exp(a_{N})-\exp(A)|\le\exp(A)\,|a_{N}-A|\exp(|a_{N}-A|)$. The sequence $a_{N}-A$ converges to $0$ (scalar multiples and sums of limits), hence so does $|a_{N}-A|$ (absolute values of limits), and by (d) there is $N_{0}$ with $|a_{N}-A|<1$ for $N\ge N_{0}$; for such $N$, $\exp(|a_{N}-A|)\le\exp(1)$ as $\exp$ is nondecreasing, so $|\exp(a_{N})-\exp(A)|\le\exp(A)\exp(1)\,|a_{N}-A|$. The right-hand side is a scalar multiple of a null sequence, hence null, and (c) gives $\exp(a_{N})\to\exp(A)$.

(g) First, for every real $u>0$ we show $|\log u|\le|u-1|/\min(u,1)$. If $u\ge1$, then $\exp(u-1)\ge1+(u-1)=u$, so $u-1=\log(\exp(u-1))\ge\log u\ge\log1=0$ by claim 2, and $|\log u|=\log u\le u-1=|u-1|$. If $0<u<1$, then $1/u>1$, so by the first case $0\le\log(1/u)\le1/u-1=(1-u)/u$; since $\log u+\log(1/u)=\log1=0$, we get $|\log u|=\log(1/u)\le|u-1|/u$. Now put $u_{N}=a_{N}/A=(1/A)\,a_{N}>0$ (legitimate as $A>0$); by the scalar-multiple limit law $u_{N}\to(1/A)A=1$, and by (d) there is $N_{0}$ with $u_{N}>\tfrac12$ for $N\ge N_{0}$, so that $|\log u_{N}|\le2|u_{N}-1|$ for $N\ge N_{0}$. As $|u_{N}-1|\to0$, (c) gives $\log u_{N}\to0$. Since $\log a_{N}=\log(Au_{N})=\log A+\log u_{N}$, the sum law gives $\log a_{N}\to\log A$, then $b\log a_{N}\to b\log A$, and (f) gives $a_{N}^{\,b}=\exp(b\log a_{N})\to\exp(b\log A)=A^{b}$. $\qquad\blacksquare$