§16 ★ Decimal Expansions of Real Numbers

We begin by recalling the brief discussion of decimals in Discussion 10.3. There we considered a decimal expansion $K.d_1 d_2 d_3 \cdots$, where $K$ is a nonnegative integer and each digit $d_j$ belongs to $\{0, 1, 2, \dots, 9\}$. This expansion represents the real number

$$K + \sum_{j=1}^\infty \frac{d_j}{10^j} = K + \sum_{j=1}^\infty d_j \cdot 10^{-j},$$

which we can also write as $\lim_{n\to\infty} s_n$, where $s_n = K + \sum_{j=1}^n d_j \cdot 10^{-j}$.

So every decimal expansion represents a nonnegative real number. We prove the converse after formalizing long division. The development here is based on suggestions by Karl Stromberg.

Why this matters

Decimals are how most of us picture real numbers, so it is worth checking that the picture is accurate: every nonnegative real has a decimal expansion, almost every real has exactly one, and the rationals are exactly the ones whose digits eventually repeat. Each of these facts is a short argument built on limits and geometric series.

16.1 Long Division.

Consider $a/b$ with $a, b$ positive integers. The $K$ term causes no difficulty, so assume $a < b$. Take $a = 3$, $b = 7$. Name the digits $d_1, d_2, \dots$ and the remainders $r_1, r_2, \dots$; the first steps give $d_1 = 4$, $d_2 = 2$ and $r_1 = 2$, $r_2 = 6$. Next we divide $7$ into $60 = 10 \cdot r_2$ and get $60 = 7 \cdot 8 + 4$. The quotient $8$ is the third digit $d_3$, and $4 = r_3$ is the new remainder. Then we multiply $r_3$ by $10$ and repeat.

In general, set $r_0 = a$. At each stage

$$\begin{aligned} & d_n \in \{0, 1, 2, \dots, 9\}, && (1) \\ & r_n = 10 \cdot r_{n-1} - b \cdot d_n, && (2) \\ & 0 \le r_n < b. && (3) \end{aligned}$$

This construction yields

$$\frac{a}{b} = \frac{d_1}{10} + \frac1{10} \frac{r_1}{b} \tag{4}$$

and, for $n \ge 2$,

$$\frac{a}{b} = \frac{d_1}{10} + \cdots + \frac{d_n}{10^n} + \frac1{10^n} \frac{r_n}{b}. \tag{5}$$

Since each $r_n < b$, $\lim_n \frac1{10^n} \frac{r_n}{b} = 0$, so (5) gives $\frac{a}{b} = \sum_{j=1}^\infty \frac{d_j}{10^j}$: thus $.d_1 d_2 d_3 \cdots$ is a decimal expansion for $\frac{a}{b}$.

Equation (4) follows from $r_1 = 10a - bd_1$. For (5) we use induction. From (2), $r_{n+1} = 10 r_n - b d_{n+1}$, so

$$\frac{r_n}{b} = \frac1{10} d_{n+1} + \frac1{10} \frac{r_{n+1}}{b}.$$

Substituting this into (5) gives

$$\frac{a}{b} = \frac{d_1}{10} + \cdots + \frac{d_n}{10^n} + \frac{d_{n+1}}{10^{n+1}} + \frac1{10^{n+1}} \frac{r_{n+1}}{b}, \tag{6}$$

which is (5) with $n$ replaced by $n+1$.

Try it: long division, step by step

Each column shows the digit $d_n$ and remainder $r_n$. The first remainder that repeats is highlighted: from there the whole computation cycles, which is the heart of Theorem 16.5.

Fraction a / b/

3/7 = 0.428571428571…

n	0	1	2	3	4	5	6
dₙ		4	2	8	5	7	1
rₙ	3	2	6	4	5	1	3

Remainder r₆ = r₀ = 3, so from then on every step repeats: period 6. There are only 7 possible remainders, so a repeat must occur within 8 steps.

Existence and uniqueness

16.2 Theorem.

Every nonnegative real number $x$ has at least one decimal expansion.

Proof

It suffices to consider $x \in [0, 1)$. We use the floor function: $\lfloor y \rfloor$ is the largest integer less than or equal to $y$. Since $10x < 10$, we can write

$$10x = d_1 + x_1, \qquad\text{where } d_1 = \lfloor 10x \rfloor \text{ and } x_1 \in [0, 1). \tag{1}$$

Note $d_1 \in \{0, 1, \dots, 9\}$ since $10x < 10$. Therefore

$$x = \frac{d_1}{10} + \frac1{10} x_1. \tag{2}$$

Suppose for some $n \ge 1$ we have chosen $d_1, \dots, d_n$ in $\{0, \dots, 9\}$ and $x_1, \dots, x_n$ in $[0, 1)$ so that

$$x = \frac{d_1}{10} + \cdots + \frac{d_n}{10^n} + \frac1{10^n} x_n. \tag{3}$$

Since $10x_n < 10$, we can write

$$10 x_n = d_{n+1} + x_{n+1}, \qquad\text{where } d_{n+1} = \lfloor 10 x_n \rfloor,\ x_{n+1} \in [0, 1). \tag{4}$$

Solving (4) for $x_n$ and substituting into (3) gives

$$x = \frac{d_1}{10} + \cdots + \frac{d_{n+1}}{10^{n+1}} + \frac1{10^{n+1}} x_{n+1},$$

completing the induction. Since $\lim_n \frac1{10^{n+1}} x_{n+1} = 0$,

$$x = \lim_n \sum_{j=1}^{n+1} \frac{d_j}{10^j} = \sum_{j=1}^\infty \frac{d_j}{10^j},$$

so $.d_1 d_2 d_3 \cdots$ is a decimal expansion for $x$.

∎

Intuitiondigits as zooming in

The first digit says which of the ten intervals $[0, 0.1), [0.1, 0.2), \dots, [0.9, 1)$ contains $x$. The second says which tenth of that interval contains it, and so on. The proof of Theorem 16.2 is this zooming-in procedure written out: each $d_{n+1} = \lfloor 10x_n \rfloor$ picks the next tenth. After $n$ digits, $x$ is pinned to an interval of length $10^{-n}$, which is why the truncations converge to $x$.

As noted in Discussion 10.3, $1.000\cdots$ and $0.999\cdots$ are decimal expansions of the same real number: the series $1 + \sum_{j=1}^\infty 0 \cdot 10^{-j}$ and $\sum_{j=1}^\infty 9 \cdot 10^{-j}$ both equal $1$. Similarly $2.75000\cdots$ and $2.74999\cdots$ both represent $\frac{11}4$ (Exercise 16.1). The next theorem says this is essentially the only way a number can have two decimal expansions.

16.3 Theorem.

A real number $x$ has exactly one decimal expansion, or else $x$ has two decimal expansions, one ending in all $0$’s and the other ending in all $9$’s.

Used in Theorem 16.5, Exercise 16.2, Exercise 16.8

Proof

Assume $x \ge 0$. If $x$ has expansion $K.000\cdots$ with $K > 0$, then it has one other, $(K-1).999\cdots$. If $x$ has expansion $K.d_1 d_2 \cdots d_r 000\cdots$ with $d_r \ne 0$, then it has one other, $K.d_1 d_2 \cdots (d_r - 1) 999\cdots$. The reader can check these claims (Exercise 16.2).

Now suppose $x$ has two distinct expansions $K.d_1 d_2 d_3 \cdots$ and $L.e_1 e_2 e_3 \cdots$.

Case $K < L$. If any $d_j < 9$, then by Exercise 16.3

$$x < K + \sum_{j=1}^\infty 9 \cdot 10^{-j} = K + 1 \le L \le x,$$

a contradiction. So $x = K + 1 = L$ and the expansions are $K.999\cdots$ and $(K+1).000\cdots$.

Case $K = L$. Let $m = \min\{j : d_j \ne e_j\}$; we may assume $d_m < e_m$. If $d_j < 9$ for some $j > m$, then by Exercise 16.3

$$\begin{aligned} x &< K + \sum_{j=1}^m d_j \cdot 10^{-j} + \sum_{j=m+1}^\infty 9 \cdot 10^{-j} = K + \sum_{j=1}^m d_j \cdot 10^{-j} + 10^{-m} \\ &= K + \sum_{j=1}^{m-1} e_j \cdot 10^{-j} + d_m \cdot 10^{-m} + 10^{-m} \le K + \sum_{j=1}^m e_j \cdot 10^{-j} \le x, \end{aligned}$$

a contradiction. Thus $d_j = 9$ for $j > m$. Likewise, if $e_j > 0$ for some $j > m$, then

$$\begin{aligned} x &> K + \sum_{j=1}^m e_j \cdot 10^{-j} = K + \sum_{j=1}^{m-1} d_j \cdot 10^{-j} + e_m \cdot 10^{-m} \\ &\ge K + \sum_{j=1}^{m-1} d_j \cdot 10^{-j} + d_m \cdot 10^{-m} + 10^{-m} = K + \sum_{j=1}^m d_j \cdot 10^{-j} + \sum_{j=m+1}^\infty 9 \cdot 10^{-j} \ge x, \end{aligned}$$

a contradiction. So $d_j = 9$ for $j > m$, $e_m = d_m + 1$, and $e_j = 0$ for $j > m$.

∎

Try it: two expansions, nested intervals

Pick a terminating decimal such as $1$ or $\frac34$ and follow either expansion. Each digit selects one tenth of the previous interval, and the number stays inside every interval, so the expansion ending in $9$’s and the one ending in $0$’s both zoom in on the same point.

Number a / b/Digits shown: n4

Expansion to follow

ends in 0’s ends in 9’s

1/1 = 1.0000… = 0.9999…

n	truncation tₙ	x − tₙ	10⁻ⁿ
1	0.9	1.00e-1	1e-1
2	0.99	1.00e-2	1e-2
3	0.999	1.00e-3	1e-3
4	0.9999	1.00e-4	1e-4

1/1 is a terminating decimal, so it has exactly two expansions (Theorem 16.3): 1.0000… and 0.9999…. Following the one that ends in 9’s, x − tₙ equals 10⁻ⁿ for large n, and tₙ → x either way.

Common pitfall0.999… is not “just below” 1

$0.999\cdots$ is not an approximation to $1$. It is the limit of $0.9, 0.99, 0.999, \dots$, and that limit equals $1$. The two expansions are two names for one number. By Theorem 16.3 this ambiguity happens only for numbers with a terminating expansion, that is, numbers of the form $\frac{m}{10^k}$.

Repeating decimals

16.4 Definition.

An expression $K.d_1 d_2 \cdots d_\ell \overline{d_{\ell+1} \cdots d_{\ell+r}}$ represents the decimal expansion in which the block $d_{\ell+1} \cdots d_{\ell+r}$ repeats forever:

$$K.d_1 d_2 \cdots d_\ell\, d_{\ell+1} \cdots d_{\ell+r}\, d_{\ell+1} \cdots d_{\ell+r}\, d_{\ell+1} \cdots d_{\ell+r} \cdots.$$

Such an expansion is a repeating decimal.

Example 1.

Every integer is a repeating decimal, e.g. $17 = 17.\overline{0} = 17.000\cdots$. Another simple example:

$$.\overline{8} = .888\cdots = \sum_{j=1}^\infty 8 \cdot 10^{-j} = \frac8{10} \sum_{j=0}^\infty 10^{-j} = \frac8{10} \cdot \frac{10}9 = \frac89.$$

Example 2.

$3.9\overline{67}$ is the repeating decimal $3.9676767\cdots$. We evaluate it:

$$\begin{aligned} 3.9\overline{67} &= 3 + 9 \cdot 10^{-1} + 6 \cdot 10^{-2} + 7 \cdot 10^{-3} + 6 \cdot 10^{-4} + 7 \cdot 10^{-5} + \cdots \\ &= 3 + 9 \cdot 10^{-1} + 67 \cdot 10^{-3} \sum_{j=0}^\infty (10^{-2})^j = 3 + 9 \cdot 10^{-1} + 67 \cdot 10^{-3} \left(\frac{100}{99}\right) \\ &= 3 + \frac9{10} + \frac{67}{990} = \frac{3{,}928}{990} = \frac{1{,}964}{495}. \end{aligned}$$

Any repeating decimal can be evaluated as a rational number this way, as the next theorem shows.

Example 3.

We find the decimal expansion of $\frac{11}7$. Long division gives $\frac{11}7 = 1.571428571428571428\cdots$, i.e. $\frac{11}7 = 1.\overline{571428}$. To check:

$$1.\overline{571428} = 1 + 571{,}428 \cdot 10^{-6} \sum_{j=0}^\infty (10^{-6})^j = 1 + \frac{571{,}428}{999{,}999} = 1 + \frac47 = \frac{11}7.$$

Many books leave the next theorem as an exercise, probably to avoid its notation. If the details seem too complicated, move on to Examples 4–7.

16.5 Theorem.

A real number $x$ is rational if and only if its decimal expansion is repeating. (Theorem 16.3 shows that if $x$ has two decimal expansions, both are repeating.)

Proof

First assume $x \ge 0$ has a repeating expansion $x = K.d_1 d_2 \cdots d_\ell \overline{d_{\ell+1} \cdots d_{\ell+r}}$. Then

$$x = K + \sum_{j=1}^\ell d_j \cdot 10^{-j} + 10^{-\ell} y, \qquad y = .\overline{d_{\ell+1} \cdots d_{\ell+r}},$$

so it suffices to show $y$ is rational. Write $y = .\overline{e_1 e_2 \cdots e_r}$. Then

$$y = \left[\sum_{j=1}^r e_j \cdot 10^{-j}\right] \left[\sum_{j=0}^\infty (10^{-r})^j\right] = \left[\sum_{j=1}^r e_j \cdot 10^{-j}\right] \frac{10^r}{10^r - 1},$$

which is rational. In fact, writing $e_1 e_2 \cdots e_r$ for the integer $\sum_{j=0}^{r-1} e_j \cdot 10^{r-1-j}$ (not the product), $y = \frac{e_1 e_2 \cdots e_r}{10^r - 1}$; see Example 3.

Next consider a positive rational $\frac{a}{b}$ with $a, b \in \mathbb{N}$, and assume $a < b$. As in 16.1, $\frac{a}{b} = .d_1 d_2 d_3 \cdots$ where $r_0 = a$ and, for $k \ge 1$,

$$d_k \in \{0, 1, \dots, 9\}, \qquad r_k = 10 \cdot r_{k-1} - d_k b, \qquad 0 \le r_k < b. \tag{1–3}$$

Since $a, b$ are integers, each $r_k$ is an integer, so

$$r_k \in \{0, 1, 2, \dots, b - 1\} \qquad\text{for } k \ge 0. \tag{4}$$

This set has $b$ elements, so the first $b + 1$ remainders cannot all be distinct: there are integers $m \ge 0$ and $p > 0$ with $0 \le m < m + p \le b$ and $r_m = r_{m+p}$.

Given $r_{k-1}$, the integers $r_k$ and $d_k$ are uniquely determined. Thus $r_j = r_k$ implies $r_{j+1} = r_{k+1}$ and $d_{j+1} = d_{k+1}$. From $r_m = r_{m+p}$ we get $r_{m+1} = r_{m+1+p}$ and $d_{m+1} = d_{m+1+p}$, and a simple induction shows “$r_k = r_{k+p}$ and $d_k = d_{k+p}$” for all $k \ge m + 1$. So the expansion of $\frac{a}{b}$ is periodic with period $p$ after the first $m$ digits:

$$\frac{a}{b} = .d_1 d_2 \cdots d_m \overline{d_{m+1} \cdots d_{m+p}}.$$∎

Intuition

The proof is a counting argument. Long division has only $b$ possible remainders, so a remainder must eventually repeat, and from that moment on the computation repeats as well. If the remainder $0$ ever appears, the expansion terminates. Otherwise only the $b - 1$ remainders $1, \dots, b - 1$ are available, so the repeating block has at most $b - 1$ digits. For $\frac17$ it has exactly $6$.

Remark.

Given $r_{k-1}$, the uniqueness of $r_k$ and $d_k$ follows from the so-called division algorithm, which is really a theorem showing that division never breaks down: if $b$ is a positive integer and $m \in \mathbb{Z}$, there are unique integers $q$ and $r$ with $m = bq + r$ and $0 \le r < b$. Here $q$ is the quotient and $r$ the remainder. With $m = 10 \cdot r_{k-1}$, naming $q = d_k$ and $r = r_k$ gives formula (2) in the proof.

Irrational numbers

Example 4.

The expansion

$$.101001000100001000001000000100000001\cdots$$

represents an irrational number. It cannot be repeating, because we arranged arbitrarily long blocks of $0$’s.

Example 5.

We do not know the complete decimal expansions of $\sqrt2$, $\sqrt3$ and many other familiar irrationals, but by the last theorem we know they cannot be repeating.

Example 6.

We have claimed $\pi$ and $e$ are irrational. These and many other facts are proved in a fascinating book by Ivan Niven.

1. $e = \sum_{k=0}^\infty \frac1{k!}$ is irrational. Assume $e = \frac{a}{b}$ with $a, b \in \mathbb{N}$. Then $(b+1)! \cdot e$ and $(b+1)! \sum_{k=0}^{b+1} \frac1{k!}$ are both integers, so the difference $(b+1)! \sum_{k=b+2}^\infty \frac1{k!}$ is a positive integer. But this number is less than $$\frac1{b+2} + \frac1{(b+2)^2} + \cdots = \frac{\frac1{b+2}}{1 - \frac1{b+2}} = \frac1{b+1} < 1,$$ a contradiction.
2. $\pi^2$ is irrational, from which the irrationality of $\pi$ follows (Exercise 16.10). The proof is sketched below.
3. It is even harder to prove $\pi$ and $e$ are not algebraic numbers (Definition 2.1); see Niven’s book, Theorems 2.12 and 9.11.

Proof that π² is irrational

The key is the sequence of integrals

$$I_n = \int_0^\pi P_n(x) \sin x\,dx, \qquad P_n(x) = \frac{(x(\pi - x))^n}{n!}.$$

Claim 1. There are polynomials $(Q_n)_{n=0}^\infty$ with integer coefficients, of degree at most $n$, such that $I_n = Q_n(\pi^2)$ for all $n$.

First we find a recursion for $I_n$. Integrating by parts (Theorem 34.2) twice,

$$I_n = \int_0^\pi P_n(x) \sin x\,dx = -\int_0^\pi P_n''(x) \sin x\,dx \qquad\text{for } n \ge 1. \tag{1}$$

Indeed,

$$\int_0^\pi P_n(x) \sin x\,dx = [-P_n(\pi)\cos\pi + P_n(0)\cos 0] + \int_0^\pi P_n'(x) \cos x\,dx = \int_0^\pi P_n'(x) \cos x\,dx,$$

since $P_n$ has a factor $x(\pi - x)$, which vanishes at $0$ and $\pi$. Therefore

$$I_n = [P_n'(\pi) \sin\pi - P_n'(0) \sin 0] - \int_0^\pi P_n''(x) \sin x\,dx = -\int_0^\pi P_n''(x) \sin x\,dx,$$

which is (1). For $n \ge 2$,

$$P_n'(x) = \frac{(x(\pi - x))^{n-1}}{(n-1)!} (\pi - 2x) = P_{n-1}(x)(\pi - 2x),$$

so, using this for $n - 1$ and the product rule,

$$P_n''(x) = P_{n-2}(x)(\pi - 2x)^2 + P_{n-1}(x)(-2) = \pi^2 P_{n-2}(x) + P_{n-2}(x)(-4\pi x + 4x^2) - 2P_{n-1}(x).$$

Since $P_{n-2}(x)(-4\pi x + 4x^2) = -4 \cdot \frac{(x(\pi-x))^{n-1}}{(n-2)!} = -4(n-1) P_{n-1}(x)$, we conclude

$$P_n''(x) = \pi^2 P_{n-2}(x) - (4n - 2) P_{n-1}(x) \qquad\text{for } n \ge 2. \tag{2}$$

Now $I_0 = 2$ is clear, and by (1) with $P_1'' = -2$, $I_1 = 4$. From (1) and (2),

$$I_n = -\pi^2 I_{n-2} + (4n - 2) I_{n-1} \qquad\text{for } n \ge 2. \tag{3}$$

Claim 1 follows by induction with $Q_0 = 2$, $Q_1 = 4$, and $Q_n(x) = -x Q_{n-2}(x) + (4n - 2) Q_{n-1}(x)$ for $n \ge 2$.

Claim 2. $\pi^2$ is irrational. Suppose $\pi^2 = a/b$. By Claim 1, each $b^n I_n = b^n Q_n(a/b)$ is an integer. Since $x(\pi - x)$ is largest at $x = \pi/2$,

$$0 < b^n I_n = b^n \int_0^\pi \frac{(x(\pi - x))^n}{n!} \sin x\,dx < b^n \int_0^\pi \frac{(\frac\pi2 \cdot \frac\pi2)^n}{n!}\,dx = \frac{\left(\frac{b\pi^2}{4}\right)^n}{n!}\,\pi.$$

The right side tends to $0$ (Exercise 9.15), so for large $n$ the integer $b^n I_n$ lies in $(0, 1)$, a contradiction.

This simplification of Niven’s famous short proof (1947) is due to Zhou and Markov, who use a similar technique to prove $\tan r$ is irrational for nonzero rational $r$, and $\cos r$ is irrational if $r^2$ is a nonzero rational.

∎

Intuitionwhy the proof that e is irrational works

The partial sums $\sum_{k=0}^{b+1} \frac1{k!}$ approximate $e$ too well for $e$ to be a fraction with denominator $b$. Multiplying by $(b+1)!$ turns those partial sums into integers, while the remaining error becomes a positive number less than $1$. A fraction $\frac ab$, multiplied by $(b+1)!$, would also be an integer, and two integers cannot differ by something strictly between $0$ and $1$.

Example 7.

Euler introduced a famous number, arising in the study of the gamma function, called Euler’s constant:

$$\gamma = \lim_{n\to\infty} \left[\left(\sum_{k=1}^n \frac1k\right) - \log_e n\right].$$

Even though $\lim \sum_{k=1}^n \frac1k = +\infty$ and $\lim \log_e n = +\infty$, the limit defining $\gamma$ exists and is finite (Exercise 16.9); $\gamma \approx 0.577216$. The amazing fact is that no one knows whether $\gamma$ is rational. Most mathematicians believe it is irrational, because it is “easier” for a number to be irrational: repeating expansions are very regular. The remark in Exercise 16.8 hints at another reason.

Reading alongside Abbott

Abbott uses decimal expansions in Section 1.6 for Cantor’s diagonal argument that $(0, 1)$ is uncountable. Here is the idea in brief. Given any list $x_1, x_2, x_3, \dots$ of numbers in $(0, 1)$, build a new number whose $n$th digit differs from the $n$th digit of $x_n$, and never use the digits $0$ or $9$. The new number cannot equal any $x_n$. Theorem 16.3 is exactly what makes the last step safe: an expansion with no $0$’s or $9$’s is the only expansion of its number, so a different digit really does mean a different number. The uncountability of the Cantor set in §13 can be proved the same way, with base-$3$ digits $0$ and $2$.

Try it: the diagonal argument

A list of numbers in (0,1) with the diagonal digits highlighted

Number	1	2	3	4	5	6	7	8	9
x1 = 0.	5	0	0	0	0	0	0	0	0
x2 = 0.	3	3	3	3	3	3	3	3	3
x3 = 0.	2	5	0	0	0	0	0	0	0
x4 = 0.	2	0	0	0	0	0	0	0	0
x5 = 0.	1	6	6	6	6	6	6	6	6
x6 = 0.	1	4	2	8	5	7	1	4	2
x7 = 0.	1	2	5	0	0	0	0	0	0
y = 0.	4	5	5	5	5	5	5	·	·

The highlighted diagonal digits are changed to 5 (or to 4 when the digit is already 5), giving y = 0.4555555…. Digit n of y differs from digit n of x_n, and y uses only the digits 4 and 5, so by Theorem 16.3 it has no second expansion: y ≠ x_n for every n. No list can contain all of (0, 1).

Check your understanding

Take the quiz (3 questions)

Which number has two different decimal expansions?

$\frac12 = 0.5000\ldots = 0.4999\ldots$$\frac13 = 0.333\ldots$$\sqrt 2$

Which number is irrational?

$0.101001000100001\ldots$ (the runs of $0$’s keep growing)$0.123123123\ldots$$0.1666\ldots$

What fraction is $0.\overline{36} = 0.363636\ldots$?

$\frac{4}{11}$$\frac{36}{100}$$\frac{9}{25}$

Exercises

Exercise 16.1

1. Show $2.74\overline{9}$ and $2.75\overline{0}$ are both decimal expansions for $\frac{11}4$.
2. Which of these expansions arises from the long division process in 16.1?

Solution

(b) $2.75\overline{0}$: long division never produces an infinite tail of $9$’s, since $d_n = 9$ forever would give $r_n - b = 10(r_{n-1} - b)$, pushing $r_n$ out of $[0, b)$.

Tags: Warm-upComputation

Exercise 16.2

Verify the claims in the first paragraph of the proof of Theorem 16.3.

Solution

Recall $\sum_{j=1}^\infty 9\cdot10^{-j} = 1$ and, more generally, $\sum_{j>r}9\cdot10^{-j} = 10^{-r}$ (geometric series).

Case $K.000\cdots$ with $K > 0$. The expansion $(K-1).999\cdots$ is a legitimate expansion (the digits $9$ are allowed and $K - 1 \ge 0$) with value $(K-1) + 1 = K$. It differs from $K.000\cdots$, so $x$ has two expansions.

Case $K.d_1\cdots d_r000\cdots$ with $d_r \ne 0$. Then $d_r - 1 \ge 0$ is a legitimate digit, and the expansion $K.d_1\cdots d_{r-1}(d_r - 1)999\cdots$ has value $K + \sum_{j<r}d_j10^{-j} + (d_r - 1)10^{-r} + 10^{-r} = K + \sum_{j \le r}d_j10^{-j} = x .$ It differs from the original in the $r$th digit, so these are two distinct expansions.

Tags: StandardFrom the definitions

Used in Theorem 16.3

Exercise 16.3

Suppose $\sum a_n$ and $\sum b_n$ are convergent series of nonnegative numbers. Show that if $a_n \le b_n$ for all $n$ and $a_n < b_n$ for at least one $n$, then $\sum a_n < \sum b_n$.

Hint

Apply Exercise 14.5 to $\sum (b_n - a_n)$, a series of nonnegative terms with at least one positive term.

Solution

Since $a_n \le b_n$, the series $\sum(b_n - a_n)$ has nonnegative terms, and it converges with sum $\sum b_n - \sum a_n$ (Exercise 14.5). Pick $N$ with $b_N - a_N > 0$. The partial sums of $\sum(b_n - a_n)$ are nondecreasing and are at least $b_N - a_N$ once $n \ge N$, so the sum is $\ge b_N - a_N > 0$. Hence $\sum b_n - \sum a_n > 0$, i.e. $\sum a_n < \sum b_n$.

Tags: StandardSeries tests

Used in Theorem 16.3

Exercise 16.4

Write the following repeating decimals as rationals, i.e. as fractions of integers.

1. $.\overline{2}$
2. $.\overline{02}$
3. $.0\overline{2}$
4. $3.\overline{14}$
5. $.\overline{10}$
6. $.\overline{1492}$

Solution

(a) $\frac29$ (b) $\frac2{99}$ (c) $\frac1{45}$ (d) $\frac{311}{99}$ (e) $\frac{10}{99}$ (f) $\frac{1492}{9999}$

Tags: Warm-upComputation

Exercise 16.5

Find the decimal expansions of the following rational numbers.

1. $\frac18$
2. $\frac1{16}$
3. $\frac23$
4. $\frac79$
5. $\frac6{11}$
6. $\frac{22}7$

Solution

(a) $.125$ (b) $.0625$ (c) $.\overline{6}$ (d) $.\overline{7}$ (e) $.\overline{54}$ (f) $3.\overline{142857}$. Check them with the widget above.

Tags: Warm-upComputation

Exercise 16.6

Find the decimal expansions of $\frac17, \frac27, \frac37, \frac47, \frac57$ and $\frac67$. Note the interesting pattern.

Solution

They are $.\overline{142857}$, $.\overline{285714}$, $.\overline{428571}$, $.\overline{571428}$, $.\overline{714285}$, $.\overline{857142}$: all cyclic shifts of the same six digits. Try them in the widget and watch the remainders.

Tags: Warm-upComputation

Exercise 16.7

Is $.1234567891011121314151617181920212223242526\cdots$ rational?

Solution

No. The expansion contains arbitrarily long blocks of $0$’s (from $10^k$), so as in Example 4 it cannot be repeating.

Tags: StandardExplain why

Exercise 16.8

Let $(s_n)$ be a sequence of numbers in $(0, 1)$. Each $s_n$ has a decimal expansion $0.d_1^{(n)} d_2^{(n)} d_3^{(n)} \cdots$. For each $n$, let $e_n = 6$ if $d_n^{(n)} \ne 6$ and $e_n = 7$ if $d_n^{(n)} = 6$. Show $.e_1 e_2 e_3 \cdots$ is the decimal expansion of some $y \in (0, 1)$ and $y \ne s_n$ for all $n$.

Remark: this shows the elements of $(0, 1)$ cannot be listed as a sequence: $(0, 1)$ is uncountable. Since $\mathbb{Q} \cap (0, 1)$ can be listed as a sequence, there are a lot of irrational numbers in $(0, 1)$!

Hint

$y$ has no $0$’s or $9$’s, so by Theorem 16.3 it has only one expansion, and it differs from that of $s_n$ in the $n$th digit.

Solution

Each $e_n$ is $6$ or $7$, so $\sum e_n10^{-n}$ converges by comparison with $\sum 7\cdot10^{-n}$; call its sum $y$. Then $y \ge \sum 6\cdot10^{-n} = \frac23 > 0$ and $y \le \sum7\cdot10^{-n} = \frac79 < 1$, so $y \in (0,1)$ and $.e_1e_2e_3\cdots$ is a decimal expansion of $y$.

The expansion of $y$ contains no $0$’s or $9$’s, so it does not end in all $0$’s or all $9$’s; by Theorem 16.3 it is the only decimal expansion of $y$. If $y = s_n$ for some $n$, then $.d^{(n)}_1d^{(n)}_2\cdots$ would be another expansion of $y$, hence identical to $.e_1e_2\cdots$, so $d^{(n)}_n = e_n$. But $e_n \ne d^{(n)}_n$ by construction. So $y \ne s_n$ for every $n$.

Tags: ChallengeContradiction

Exercise 16.9

Let $\gamma_n = \left(\sum_{k=1}^n \frac1k\right) - \log_e n = \sum_{k=1}^n \frac1k - \int_1^n \frac1t\,dt$.

1. Show $(\gamma_n)$ is a decreasing sequence.
2. Show $0 < \gamma_n \le 1$ for all $n$.
3. Observe that $\gamma = \lim_n \gamma_n$ exists and is finite.

Hint

For (a), look at $\gamma_n - \gamma_{n+1} = \int_n^{n+1} \frac1t\,dt - \frac1{n+1}$. For (b), compare with the rectangles in Example 1 of §15.

Solution

1. $\gamma_n - \gamma_{n+1} = \left[\sum_{k \le n}\frac1k - \log n\right] - \left[\sum_{k \le n+1}\frac1k - \log(n+1)\right] = \int_n^{n+1}\frac{dt}t - \frac1{n+1}$. On $[n, n+1]$ we have $\frac1t \ge \frac1{n+1}$, so the integral is at least $\frac1{n+1}$, and $\gamma_n - \gamma_{n+1} \ge 0$. So $(\gamma_n)$ is decreasing.
2. By Example 1 of §15, $\sum_{k=1}^n\frac1k \ge \log(n+1) > \log n$, so $\gamma_n > 0$. For the upper bound, on $[k-1, k]$ we have $\frac1k \le \frac1t$, so $\sum_{k=2}^n\frac1k \le \int_1^n\frac{dt}t = \log n$. Hence $\sum_{k=1}^n\frac1k \le 1 + \log n$ and $\gamma_n \le 1$.
3. A decreasing sequence that is bounded below converges (Theorem 10.2), so $\gamma = \lim\gamma_n$ exists, and $0 \le \gamma \le 1$.

Tags: ChallengeMonotone & CauchySeries tests

Exercise 16.10

In Example 6(b) we showed $\pi^2$ is irrational. Use this to show $\pi$ is irrational. What can you say about $\sqrt\pi$ and $\sqrt[3]\pi$? $\pi^4$?

Solution

If $\pi$ were rational so would be $\pi^2$. Likewise $\sqrt\pi$ and $\sqrt[3]\pi$ are irrational (their squares or cubes are $\pi$). Nothing follows about $\pi^4$ from these facts alone.

Tags: Warm-upContradiction