1.5 Roots and Rational Powers

Number Systems, Chapter 6 found $\sqrt a$ as the supremum of the numbers whose square stays below $a$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 6.1.4). The same construction gives every $n$ -th root, and the factorization of $t^n - s^n$ from Section 1.1 makes it work for every exponent. Roots then give powers with rational exponents, $a^{m/n}$ , once we check that the value does not depend on how the fraction $\frac{m}{n}$ is written. These are the “real powers” of the chapter title as far as this chapter goes: rational exponents of positive reals. Chapter 8 extends them to every real exponent. The section ends with two estimates that make the limits $\sqrt[n]{a} \to 1$ and $\sqrt[n]{n} \to 1$ routine.

1.5.1. n-th roots

For $n = 2$ , Number Systems, Chapter 6 built the square root as the supremum of a set (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 6.1.4). An exercise there did the same for cube roots (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Exercise (B2: Cube roots)). For a general exponent the construction is the same. Only the estimate for $(s + h)^n$ needs a new tool, and Lemma 1.1.8 supplies it. Here $n \cdot 1_{\mathbb{R}}$ is written $n$ . Powers in $\mathbb{R}$ are those of Proposition 1.1.1 with $R = \mathbb{R}$ . They follow the same recursion as Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 4.2.1 (b).

For example, $2$ is the cube root of $8$ : it is non-negative, and $2^3 = 8$ . No other non-negative number has cube $8$ . The lemma says the same for every $a \ge 0$ and every exponent $n \ge 1$ .

Lemma 1.5.1 (Real n-th Roots)

Let $n \in \mathbb{N}$ with $n \ge 1$ .

(a) If $s, t \in \mathbb{R}$ and $0 \le s < t$ , then $s^n < t^n$ .
(b) For every $a \in \mathbb{R}$ with $a \ge 0$ there is exactly one $b \in \mathbb{R}$ with $b \ge 0$ and $b^n = a$ . We write $\sqrt[n]{a} \mathrel{:=} b$ . For $n = 2$ it is the square root $\sqrt a$ of Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 6.1.4.

Idea. Part (a) is the factorization $t^n - s^n = (t - s)(t^{n-1} + t^{n-2}s + \dots + s^{n-1})$ : both factors are positive. For (b), let $s$ be the supremum of the set $S$ of non-negative $y$ with $y^n \le a$ , and rule out $s^n < a$ and $s^n > a$ .

If $s^n < a$ , we want $h > 0$ with $(s + h)^n \le a$ . Working backwards: for $0 < h \le 1$ , the same factorization writes $(s + h)^n - s^n$ as $h$ times a sum of $n$ terms, each at most $(s + 1)^{n-1}$ . So $(s + h)^n - s^n \le h \cdot n(s + 1)^{n-1}$ , and it is enough that this be at most half the room $a - s^n$ . That gives $h \mathrel{:=} \min\{1, (a - s^n)/(2n(s + 1)^{n-1})\}$ .

If $s^n > a$ , we want $h > 0$ with $(s - h)^n \ge a$ still, which makes $s - h$ a smaller upper bound. Now $s^n - (s - h)^n \le h \cdot n s^{n-1}$ , and it is enough that this equal the excess $s^n - a$ . That gives $h \mathrel{:=} (s^n - a)/(n s^{n-1})$ .

Before reading on: where is completeness used, and why must $h \le 1$ in the first case?

Proof. Three facts are used throughout. All three rest on Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c), or equality.

(i) A power of a non-negative real number is non-negative, and a power of a positive one is positive. This is induction on the exponent.
(ii) If $0 \le y \le 1$ , then $y^k \le y$ for every $k \ge 1$ . If $y \ge 1$ , then $y^k \ge y$ . Both are induction, with $y^{k+1} = y^k y$ .
(iii) If $0 \le x \le x'$ and $0 \le y \le y'$ , then $xy \le x'y \le x'y'$ . Each step multiplies by a non-negative number.

(a) By Lemma 1.1.8 in the commutative ring $\mathbb{R}$ , $t^n - s^n = (t - s) \sum_{k=0}^{n-1} t^k s^{n-1-k} .$ Every term of the sum is $\ge 0$ , and the term with $k = n - 1$ is $t^{n-1} > 0$ , since $t > s \ge 0$ . So the sum is $\ge t^{n-1} > 0$ . Also $t - s > 0$ . Hence $t^n - s^n > 0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c)).
(b) Uniqueness. If $b, c \ge 0$ and $b \ne c$ , say $b < c$ , then $b^n < c^n$ by (a); so at most one non-negative $b$ has $b^n = a$ .

Existence. If $a = 0$ , then $b \mathrel{:=} 0$ works, since $0^n = 0^{n-1} \cdot 0 = 0$ . Let $a > 0$ , and let $S \mathrel{:=} \{ y \in \mathbb{R} : y \ge 0 \text{ and } y^n \le a \} .$ Put $m \mathrel{:=} \min\{1, a\} > 0$ . Since $0 < m \le 1$ , $m^n \le m \le a$ , so $m \in S$ . Put $M \mathrel{:=} \max\{1, a\}$ . If $y > M$ , then $y > 1$ , so $y^n \ge y > M \ge a$ and $y \notin S$ ; hence $M$ is an upper bound of $S$ . Since $\mathbb{R}$ is a complete ordered field (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.5.2), $S$ has a supremum; let $s \mathrel{:=} \sup S$ . Then $s \ge m > 0$ .

Case $s^n < a$ . Let $h \mathrel{:=} \min\Bigl\{ 1, \frac{a - s^n}{2n(s + 1)^{n-1}} \Bigr\} .$ It depends only on $a$ , $n$ and $s$ , and $h > 0$ , since $a - s^n > 0$ and $2n(s + 1)^{n-1} > 0$ . By Lemma 1.1.8, $(s + h)^n - s^n = h \sum_{k=0}^{n-1} (s + h)^k s^{n-1-k} .$ Since $0 \le s < s + h \le s + 1$ , each term satisfies $(s + h)^k s^{n-1-k} \le (s + 1)^k (s + 1)^{n-1-k} = (s + 1)^{n-1}$ . This uses (a) or equality, the third fact, and the rule for adding exponents. A sum of $n$ terms, each at most $(s + 1)^{n-1}$ , is at most $n (s + 1)^{n-1}$ . Hence $(s + h)^n \le s^n + h \cdot n(s + 1)^{n-1} \le s^n + \frac{a - s^n}{2} < a .$ So $s + h \in S$ , and $s + h > s$ , against $s$ being an upper bound of $S$ .

Case $s^n > a$ . Let $h \mathrel{:=} \frac{s^n - a}{n s^{n-1}} ,$ which depends only on $a$ , $n$ and $s$ , and is positive. Since $a > 0$ , we have $s^n - a < s^n \le n s^n$ , so $h < s^n/(n s^{n-1}) = s/n \le s$ ; thus $0 < s - h < s$ . By Lemma 1.1.8 and the same bound, with $0 < s - h < s$ , each of the $n$ terms of $\sum_{k=0}^{n-1} s^k (s - h)^{n-1-k}$ is at most $s^{n-1}$ , so $s^n - (s - h)^n \le h \cdot n s^{n-1} = s^n - a ,$ that is, $(s - h)^n \ge a$ . Now let $y \in S$ . If $y > s - h$ , then (a) gives $y^n > (s - h)^n \ge a$ , against $y \in S$ . So $y \le s - h$ , and $s - h$ is an upper bound of $S$ smaller than $s$ , against $s$ being the least upper bound.

Both cases are impossible, so $s^n = a$ by trichotomy, and $b \mathrel{:=} s \ge 0$ works.

For $n = 2$ , the number $\sqrt[2]{a}$ is non-negative and squares to $a$ , so it is $\sqrt a$ , by the uniqueness in Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 6.1.4.

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 5.5.1 (bounded sets have suprema); Proposition 1.1.1 (a) (adding exponents); Proposition 1.1.5 (b), (c) (one term of a sum of non-negative terms; termwise bound).

The two values of $h$ were found by working backwards from what we wanted. Each depends on $a$ , $n$ and $s$ only. Completeness was used once, to produce $s$ . For $a > 0$ the root is positive, since $0^n = 0 \ne a$ . For instance $\sqrt[3]{8} = 2$ and $\sqrt[4]{4} = \sqrt 2$ , since $2^3 = 8$ and $(\sqrt 2)^4 = \bigl((\sqrt 2)^2\bigr)^2 = 4$ , and both are non-negative. In this chapter the symbol $\sqrt[n]{a}$ is used only for non-negative real $a$ . Chapter 8 adds odd roots of negative reals, and Chapter 3 shows why it could not mean “the” $n$ -th root of a complex number.

The last example already used the working principle of this section. To identify a root, we do not compute it: we take a candidate, check that it is non-negative, and check that its $n$ -th power is $a$ . Uniqueness in part (b) does the rest. Each equation between roots follows this way from a rule for powers.

Lemma 1.5.2 (Rules for n-th Roots)

Let $n, m \ge 1$ be integers and $a, b \in \mathbb{R}$ with $a, b \ge 0$ .

(a) $\sqrt[n]{ab} = \sqrt[n]{a} \, \sqrt[n]{b}$ .
(b) $\sqrt[n]{a^m} = \bigl( \sqrt[n]{a} \bigr)^m$ .
(c) $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$ .
(d) If $a < b$ , then $\sqrt[n]{a} < \sqrt[n]{b}$ .

Idea. Each of (a)–(c) says that some non-negative number is a certain root. So we raise the right side to the relevant power, compute with the laws of powers, and let the uniqueness in Lemma 1.5.1 (b) identify it. Number Systems identified inverses the same way (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Lemma 4.1.1): name the defining property, check it, use uniqueness. Part (d) is the converse of part (a) of Lemma 1.5.1, proved by contradiction.

Proof. Write $x \mathrel{:=} \sqrt[n]{a}$ and $y \mathrel{:=} \sqrt[n]{b}$ , so $x, y \ge 0$ , $x^n = a$ and $y^n = b$ . Powers and products of non-negative reals are non-negative (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c), by induction for powers).

(a) The number $xy$ is $\ge 0$ , and $(xy)^n = x^n y^n = ab$ by Proposition 1.1.1 (c). By the uniqueness in Lemma 1.5.1 (b), $xy = \sqrt[n]{ab}$ .
(b) The number $x^m$ is $\ge 0$ , and by Proposition 1.1.1 (b), used twice with $mn = nm$ , $(x^m)^n = x^{mn} = (x^n)^m = a^m .$ By uniqueness, $x^m = \sqrt[n]{a^m}$ .
(c) Let $z \mathrel{:=} \sqrt[m]{x}$ , so $z \ge 0$ and $z^m = x$ . By Proposition 1.1.1 (b), $z^{mn} = (z^m)^n = x^n = a$ . By uniqueness with the exponent $mn \ge 1$ , $z = \sqrt[mn]{a}$ .
(d) Let $a < b$ , and suppose $y \le x$ . If $y = x$ , then $b = y^n = x^n = a$ . If $y < x$ , then $b = y^n < x^n = a$ by Lemma 1.5.1 (a), since $0 \le y$ . Both contradict $a < b$ . So $x < y$ .

Part (d) and Lemma 1.5.1 (a) together say that for non-negative reals, $a < b$ holds if and only if $\sqrt[n]{a} < \sqrt[n]{b}$ : taking an $n$ -th root and raising to the $n$ -th power both preserve strict order. That two-way comparison is how roots are compared without computing them, as in Example (Rational powers by hand) below.

Quick check. Simplify $\sqrt[3]{\sqrt{64}}$ in two ways.

Solution. By Lemma 1.5.2 (c) with $m = 3$ , $n = 2$ , it is $\sqrt[6]{64} = 2$ , since $2 \ge 0$ and $2^6 = 64$ . Directly, $\sqrt{64} = 8$ (as $8 \ge 0$ and $8^2 = 64$ ), and $\sqrt[3]{8} = 2$ .

1.5.2. Integer powers

A rational exponent $\frac{m}{n}$ may have a negative numerator, so before we can define $a^{m/n}$ we need $a^m$ for negative integers $m$ . There is only one sensible choice. If the law $a^{m+k} = a^m a^k$ is to survive, then $a^{-k} a^k = a^0 = 1$ , so $a^{-k}$ has to be the inverse of $a^k$ . For instance, $2^{-3}$ must be the inverse of $2^3 = 8$ , which is $\frac18 = \bigl(\frac12\bigr)^3$ . The definition makes this so, and it works in every field. Chapter 3 meets it again in $\mathbb{C}$ , in an exercise.

Definition 1.5.1 (Integer Powers)

Let $F$ be a field, $a \in F$ with $a \ne 0$ , and $n \in \mathbb{N}$ . Put $a^{-n} \mathrel{:=} \bigl( a^{-1} \bigr)^n ,$ where the right side is the power of Proposition 1.1.1.

Only $n = 0$ could clash with the earlier meaning: $-0 = 0$ , and both readings give $1$ , since $a^0 = 1 = (a^{-1})^0$ . For $n \ge 2$ the symbol $a^{-n}$ is new. At $n = 1$ it already meant the inverse, and the definition agrees with it: $(a^{-1})^1 = a^{-1}$ . The base must be non-zero, since $0$ has no inverse. For instance, in $\mathbb{R}$ , $2^{-3} = \bigl(\frac12\bigr)^3 = \frac18$ .

Proposition 1.5.3 (Laws of Integer Powers)

Let $F$ be a field, let $a, b \in F$ with $a \ne 0$ and $b \ne 0$ , and let $m, n \in \mathbb{Z}$ .

(a) $a^{m+n} = a^m a^n$ .
(b) $(a^m)^n = a^{mn}$ .
(c) $(ab)^n = a^n b^n$ .
(d) $a^n \ne 0$ and $a^{-n} = (a^n)^{-1}$ .
(e) If $F$ is an ordered field and $a > 0$ , then $a^n > 0$ .
(f) If $F$ is an ordered field, $a > 1$ and $m < n$ , then $a^m < a^n$ .

Idea. Each law is known for exponents in $\mathbb{N}$ (Proposition 1.1.1). A negative exponent is a natural one applied to $a^{-1}$ , and $(a^{-1})^k = (a^k)^{-1}$ lets us move the inverse wherever it is needed. So each law splits into sign cases, and each case reduces to $\mathbb{N}$ .

The other cases reduce to natural exponents in the same way. One case shows the pattern: (a) with $m \ge 0 > n$ . Write $n = -k$ with $k \ge 1$ . We use (d) for natural exponents, $(a^{-1})^k = (a^k)^{-1}$ , proved first among the sign cases below. If $m \ge k$ , then $a^m = a^{m-k} a^k$ , so $a^m a^{-k} = a^{m-k} \, a^k (a^k)^{-1} = a^{m-k} = a^{m+n} .$ If $m < k$ , then $a^{-k} = (a^{-1})^{m} (a^{-1})^{k-m}$ and $a^m (a^{-1})^m = (a a^{-1})^m = 1$ , so $a^m a^{-k} = (a a^{-1})^m \, (a^{-1})^{k-m} = (a^{-1})^{k-m} = a^{-(k-m)} = a^{m+n} .$ Both computations use only the laws for natural exponents, parts (a) and (c) of Proposition 1.1.1, together with that identity and $1^m = 1$ .

Second readingThe sign cases for integer powers

Proof. Throughout, exponents written $j, k, \ell$ are in $\mathbb{N}$ , and laws “for $\mathbb{N}$ ” are Proposition 1.1.1. Note $1^k = 1$ for every $k$ (induction: $1^{k+1} = 1^k \cdot 1$ ).

(d) Natural exponents. For $k \in \mathbb{N}$ , $a^k \ne 0$ by part (d) for $\mathbb{N}$ , and $a^k (a^{-1})^k = (a a^{-1})^k = 1^k = 1$ by part (c) for $\mathbb{N}$ . By Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Lemma 4.1.1 (a), $a^{-k} = (a^{-1})^k = (a^k)^{-1}$ , and it is non-zero, since its product with $a^k$ is $1 \ne 0$ . Negative exponents. For $n = -k$ with $k \ge 1$ , $a^n = (a^k)^{-1} \ne 0$ , and $(a^n)^{-1} = ((a^k)^{-1})^{-1} = a^k = a^{-n}$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Lemma 4.1.1 (a). So (d) holds for every $n \in \mathbb{Z}$ . In particular, for every $n \in \mathbb{Z}$ , $a^{-n} = \bigl( a^{-1} \bigr)^{n} , \tag{$\ast$}$ by the definition for $n \ge 0$ , and for $n = -k < 0$ because both sides equal $a^k$ : the right side is $((a^{-1})^{-1})^k = a^k$ .
(e) Both $m, n \ge 0$ : part (a) for $\mathbb{N}$ . Both negative: let $m = -j$ and $n = -k$ . By the definition and part (a) for $\mathbb{N}$ applied to $a^{-1}$ , $a^{m+n} = (a^{-1})^{j+k} = (a^{-1})^j (a^{-1})^k = a^m a^n .$ Mixed signs. The case $m \ge 0 > n$ is the computation above. The case $m < 0 \le n$ is the same with $m$ and $n$ exchanged, since $F$ is commutative.
(f) $n = k \ge 0$ . If $m \ge 0$ , this is part (b) for $\mathbb{N}$ . If $m = -j < 0$ , then $(a^m)^k = ((a^{-1})^j)^k = (a^{-1})^{jk} = a^{-jk} = a^{mk}$ , by part (b) for $\mathbb{N}$ applied to $a^{-1}$ . $n = -k < 0$ . Since $a^m \ne 0$ by (d), the definition and (d) give $(a^m)^{-k} = ((a^m)^{-1})^k = (a^{-m})^k$ , and the case just proved, with exponents $-m \in \mathbb{Z}$ and $k \ge 0$ , gives $(a^{-m})^k = a^{-mk} = a^{mn}$ .
(g) $n = k \ge 0$ : part (c) for $\mathbb{N}$ . $n = -k < 0$ : $ab \ne 0$ and $(ab)^{-1} = a^{-1} b^{-1}$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Lemma 4.1.1 (a)), so $(ab)^{-k} = (a^{-1} b^{-1})^k = (a^{-1})^k (b^{-1})^k = a^{-k} b^{-k} .$
(h) For $k \in \mathbb{N}$ , $a^k > 0$ by induction: $a^0 = 1 > 0$ , and if $a^k > 0$ , then $a^{k+1} = a^k a > 0 \cdot a = 0$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c). (This is the proof of Sets and Functions, Proposition 3.3.4 (e), which works in any ordered field.) For $n = -k < 0$ , $a^{-1} > 0$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (a), so $a^n = (a^{-1})^k > 0$ by the case just proved.
(i) First, $a^\ell > 1$ for every $\ell \ge 1$ , by induction: $a^1 = a > 1$ , and if $a^\ell > 1$ , then $a^{\ell+1} = a^\ell a > 1 \cdot a = a > 1$ , by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c) with the positive factor $a$ . Now let $m < n$ , and put $\ell \mathrel{:=} n - m \ge 1$ . By (a), $a^n = a^m a^\ell$ , and $a^m > 0$ by (e). Multiplying $a^\ell > 1$ by $a^m > 0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c)) gives $a^n > a^m$ .

Quick check. Compute $2^{-3} \cdot 4^{2}$ and $\bigl(\frac23\bigr)^{-2}$ in $\mathbb{R}$ .

Solution. $2^{-3} = \bigl(\frac12\bigr)^3 = \frac18$ and $4^2 = 16$ , so the product is $\frac{16}{8} = 2$ . (Alternatively, $4^2 = (2^2)^2 = 2^4$ and $2^{-3} 2^4 = 2^1$ by Proposition 1.5.3 (b), (a).) And $\bigl(\frac23\bigr)^{-2} = \bigl(\frac32\bigr)^2 = \frac94$ , since the inverse of $\frac23$ is $\frac32$ .

1.5.3. Rational powers

Which number should $8^{2/3}$ be? Reading the exponent as “square of the cube root” gives $(\sqrt[3]{8})^2 = 2^2 = 4$ . But $\frac23 = \frac46$ , and reading $\frac46$ the same way gives $(\sqrt[6]{8})^4$ . That is $4$ again, as it must be if the notation is to mean anything: its sixth power is $8^4 = 4096 = 4^6$ . A definition that picks one way of writing the exponent is a definition by representatives, and it is only a definition if every representative gives the same value.

A rational power is a power of a root, and it does not matter which fraction names the exponent.

Definition 1.5.2 (Rational Powers)

Let $a \in \mathbb{R}$ with $a > 0$ , and let $r \in \mathbb{Q}$ . Write $r = \frac{m}{n}$ with $m \in \mathbb{Z}$ and $n \in \mathbb{Z}_{>0}$ , and put $a^r \mathrel{:=} \bigl( \sqrt[n]{a} \bigr)^m .$

In words: take the $n$ -th root of the base, which is positive since $a > 0$ , and raise it to the integer power $m$ of Definition 1.5.1. The base is strictly positive, so that negative $m$ is allowed. Negative bases cannot be admitted at all; an example after the next proposition shows why.

The definition chose one fraction for $r$ , and every rational number has infinitely many. So there is something to check before $a^r$ means anything.

Proposition 1.5.4 (Rational Powers Are Well Defined)

Let $a \in \mathbb{R}$ with $a > 0$ . If $m, p \in \mathbb{Z}$ and $n, q \in \mathbb{Z}_{>0}$ satisfy $\frac{m}{n} = \frac{p}{q}$ , then $\bigl( \sqrt[n]{a} \bigr)^m = \bigl( \sqrt[q]{a} \bigr)^p .$ For $r = m \in \mathbb{Z}$ , the number $a^r$ is the integer power $a^m$ of Definition 1.5.1, and for $r = \frac1n$ it is $\sqrt[n]{a}$ .

Try it first (ten minutes): show $(\sqrt[2]{a})^1 = (\sqrt[4]{a})^2$ , then say what the general proof must compare.

Idea. The attempt runs into the fact that we cannot compute either root, so we cannot compare them directly. In the trial case, squaring the right side gives $((\sqrt[4]{a})^2)^2 = a$ , so it is the non-negative number with square $a$ : uniqueness of square roots identifies it. In general, raise both candidates to the power $nq$ , which clears both roots: the left side becomes $a^{mq}$ and the right side $a^{pn}$ , and $mq = pn$ is what $\frac{m}{n} = \frac{p}{q}$ means. Two positive numbers with the same $nq$ -th power are equal by Lemma 1.5.1 (b). This is the method of Sets and Functions, Example (Method: checking that a definition on classes is well defined): two representatives, their relation unwound, equal outputs.

Proof. Let $u \mathrel{:=} (\sqrt[n]{a})^m$ and $v \mathrel{:=} (\sqrt[q]{a})^p$ . The roots are positive since $a > 0$ , so $u, v > 0$ by Proposition 1.5.3 (e). By Proposition 1.5.3 (b), used twice with $m \cdot nq = n \cdot mq$ , $u^{nq} = \bigl( \sqrt[n]{a} \bigr)^{mnq} = \Bigl( \bigl( \sqrt[n]{a} \bigr)^{n} \Bigr)^{mq} = a^{mq} ,$ and in the same way $v^{nq} = \bigl( (\sqrt[q]{a})^{q} \bigr)^{pn} = a^{pn}$ . Multiplying $\frac{m}{n} = \frac{p}{q}$ by $nq$ gives $mq = pn$ . Hence $u^{nq} = v^{nq}$ , with $nq \ge 1$ , and the uniqueness in Lemma 1.5.1 (b) with exponent $nq$ gives $u = v$ .

For $r = m \in \mathbb{Z}$ , use the representative $\frac{m}{1}$ : $\sqrt[1]{a} = a$ , since $a \ge 0$ and $a^1 = a$ , so $a^r = a^m$ . For $r = \frac1n$ , use $\frac{1}{n}$ : $a^r = (\sqrt[n]{a})^1 = \sqrt[n]{a}$ .

So $a^r$ depends on $r$ alone, and the two special cases say that the new notation extends the old: $a^{3}$ , $a^{-2}$ and $a^{1/n} = \sqrt[n]{a}$ all mean what they meant before.

Warning. Negative bases do not have rational powers. The real number $-2$ has cube $-8$ , so one might want $(-8)^{1/3} = -2$ . But $\frac13 = \frac26$ , and read through $\frac26$ the power would be the square of a sixth root of $-8$ , and no real number has sixth power $-8$ , since an even power is $\ge 0$ . Reading $\frac26$ as “sixth root of the square” gives $((-8)^2)^{1/6} = 64^{1/6} = 2$ , positive. No value is independent of the representative, which is why Definition 1.5.2 asks for $a > 0$ .

With well-definedness in hand, the laws come almost for free. For instance, $8^{1/3} \cdot 8^{1/3} = 2 \cdot 2 = 4$ , and $8^{2/3} = 4$ as computed above: the first law below says this always works. Any two rational exponents can be written over one denominator. Then most laws become statements about integer powers of one root $\sqrt[n]{a}$ .

Proposition 1.5.5 (Laws of Rational Powers)

Let $a, b \in \mathbb{R}$ with $a > 0$ and $b > 0$ , and let $r, s \in \mathbb{Q}$ .

(a) $a^{r+s} = a^r a^s$ .
(b) $(a^r)^s = a^{rs}$ .
(c) $(ab)^r = a^r b^r$ .
(d) $a^r > 0$ .
(e) If $a > 1$ and $r < s$ , then $a^r < a^s$ .
(f) If $a < b$ and $r > 0$ , then $a^r < b^r$ .

Idea. Write $r = \frac{m}{n}$ and $s = \frac{p}{n}$ over a common denominator $n > 0$ ; Proposition 1.5.4 allows any representative. Then $a^r = c^m$ and $a^s = c^p$ with $c \mathrel{:=} \sqrt[n]{a}$ , and (a), (d), (e) become laws of integer powers of $c$ (Proposition 1.5.3).

Proof. Write $r = \frac{m'}{n'}$ and $s = \frac{p'}{q'}$ with $n', q' > 0$ , and put $n \mathrel{:=} n'q' > 0$ , $m \mathrel{:=} m'q'$ , $p \mathrel{:=} p'n'$ . Then $r = \frac{m}{n}$ and $s = \frac{p}{n}$ . Let $c \mathrel{:=} \sqrt[n]{a} > 0$ . By Proposition 1.5.4, $a^r = c^m$ and $a^s = c^p$ .

(a) Since $r + s = \frac{m + p}{n}$ , Proposition 1.5.4 gives $a^{r+s} = c^{m+p}$ . Adding integer exponents, $c^{m+p} = c^m c^p = a^r a^s$ .
(b) Let $a > 1$ . Then $c > 1$ : otherwise $c \le 1$ , and either $c = 1$ and $a = 1^n = 1$ , or $0 < c < 1$ and $a = c^n < 1^n = 1$ by Lemma 1.5.1 (a); both contradict $a > 1$ . From $r < s$ , multiplying by $n > 0$ , we get $m < p$ . By Proposition 1.5.3 (f), $a^r = c^m < c^p = a^s$ .

Uses: Proposition 1.5.3 (a) (adding integer exponents).

Second readingProofs of (b), (c), (d), (f)

Proof. Keep $r = \frac{m}{n}$ , $c = \sqrt[n]{a}$ as above.

(d) $a^r = c^m > 0$ by Proposition 1.5.3 (e), since $c > 0$ .
(e) Here it is convenient to write $s = \frac{p}{q}$ with its own denominator $q > 0$ . Let $w \mathrel{:=} a^r$ , which is positive by (d). Let $u \mathrel{:=} w^s = (a^r)^s$ and $v \mathrel{:=} a^{rs}$ , where $rs = \frac{mp}{nq}$ . Both $u$ and $v$ are positive by (d). Every change of the form $(x^i)^j = x^{ij} = (x^j)^i$ below is Proposition 1.5.3 (b). By the definition, $w^n = (c^m)^n = (c^n)^m = a^m$ , and $u^q = \Bigl( \bigl( \sqrt[q]{w} \bigr)^p \Bigr)^q = \Bigl( \bigl( \sqrt[q]{w} \bigr)^q \Bigr)^p = w^p .$ Hence $u^{nq} = (u^q)^n = (w^p)^n = (w^n)^p = (a^m)^p = a^{mp} .$ Also $v = (\sqrt[nq]{a})^{mp}$ , so $v^{nq} = \bigl( (\sqrt[nq]{a})^{nq} \bigr)^{mp} = a^{mp}$ . So $u^{nq} = v^{nq}$ , and the uniqueness in Lemma 1.5.1 (b) with exponent $nq$ gives $u = v$ . This covers $m \le 0$ and $p \le 0$ as well, because only integer-power laws were used.
(f) By Lemma 1.5.2 (a), $\sqrt[n]{ab} = \sqrt[n]{a} \, \sqrt[n]{b}$ , and both factors are positive. By Proposition 1.5.3 (c), $(ab)^r = \bigl( \sqrt[n]{a} \, \sqrt[n]{b} \bigr)^m = \bigl( \sqrt[n]{a} \bigr)^m \bigl( \sqrt[n]{b} \bigr)^m = a^r b^r .$
(g) Let $a < b$ and $r > 0$ . Since $r = \frac{m}{n} > 0$ and $n > 0$ , $m \ge 1$ . Let $d \mathrel{:=} \sqrt[n]{b}$ . By Lemma 1.5.2 (d), $0 < c < d$ . By Lemma 1.5.1 (a) with the exponent $m \ge 1$ , $a^r = c^m < d^m = b^r$ .

Part (e) needs $a > 1$ : for $a = 1$ every power is $1$ , and for $0 < a < 1$ the order reverses, as Exercise C1 shows. Part (f) needs $r > 0$ : for $r = 0$ both sides are $1$ .

Example (Rational powers by hand)

Compute $27^{2/3} \cdot 9^{-1/2}$ and $\bigl(\frac{16}{81}\bigr)^{-3/4}$ . Which is larger, $2^{1/2}$ or $3^{1/3}$ ?

Solution. Since $3 \ge 0$ and $3^3 = 27$ , $\sqrt[3]{27} = 3$ , so $27^{2/3} = 3^2 = 9$ . Since $\sqrt{9} = 3$ , $9^{-1/2} = 3^{-1} = \frac13$ . The product is $9 \cdot \frac13 = 3$ .

Since $\frac23 \ge 0$ and $\bigl(\frac23\bigr)^4 = \frac{16}{81}$ , $\sqrt[4]{16/81} = \frac23$ . So $\bigl(\frac{16}{81}\bigr)^{-3/4} = \bigl(\frac23\bigr)^{-3} = \bigl(\frac32\bigr)^3 = \frac{27}{8}$ .

For the comparison, raise both to the sixth power, which clears both roots: by Proposition 1.5.5 (b), $(2^{1/2})^6 = 2^3 = 8$ and $(3^{1/3})^6 = 3^2 = 9$ . Both numbers are positive. If $2^{1/2} \ge 3^{1/3}$ , then $8 \ge 9$ by Lemma 1.5.1 (a) (or equality), which is false. So $2^{1/2} < 3^{1/3}$ .

1.5.4. Means and estimates with roots

Section 1.3 proved the inequality between the means in the form $a_1 \cdots a_n \le A^n$ , without roots, because none were available. With $n$ -th roots it takes its familiar shape. For the two numbers $2$ and $8$ , the geometric mean is $\sqrt{2 \cdot 8} = 4$ , and the arithmetic mean is $\frac{2 + 8}{2} = 5$ .

Corollary 1.5.6 (Geometric Mean Below Arithmetic Mean)

Let $n \ge 1$ be an integer and let $a_1, \dots, a_n \in \mathbb{R}$ with $a_k \ge 0$ for every $k$ . Then $\sqrt[n]{a_1 a_2 \cdots a_n} \le \frac{1}{n} \sum_{k=1}^{n} a_k ,$ with equality if and only if $a_1 = a_2 = \dots = a_n$ .

Idea. This is Theorem 1.3.3 with an $n$ -th root taken on both sides. The theorem bounds the product $G^n$ , and the root turns that bound into one on $G$ . Taking roots preserves order by Lemma 1.5.2 (d).

Uses: Example (Method: bound a product by a power) (bounding the product).

Proof. Let $A \mathrel{:=} \frac1n \sum a_k \ge 0$ and $G \mathrel{:=} \sqrt[n]{a_1 \cdots a_n} \ge 0$ . By Theorem 1.3.3 (a), $G^n = a_1 \cdots a_n \le A^n$ . If $G > A$ , then $G^n > A^n$ by Lemma 1.5.1 (a), since $A \ge 0$ ; so $G \le A$ . If $G = A$ , then $a_1 \cdots a_n = A^n$ , and Theorem 1.3.3 (b) gives $a_1 = \dots = a_n$ . Conversely, if all $a_k$ are equal, then $a_1 \cdots a_n = A^n$ by Theorem 1.3.3 (b), so $G = \sqrt[n]{A^n} = A$ by uniqueness, as $A \ge 0$ .

The last results of this chapter turn roots into estimates. How close to $1$ is $\sqrt[n]{a}$ for large $n$ ? Write the root as $1 + x$ with $x \ge 0$ . Then $a = (1 + x)^n$ , and a power of $1 + x$ is exactly what Example (Method: bound a power from below) bounds from below by one term. A lower bound on $(1 + x)^n$ is an upper bound on $x$ . For instance, $\sqrt[3]{8} = 2$ , and part (a) below gives the bound $\sqrt[3]{8} \le 1 + \frac{8 - 1}{3} = \frac{10}{3}$ .

Proposition 1.5.7 (Two Root Estimates)

(a) If $a \in \mathbb{R}$ , $a \ge 1$ and $n \ge 1$ is an integer, then $1 \le \sqrt[n]{a} \le 1 + \dfrac{a - 1}{n}$ .
(b) If $n \ge 2$ is an integer, then $1 \le \sqrt[n]{n} \le 1 + \sqrt{\dfrac{2}{n - 1}}$ .

Idea. The method of Example (Method: bound a power from below), read for a root: write the root as $1 + x$ , keep one term of $(1 + x)^n$ , solve for $x$ . For (a), Bernoulli’s term $nx$ suffices, because the left side $a$ is fixed. For (b) the left side is $n$ , which grows, so the linear term gives only $x \le \frac{n-1}{n}$ ; the quadratic term $\binom{n}{2} x^2$ is needed.

Proof. In both parts let $y$ be the number in question, $y \mathrel{:=} \sqrt[n]{a}$ or $y \mathrel{:=} \sqrt[n]{n}$ , and let $x \mathrel{:=} y - 1$ . Since $\sqrt[n]{1} = 1$ (as $1 \ge 0$ and $1^n = 1$ ), Lemma 1.5.2 (d) gives $y \ge 1$ whenever the radicand is $\ge 1$ , with equality if it is $1$ . So $x \ge 0$ , which is the left inequality in both parts.

(a) By Sets and Functions, Theorem 3.3.7 (with $x \ge 0 \ge -1$ ), $a = (1 + x)^n \ge 1 + nx$ . So $nx \le a - 1$ , and dividing by $n > 0$ , $x \le \frac{a - 1}{n}$ .
(b) Recall $\binom{n}{2} = \frac{n(n-1)}{2}$ . By Corollary 1.2.4 (b) with $k = 2 \le n$ , $n = (1 + x)^n \ge \frac{n(n - 1)}{2} x^2 .$ Dividing by $\frac{n(n-1)}{2} > 0$ gives $x^2 \le \frac{2}{n - 1}$ . Let $w \mathrel{:=} \sqrt{2/(n - 1)} \ge 0$ . If $x > w$ , then $x^2 > w^2 = \frac{2}{n-1}$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (g), since $x, w \ge 0$ ; so $x \le w$ .

Uses: Proposition 1.2.2 (c) (the value of $\binom{n}{2}$ ).

Both bounds are explicit for every $n$ , so a threshold for any tolerance can be read off. The limit statements follow at once by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Method: proving a limit from the definition).

Example (Roots tend to one)

Show that $\sqrt[n]{a} \to 1$ for every real $a \ge 1$ , and that $\sqrt[n]{n} \to 1$ .

Solution. First limit. By Proposition 1.5.7 (a), $\left| \sqrt[n]{a} - 1 \right| = \sqrt[n]{a} - 1 \le \frac{a - 1}{n}$ for $n \ge 1$ . If $a = 1$ , the sequence is constant. If $a > 1$ , let $\varepsilon > 0$ , choose an integer $N > (a - 1)/\varepsilon$ , so $N \ge 1$ ; it depends on $a$ and $\varepsilon$ only. For $n \ge N$ , $\left| \sqrt[n]{a} - 1 \right| \le \frac{a-1}{n} \le \frac{a-1}{N} < \varepsilon$ .

Second limit. By Proposition 1.5.7 (b), $\left| \sqrt[n]{n} - 1 \right| \le \sqrt{2/(n - 1)}$ for $n \ge 2$ . Let $\varepsilon > 0$ and choose an integer $N > 1 + 2/\varepsilon^2$ ; then $N \ge 2$ , and $N$ depends on $\varepsilon$ only. For $n \ge N$ , $n - 1 > 2/\varepsilon^2 > 0$ , so $\frac{2}{n - 1} < \varepsilon^2$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (b)). Square roots preserve strict order between non-negative numbers (Lemma 1.5.2 (d) with $n = 2$ ). So $\sqrt{2/(n-1)} < \sqrt{\varepsilon^2} = \varepsilon$ . Hence $\left| \sqrt[n]{n} - 1 \right| < \varepsilon$ .

Quick check. Find an $N$ with $\sqrt[n]{5} < 1.01$ for every $n \ge N$ .

Solution. By Proposition 1.5.7 (a) with $a = 5$ , $\sqrt[n]{5} \le 1 + \frac4n$ . And $1 + \frac4n < 1.01$ exactly when $\frac4n < \frac{1}{100}$ , that is, $n > 400$ . So $N = 401$ works. (The bound is generous: already $\sqrt[162]{5} < 1.01$ . The definition asks for some threshold, not the least.)

Remark (Looking ahead). Chapter 8 defines $a^x$ for every real $x$ through the exponential function and checks that it agrees with the $a^r$ of this section; the analysis volumes construct the exponential itself.

1.5.5. Exercises

A. Check your understanding

Exercise (A1)

(a) Where does the proof of Lemma 1.5.1 use that $\mathbb{R}$ is complete?
(b) Why is $h \le 1$ imposed in the case $s^n < a$ of that proof?
(c) Run the first steps of the existence proof for $n = 3$ , $a = 8$ : what are $m$ , $M$ and $s$ ?
(d) Compute $8^{-2/3}$ .
(e) True or false: $(a^r)^s = a^{rs}$ for all real $a > 0$ and all $r, s \in \mathbb{Q}$ .

Solution.

(a) Once: to produce $s \mathrel{:=} \sup S$ , from a non-empty set bounded above. Nothing else in the proof needs it.
(b) So that $s + h \le s + 1$ . Then each of the $n$ terms $(s + h)^k s^{n-1-k}$ is at most $(s + 1)^{n-1}$ , a bound that does not involve $h$ , and $h$ can be solved for. Without $h \le 1$ the terms would grow with $h$ .
(c) $m = \min\{1, 8\} = 1$ , $M = \max\{1, 8\} = 8$ , and $s = \sup\{ y \ge 0 : y^3 \le 8 \} = 2$ : the proof shows $s^3 = 8$ , and $2$ is the non-negative number with cube $8$ .
(d) $\sqrt[3]{8} = 2$ , so $8^{-2/3} = 2^{-2} = \frac14$ .
(e) True: Proposition 1.5.5 (b).

B. Practice

Exercise (B1: Simplify)

Simplify, for real $a, b > 0$ :

(a) $32^{3/5}$ ;
(b) $\bigl( a^{1/2} b^{-1/3} \bigr)^6$ ;
(c) $\dfrac{4^{3/2}}{8^{2/3}}$ .

Solution.

(a) $\sqrt[5]{32} = 2$ , since $2^5 = 32$ ; so $32^{3/5} = 2^3 = 8$ .
(b) By Proposition 1.5.5 (c) and (b), $(a^{1/2} b^{-1/3})^6 = (a^{1/2})^6 (b^{-1/3})^6 = a^{3} b^{-2}$ .
(c) $4^{3/2} = (\sqrt 4)^3 = 2^3 = 8$ and $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$ , so the quotient is $\frac84 = 2$ .

Exercise (B2: Compare without a calculator)

Put $3^{1/3}$ , $2^{1/2}$ and $5^{1/5}$ in increasing order. Justify your answer.

Solution. Raise all three to the power $30$ , a common multiple of $3$ , $2$ , $5$ . By the law $(a^r)^s = a^{rs}$ , $(3^{1/3})^{30} = 3^{10} = 59049 , \quad (2^{1/2})^{30} = 2^{15} = 32768 , \quad (5^{1/5})^{30} = 5^{6} = 15625 .$ All three numbers are positive, and for positive $x, y$ , $x^{30} < y^{30}$ forces $x < y$ : otherwise $y \le x$ and $y^{30} \le x^{30}$ by Lemma 1.5.1 (a) (or equality). Hence $5^{1/5} < 2^{1/2} < 3^{1/3}$ .

Uses: Proposition 1.5.5 (b) (powers of powers).

Exercise (B3: An explicit N)

Find an $N$ such that $\sqrt[n]{n} < 1.1$ for every $n \ge N$ .

Solution. By Proposition 1.5.7 (b), $\sqrt[n]{n} \le 1 + \sqrt{2/(n - 1)}$ for $n \ge 2$ , so it suffices that $\sqrt{2/(n - 1)} < 0.1$ . Both sides are non-negative. Squaring and taking square roots both preserve strict order between non-negative numbers. So this holds exactly when $\frac{2}{n - 1} < 0.01$ , that is, when $n - 1 > 200$ . So $N = 202$ works: for $n \ge 202$ , $\sqrt[n]{n} \le 1 + \sqrt{2/(n-1)} < 1.1$ .

Uses: Lemma 1.5.1 (a) (squaring preserves order); Lemma 1.5.2 (d) (square roots preserve order).

C. Going deeper

Exercise (C1: Chapter check: powers of a base below one)

Let $a \in \mathbb{R}$ with $0 < a < 1$ , and let $r, s \in \mathbb{Q}$ . Prove:

(i) if $r < s$ , then $a^r > a^s$ ;
(ii) if $r > 0$ , then $a^r < 1$ .

Solution. Put $b \mathrel{:=} \frac1a$ . Since $0 < a < 1$ , Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (b) gives $b > \frac11 = 1$ . For every $t \in \mathbb{Q}$ , $1^t = 1$ : with $t = \frac{m}{n}$ , $\sqrt[n]{1} = 1$ by uniqueness (as $1 \ge 0$ and $1^n = 1$ ), and $1^m = 1$ for every $m \in \mathbb{Z}$ by Proposition 1.5.3 (d). So, by Proposition 1.5.5 (c), $a^t b^t = (ab)^t = 1^t = 1 ,$ and $a^t = (b^t)^{-1}$ for every $t \in \mathbb{Q}$ .

(i) Let $r < s$ . By Proposition 1.5.5 (e), applied to the base $b > 1$ , $b^r < b^s$ , and both are positive by part (d). Reciprocals reverse the order between positive numbers (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (b)), so $a^r = (b^r)^{-1} > (b^s)^{-1} = a^s .$
(ii) Let $r > 0$ . By (i) with the exponents $0 < r$ , $a^r < a^0$ , and $a^0 = 1$ , since by Proposition 1.5.4 $a^0$ is the integer power $a^0 = 1$ .

What a complete answer needs. (10 points) - 2: reduces to a base above one, $b = \frac1a > 1$ , with the reason. - 3: the link to that base, $a^t = (b^t)^{-1}$ from (c) of the laws and $1^t = 1$ . - 2: the law (e) applied to a base above one, not to $a$ . - 2: the reversal of order under reciprocals, with positivity from (d). - 1: (ii) from (i), with $a^0 = 1$ .

Alternatively, over a common denominator $r = \frac{m}{n}$ , $s = \frac{p}{n}$ :

3: $c = \sqrt[n]{a} \in (0, 1)$ with the reason, and $m < p$ .
2: $c^{p-m} < 1$ , with the reason (for instance $(1/c)^{p-m} > 1$ ).
2: $c^p = c^m c^{p-m} < c^m$ , using $c^m > 0$ .
2: the conclusion $a^r = c^m > c^p = a^s$ .
1: (ii) from (i), with $a^0 = 1$ . Slips that lose the step: applying (e) to a base $a < 1$ .

Exercise (C2: The compound sequence is bounded)

Show that $\bigl( 1 + \frac1n \bigr)^n < 3$ for every integer $n \ge 1$ .

Solution (Hint 1). Expand by the binomial theorem and compare the term $\binom{n}{k} \frac{1}{n^k}$ with $\frac{1}{k!}$ .

Solution (Hint 2). By Proposition 1.2.2 (c), $\binom{n}{k} \dfrac{1}{n^k} = \dfrac{1}{k!} \displaystyle\prod_{j=0}^{k-1} \frac{n - j}{n}$ .

Solution. Let $n \ge 1$ . By Theorem 1.2.3 with $x = \frac1n$ , $y = 1$ , $\Bigl( 1 + \frac1n \Bigr)^n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} .$ The term $k = 0$ is $1$ . Let $1 \le k \le n$ . By Proposition 1.2.2 (c), $\binom{n}{k} k! = \prod_{j=0}^{k-1} (n - j)$ , and $n^k = \prod_{j=0}^{k-1} n$ , so $\binom{n}{k} \frac{1}{n^k} = \frac{1}{k!} \prod_{j=0}^{k-1} \frac{n - j}{n} .$ For $0 \le j \le k - 1 \le n - 1$ , the factor $\frac{n - j}{n}$ lies in $(0, 1]$ . So the product lies in $(0, 1]$ , by induction on the number of factors. Hence the term is at most $\frac{1}{k!}$ , and $\frac{1}{k!} \le \frac{1}{2^{k-1}}$ by Example (Factorials beat powers of two). Comparing termwise and summing the geometric series with ratio $\frac12$ , $\Bigl( 1 + \frac1n \Bigr)^n \le 1 + \sum_{k=1}^{n} \Bigl( \frac12 \Bigr)^{k-1} = 1 + 2\Bigl( 1 - \frac{1}{2^n} \Bigr) = 3 - \frac{2}{2^n} < 3 ,$ where the last step uses $\frac{2}{2^n} > 0$ .

Together with Example (The compound-interest sequence increases), which showed that the sequence increases, this says the compound sequence is increasing and bounded above by $3$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (d) (products of factors in $(0, 1]$ ); Proposition 1.1.5 (a) (termwise comparison); Sets and Functions, Proposition 3.3.6 (the geometric sum).

What a complete answer needs. (10 points) - 3: the binomial expansion, with each term written in the factorial form. - 3: the termwise bound $\binom{n}{k} n^{-k} \le \frac{1}{k!}$ , with the factors $\frac{n-j}{n}$ shown to lie in $(0, 1]$ , and $\frac{1}{k!}$ bounded by a summable comparison sequence (for instance $\frac{1}{2^{k-1}}$ ), with reason. - 2: the comparison sum evaluated exactly (geometric or telescoping). - 2: the strict inequality $< 3$ , with its reason. Slips that lose the step: bounding $\binom{n}{k} n^{-k}$ by $\frac{1}{k!}$ without the factor-by-factor reason, or summing the comparison series to infinity.

R. Review

Exercise (R1)

Compute $\displaystyle\sum_{k=1}^{n} \frac{1}{\sqrt{k} + \sqrt{k + 1}}$ for $n \ge 1$ .

Solution. The tool is telescoping (Example (Method: telescoping)). Since $k + 1 > 0$ , $\sqrt{k+1} > 0$ , so $\sqrt k + \sqrt{k+1} > 0$ . Since $(\sqrt{k+1} - \sqrt k)(\sqrt{k+1} + \sqrt k) = (k + 1) - k = 1$ , $\frac{1}{\sqrt k + \sqrt{k + 1}} = \sqrt{k + 1} - \sqrt k .$ With $b_j \mathrel{:=} \sqrt{j + 1}$ and $j = k - 1$ , Theorem 1.1.7 (a) gives $\sum_{k=1}^{n} \bigl( \sqrt{k + 1} - \sqrt k \bigr) = \sum_{j=0}^{n-1} (b_{j+1} - b_j) = b_n - b_0 = \sqrt{n + 1} - 1 .$

Exercise (R2)

Compute $\displaystyle\sum_{k=0}^{n} \binom{n}{k} (-2)^k$ for $n \in \mathbb{N}$ .

Solution. By the binomial theorem (Theorem 1.2.3) in $\mathbb{R}$ with $x = -2$ and $y = 1$ , the sum is $(-2 + 1)^n = (-1)^n$ .

Exercise (R3)

Prove that $\dfrac{n + \sqrt n}{n^2 + 1} \to 0$ .

Solution. The first tool is keeping the dominant term (Example (Method: keep the dominant term)). The second is proving a limit from the definition (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Method: proving a limit from the definition)). For $n \ge 1$ , $n \le n^2$ , so $\sqrt n \le \sqrt{n^2} = n$ by Lemma 1.5.2 (d) (or equality). Here $\sqrt{n^2} = n$ by uniqueness, since $n \ge 0$ . Hence the numerator is at most $2n$ . The denominator is at least $n^2 > 0$ , so $0 < \frac{n + \sqrt n}{n^2 + 1} \le \frac{2n}{n^2} = \frac{2}{n} \qquad (n \ge 1) .$ Let $\varepsilon > 0$ and choose an integer $N > 2/\varepsilon$ ; then $N \ge 1$ , and $N$ depends on $\varepsilon$ only. For $n \ge N$ the distance to $0$ is at most $2/n \le 2/N < \varepsilon$ .

The look-alike tool, the quotient rule for limits, does not apply to the fraction as written: neither the numerator nor the denominator converges.