Rudin 【Principle of Mathematical Analysis】Notes & Solutions

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1.17 Definition An ordered field is a field $F$ with an order
$\quad$associated such that.
$\quad$(i) $\ \forall x,y,z\in F\;(x < y)\implies (x+z < y+z)$
$\quad$(ii) $\forall x,y\in F\; (x>0)\wedge (y>0)\implies (xy > 0)$
If $x>0$, we call $x$ possitive; if $x < 0$, we call $x$ negative.

$\mathbb{Q}$ is an example of ordered field.

1.18 Proposition Let $F$ be an ordered field,
$\quad x,y,z\in F$ then
(a) $(x > 0) \Leftrightarrow (-x < 0)$
(b) $(x > 0)\wedge (y < z)\implies (xy < xz)$
(c) $(x < 0)\wedge (y < z)\implies (xy > xz)$
(d) $(x\ne 0)\implies (x^2 > 0)$. In particular, $1 > 0$
(e) $(0 < x < y)\implies (0 < 1/y < 1/x)$
Proof
(a) $(x > 0)\implies (0 = -x + x > -x + 0 = -x)$,
$\quad$ inversely,$\;(x < 0)\implies (0 = -x+x < -x+0)$
(b) $(x>0)\wedge (z > y)\implies (x>0)\wedge (z-y > y-y = 0)$
$\quad\;$ thus $xy + 0 < xy + x(z-y) = xz$
(c) By (a),(b) and Proposition 1.16c,
$\quad\;\;-(x(z-y)) = (-x)(z-y) > 0$
$\quad$ so $x(z-y)< 0,\;xz < xy.$
(d) $x > 0\implies x^2>0\,$(Def. 1.17(ii)), if $x < 0$ then
$\quad -x > 0,\; x^2 = (-x)^2 > 0$ by (a) & 1.16(d)
(e) since $((x>0)\wedge (1/x\le 0)) \implies 1 = x(1/x) \le 0$
$\quad$ contradicts to (d), $1/x > 0.$ likewise $1/y > 0$
$\quad$ therefore $((1/x)(1/y) > 0) \wedge (x < y)\implies$
$\quad\;1/y = x(1/x)(1/y) < y(1/x)(1/y) = 1/x.$

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The Real Field
The core of the chapter: the existence theorem (of Real Field)
1.19 Theorem There exists an ordered field $\mathbb{R}$ which has the
$\quad$least-upper-bound propery. And $\mathbb{R}$ contains $\mathbb{Q}$ as a subfield.

The 2nd statement means that $\mathbb{Q}\subset \mathbb{R}$ and the addition and multiplication in $\mathbb{R}$, when applied to members of $\mathbb{Q}$, coincide with the usual operations on $\mathbb{Q}$; also, the positive rational numbers are positive elements of $\mathbb{R}$

The members of $\mathbb{R}$ are called real numbers. The proof of the theorem, actually constructs $\mathbb{R}$ from $\mathbb{Q}$, will presented as an Appendix to Chap.1.

We'd like to show how the least-upper-bound property implies the following:

1.20 Theorem
(a) $\forall x\in\mathbb{R}^+,\forall y\in\mathbb{R}\; \exists n\in\mathbb{N}\quad (nx > y)$
(b) $ (x,y \in\mathbb{R})\wedge (x < y) \implies (x,y)\cap \mathbb{Q} \ne \varnothing $
Proof
$\quad$(a) If no such an $n$, then $A = {nx | n\in\mathbb{N}}$ has upper bound $y$ thus $\alpha = \sup A$ exists in $\mathbb{R}$. Since $x>0$, $\alpha - x$ is not an upper bound of A hence $\alpha -x < mx \le \alpha$ for some $m\in\mathbb{N}$, but then $\alpha < (m+1)x$ and so $\alpha$ cannot be an upper bound of $A$.
$\quad$(b) Take(by(a)) $n\in\mathbb{N}$ such that $n(y-x) > 1$. Likewise we have $m_1,m_2\in\mathbb{N}$ such that $m_1>nx,\; m_2 > -nx$ i.e. $-m_2 < nx < m_1$. Therefore there is an integer $m$ such that $m-1\le nx < m$ (with $-m_2 < m \le m_1$) Combine all these we see that
$nx < m\le 1+nx < n(y-x)+nx = ny$. Since $n$ is positive, we get $x < \frac{m}{n} < y\quad\square$

Now the existence of $n$th roots of positive reals

1.21 Theorem $\forall x\in\mathbb{R}^+ \ \forall n\in\mathbb{N}^+\; \exists ! y\in\mathbb{R}^+\; (y^n = x)\quad$ [We write $y = \sqrt[n]{x}$ or $y = x^{1/n}$]
Proof The uniqueness of $n$th root is obvious since $0< y_1 < y_2 \implies y_1^n < y_2^n$
Let $t = x/(1+x)$, then $0 < t < \min(x, 1)$ thus $t^n \le t < x$ and so $E = \{t \in\mathbb{R}: t > 0, t^n < x \} \ne \varnothing$
Clearly that $1+x$ is an upper bound of $E$ and so $y = \sup E$ exists.
We need the inequality $b^n - a^n < n(b - a)b^{n - 1}\quad (0 < a < b)$.
If $y^n < x$, choose $h\in (0,1)$ such that $h < \frac{x - y^n}{n(y+1)^{n -1}}$, then for $a = y, b= y +h$ the inequality give
$\;\small(y +h)^n - y^n < nh(y +h)^{n -1} < nh(y+1)^{n -1} < x -y^{n -1}$
and so $y$ cannot be $\sup E$
If $y^n > x$, let $k = \frac{y^n -x}{ny^{n -1}}$, then $0< k< y$ and for $t \ge y -k$, we have
$\quad\quad y^n - t^n \le y^n -(y -k)^n < nky^{n -1}=y^n -x$ and so $(t\ge y -k)\implies (t^n > x)) $
$\quad\quad y -k (< y)$ is an upper bound of $E$ thus $y\ne \sup E$
Now we see that $y = x^n\quad\quad\square$

Corollary If $a$ and $b$ are positive real numbers and $n$ is a positive integer, then\[(ab)^{1/n} = a^{1/n} b^{1/n}\quad\quad\quad \]Proof Put $\alpha = a^{1/n}, \beta = b^{1/n}$, since multiplication is commutative, we have $ab = \alpha^n \beta^n = (\alpha\beta)^n$
Now the uniqueness of $n$th root implies that $(ab)^{1/n} = \alpha\beta = a^{1/n} b^{1/n}$

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1.22 Decimals Let $0< x\in\mathbb{R}$, let $n_0$ be the largest integer no more than $x$. Notice that the existence of $n_0$ relies on the archimedean property of $\mathbb{R}$. Having chosen $n_0,\cdots,n_{k-1}$, let $n_k$ be the largest integer such that $\displaystyle{n_0 + \frac{n_1}{10}+\cdots +\frac{n_k}{10^k}\le x} $ and let $\displaystyle{E = \{n_0 + \frac{n_1}{10}+\cdots +\frac{n_k}{10^k} \; | \; k\in \mathbb{N}\}}$
The decimal expansion of x is $n_0.n_1 n_2 n_3 \cdots.$ (Notice that $0\le n_k < 10\quad(\forall k>0)$)
Conversely, for any infinite decimal, the corresponding $E$ is bounded above and the infinite decimal is the decimal expansion of $\sup E$.

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1.23 Definition The extended real number system consists of the real field $\mathbb{R}$ and two symbols, $+\infty$ and $-\infty$. Rreserving the original order in $\mathbb{R}$ and define\[-\infty < x < +\infty\quad (\forall x\in\mathbb{R}) \] $\quad\quad$ It's clear that $+\infty$ is an upper bound of every subset of the extended real number system, and every nonempty subset has a least upper bound. We have similar remarks for lower bounds in this system.

The extended real number system does not form a field. But we have the following conventions:
(a) $\displaystyle{x + \infty = +\infty,\;\;x-\infty = -\infty,\;\;\frac{x}{+\infty} = \frac{x}{-\infty} = 0\;(\forall x\in\mathbb{R})}$
(b) $ x\cdot (+\infty) = +\infty,\quad x\cdot (-\infty) = -\infty\quad (0 < x\in\mathbb{R})$
(c) $ x\cdot (+\infty) = - \infty,\quad x\cdot (-\infty) = +\infty\quad (0 > x\in\mathbb{R})$

When it is desired to make the distinction between real numbers on the one hand and the symbols $\pm \infty$ on the other quite explicit, the former are called finite.

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1.24 Definition A complex number is an ordered pair $(a,b)$
$\quad$ of real numbers. "Ordered" means that $(a,b)$ and $(b,a)$
$\quad$ are regarded as distinct if $a\ne b$
$\quad$ Let $x=(a,b),\; y=(c,d)$ be two complex numbers. We
$\quad$ write $x=y$ iff $(a=c)\wedge(b=d)$. (It's not so superfluous:
$\quad$ think of equality of rational numbers, represented as
$\quad$ quotients of integers.) We define \[x+y=(a+c,b+d),\; xy = (ac-bd,ad+bc) \] 1.25 Theorem These definition of addition and multiplication turn the set of all complex numbers $\mathbb{C}$ into a field, with $(0,0)$ and $(1,0)$ in the rule of $0$ and $1$.
Proof We simply verify the field axioms in Definition 1.12 using the field structure of $\mathbb{R}$ of course.
Let $x = (a,b),\; y=(c,d),\; z=(e,f)$, for example, if $x \ne 0 = (0,0)$, then $a^2+b^2 > 0$ by 1.18d, and we can define $\displaystyle{\frac{1}{x} = \left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2} \right )}$ since $\displaystyle{(a,b)\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right ) = (1,0) = 1}$. That's $(M_5)$

For verify (D) in Definition 1.12, we have \[\begin{align}
x(y+z) & = (a,b)(c+e,d+f) \\
& = (ac+ae-bd-bf,ad+af+bc+be) \\
& = (ac-bd,ad+bc)+(ae-bf,af+be) \\
& = xy + xz. \end{align} \] 1.26 Theorem $(a,0)+(b,0) = (a+b,0),\;(a,0)(b,0) = (ab,0)\,\small(\forall x,b\in\mathbb{R}).\;\square$

Theorem 1.26 shows that the complex numbers of the form $(t,0)$ have the same arithmetic properties as the corresponding real numbers $t$. We can therefore identify $(t,0)$ with $t$. This identification gives us the real field as a subfield of the complex field.

1.27 Definition $i = (0,1)$

1.28 Theorem $i^2 = -1$
Proof $i^2 = (0,1)(0,1) = (-1,0) = -1.\quad\quad\square$

1.29 Theorem $(a,b) = a + bi\quad (\forall a,b\in\mathbb{R})$
Proof $a+bi = (a,0)+(b,0)(0,1)=(a,0)+(0,b)=(a,b).\;\;\square$

1.30 Definition If $a,b\in\mathbb{R}$ and $z = a+bi$, then $\overline{Z} = a - bi$ is called the conjugate of $z$. The numbers $a$ and $b$ are the real part and the imaginary part of $z$, respectively.
$\quad\quad$ We shall occasionally write $a = \operatorname{Re} (z),\quad b = \operatorname{Im}(z).$

1.31 Theorem If $z$ and $w$ are complex, then
$\quad\quad$(a) $\overline{z+w} = \overline{z} + \overline{w}$
$\quad\quad$(b) $\overline{zw} = \overline{z} \overline{w}$
$\quad\quad$(c) $z + \overline{z} = 2\operatorname{Re}(z),\; z -\overline{z} = 2\operatorname{Im}(z)$
$\quad\quad$(d) $z\overline{z}$ is real and positive (except when $z = 0$)$\quad\square$

1.32 Definition The absolute value $|z|\; (z\in\mathbb{C})$ is a non-negative square root: $|z|=(z\overline{z})^{1/2}$
$\quad\quad$ The existence (and uniqueness) of $|z|$ follows from Theorem 1.21 and Theorem 1.31d.
$\quad\quad$ Note that when $x \in \mathbb{R}\subset \mathbb{C}$, then $\overline{x} = x$, hence $|x| = \sqrt{x^2} = x$ if $x\ge 0$, $|x| = -x$ if $x < 0$

1.33 Theorem Let $z$ and $w$ be complex numbers. Then
$\quad\quad$ (a) $|z| > 0$ unless $z = 0,\; |0| = 0$
$\quad\quad$ (b) $|\overline{z}| = |z|$
$\quad\quad$ (c) $|zw| = |z||w|$
$\quad\quad$ (d) $|\operatorname{Re} z| \le |z|$
$\quad\quad$ (e) $|z+w| < |z|+|w|$
Proof (a) and (b) are trivial. Put $z = a+bi,\; w = c+di$, with $a,b,c,d\in\mathbb{R}$.
$\quad$ Then $|zw|^2 = (ac-bd)^2+(ad+bc)^2 = (a^2+b^2)(c^2+d^2)$
$\qquad=|z|^2|w|^2 = (|z||w|)^2$. Now (c) follows from the uniqueness assertion of Theorem 1.21.
$\quad\quad$ For (d) we have $(a^2 \le a^2+b^2) \implies (|a|\le \sqrt{a^2+b^2})$
$\quad\quad$ For (e) we have \[\begin{align}|z+w|^2 &=(z+w)(\overline{z}+\overline{w})\\
&= z\overline{z}+z\overline{w}+\overline{z}w+w\overline{w}\\
&= |z|^2 + 2\operatorname{Re}(z\overline{w}) + |w|^2\\
&\le |z|^2 + 2|z\overline{w}| + |w|^2\\
&= |z|^2 + 2|z||w| + |w|^2 = (|z|+|w|)^2. \end{align}\] 1.34 Notation If $x_1,\cdots,x_n$ are complex numbers, we write \[x_1+x_2+\cdots +x_n = \sum_{j=1}^n x_j\] $\quad\quad$ We conclude the section with an important inequality, usually known as the Schwarz inequality.

1.35 Theorem If $a_1,\cdots,a_n,\; b_1,\cdots,b_n \in \mathbb{C}$ then \[ \left|\sum_{j=1}^n a_j \overline{b}_j\right|^2 \le \sum_{j=1}^n |a_j|^2\sum_{j=1}^n |b_j|^2 \] Proof$\quad$ Put $A = \sum |a_j|^2,\; B = \sum |b_j|^2,\; C = \sum a_j\overline{b}_j$ (for this proof, $j = \overline{1,n}$ in all sums). If $B = 0$, then $b_j = 0\quad (j=\overline{1,n})$ and the conclusion is trivial. Assume $B > 0$. By Theorem 1.31 we then have
$\quad\sum |Ba_j - Cb_j|^2 = \sum (Ba_j - Cb_j)(B\overline{a}_j - \overline{C b_j})$
$\qquad= B^2 \sum |a_j|^2 - B\overline{C}\sum a_j\overline{b}_j - BC\sum \overline{a}_j b_j +|C|^2 \sum |b_j|^2$
$\qquad= B^2 A - B|C|^2 = B(BA - |C|^2)$
$\quad\quad$ Since $B > 0$, we see that $AB - |C|^2 \ge 0$. this is the desired inequality. $\quad\square$

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1.36 Definitions For each positive integer $k$, let $\mathbb{R}^k$ be the set of all ordered $k$-tuples \[\mathbf{x} = (x_1,x_2,\cdots,x_k)\] Where $x_1,\cdots,x_k\in\mathbb{R}$, called the coordinates of $\mathbf{x}$. The elements of $\mathbb{R}^k$ are called points, or vectors, especially when $k>1$. We shall denote vectors by boldfaced letters. If $\mathbf{y} = (y_1,\cdots,y_k)$ and $\alpha \in\mathbb{R}$, put \[\mathbf{x} + \mathbf{y} = (x_1+y_1,\cdots,x_k+y_k), \quad
\alpha \mathbf{x} = (\alpha x_1,\cdots,\alpha x_k) \] So $\mathbf{x}+\mathbf{y},\; \alpha \mathbf{x} \in\mathbb{R}^k$. This defines addition of vectors, as well as multiplication of a vector by a real number (a scalar). These two operations satisfy the commutative, associative and distributive laws (trivial) and make $\mathbb{R}^k$ into a vector space over the real field. The zero element of $\mathbb{R}^k$ (sometimes called the origin or the null vector) is the point $\mathbf{0}$ with all the coordinates are $0$
$\quad\quad$ We shall define the so called "inner product" (or scalar product) of $\mathbf{x}$ and $\mathbf{y}$ by \[\mathbf{x}\cdot \mathbf{y}=\sum_{i=1}^k x_i y_i \] and the norm of $\mathbf{x}$ by \[ |\mathbf{x}| = (\mathbf{x}\cdot \mathbf{x})^{1/2} = \left(\sum_{i=1}^k x_i^2 \right )^{1/2}\] $\quad\quad$ The structure now defined (the vector space $\mathbb{R}^k$ with the above inner product and norm) is called euclidean $k$-space.

1.37 Theorem Let $\mathbf{x,y,z} \in \mathbb{R}^k$ and $\alpha \in\mathbb{R}$, Then
$\quad\quad$ (a) $|\mathbf{x}| \ge 0$
$\quad\quad$ (b) $|\mathbf{x}| = 0 \Longleftrightarrow \mathbf{x} = \mathbf{0} $
$\quad\quad$ (c) $|\alpha \mathbf{x}| = |\alpha||\mathbf{x}|$
$\quad\quad$ (d) $|\mathbf{x}\cdot \mathbf{y}| \le |\mathbf{x}||\mathbf{y}|$
$\quad\quad$ (e) $|\mathbf{x}+\mathbf{y}| \le |\mathbf{x}|+|\mathbf{y}|$
$\quad\quad$ (f) $|\mathbf{x}-\mathbf{z}| \le |\mathbf{x}-\mathbf{y}|+|\mathbf{y}-\mathbf{z}|$
Proof (a),(b) and (c) are obvious and (d) is an immediate consequence of the Schwarz inequality. Then by (d) we have \[ \begin{align} |\mathbf{x}+\mathbf{y}|^2 & = (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y})\\
& =\mathbf{x\cdot x} + 2\mathbf{x}\cdot \mathbf{y}+\mathbf{y}\cdot\mathbf{y}\\
& \le |\mathbf{x}|^2 +2|\mathbf{x}||\mathbf{y}|+|\mathbf{y}|^2\\
& = (|\mathbf{x}|+|\mathbf{y}|)^2 \end{align}\] and so (e) is proved. Finally, (f) follows from (e) when $\mathbf{x,\;y}$ are replaced by $\mathbf{x-y,\; y-z}$ respectively.

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1.38 Remarks Theorem 1.37(a),(b) and (f) will allow us (see Chap.2) to regard $\mathbb{R}^k$ as a metric space.
$\quad\quad \mathbb{R}^1$ (the set of all real numbers) is usually called the line, or the real line. Likewise $\mathbb{R}^2$ is called the plane, or the complex plane (compare Definitions 1.24 and 1.36). In these two cases the norm is just the absolute value of the corresponding real or complex number.

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APPENDIX The proof of Theorem 1.19, i.e. constructing $\mathbb{R}$ from $\mathbb{Q}$ will presented in several steps.

Step 1 A set $\alpha$ is called a cut of $\mathbb{Q}$ if it is contained in $\mathbb{Q}$ with the following properties:
$\quad\quad \begin{align} \text{(I)} & \varnothing \ne \alpha \ne \mathbb{Q}\\
\text{(II)} & (p\in\alpha)\wedge (p> q\in\mathbb{Q}) \implies (q\in\alpha) \\
\text{(III)} & \forall p\in\alpha \exists r\in\alpha\; (p < r) \end{align}$
Let $\mathbb{R} = \{\alpha \mid \alpha \text{ is a cut of } \mathbb{Q}\}$, and assume that in this appendix, the letters $p,q,r,\cdots$ denote rational numbers and $\alpha,\beta,\gamma,\cdots$ denote cuts.
$\quad\quad$ (II) implies two facts which will be used freely:
$\quad\quad\quad\quad\quad (p\in\alpha)\wedge (q\not\in\alpha)\implies (p < q)\;$, and $(s > r\not\in\alpha) \implies (s\not\in\alpha)$
$\quad\quad$ The the axioms (M) and (D) of Definition 1.12 hold with $\mathbb{R}^+$ in place of $F$ and $1^*$ in the role of $1$.
$\quad\quad$ The proofs are so similar to the ones given in detail in Step 4 that we omit them.
$\quad\quad$ Note, in particular, the multiplication property of Definition 1.17 holds: $\alpha,\beta\in 0^* \implies \alpha\beta > 0^*$

Step 2 Define "$\alpha < \beta$" to mean: $\alpha \subsetneq \beta$ ($\alpha$ ia a proper subset of $\beta$)
$\quad\quad$ We need to show this meets the requirements of Definition 1.5.
$\quad\quad$ Clearly $(\alpha < \beta < \gamma \implies \alpha < \gamma)$ and at most one of $3$ relations $\alpha < \beta, \alpha = \beta, \beta < \alpha$ can hold for any pair of $\alpha,\beta$. To show at least one such relation holds, assume that first 2 relations fail, then $\alpha$ is not a subset of $\beta$ hence there is a $p\in\alpha$ with $p\not\in\beta$. So $q\in\beta \implies q < p \implies q\in\alpha$, i.e. $\beta < \alpha$
$\quad\quad$ Thus $\mathbb{R}$ is now an ordered set.

Step 3 $\mathbb{R}$ has the least-upper-bound property.
$\quad\quad$ Let $\varnothing \ne A \subset \mathbb{R}$, and $\beta\in\mathbb{R}$ is an upper bound of $A$. Let $\gamma = \bigcup A$, then $\varnothing \ne \gamma \subset \beta$ hence $\gamma \ne \mathbb{Q}$. It's easy to further prove that $\gamma$ is a cut (a member of $\mathbb{R}$) and an upper bound of $A$.
$\quad\quad$ Suppose $\delta < \gamma$, then $s\not\in\delta$ for some $s \in\gamma$. So $s\in\alpha$ for some $\alpha\in A$ and $\delta < \alpha$. Therefore $\delta$ is not an upper bound of $A$. We conclude that $\gamma = \sup A$.

Step 4 Define $\alpha + \beta = \{r+s \mid r\in\alpha,\; s\in\beta \}, \; 0^* = \{r \in\mathbb{Q}: r < 0 \}$. Clearly $0^*$ is a cut. We shall verify the axioms for addition (in Definition 1.12) hold in $\mathbb{R}$ with $0^*$ playing the role of $0$.
$\quad(A_1)$ Take ${r}',{s}' \in\mathbb{Q}\setminus (\alpha \cup \beta)$, then $\forall r\in\alpha\,\forall s\in\beta\,(r+s < {r}'+{s}') \implies ({r}'+{s}' \not\in \alpha+\beta \ne \varnothing)$ So $\alpha+\beta$ has cut property (I).
$\quad\quad$ If $p\in\alpha+\beta$ then $p = r+s$ for some $r\in\alpha,\;s \in\beta$. To see (II) holds, for $q < p$, we have $(q-s < r)\implies q-s\in\alpha \implies q = (q-s)+s \in \alpha +\beta$; Now choose $t\in\alpha$ so that $t > r$ then $p < t+s \in \alpha+\beta$. Thus (III) holds and so $\alpha + \beta \in\mathbb{R} (\forall \alpha,\beta\in\mathbb{R})$
$\quad(A_2) \;\alpha +\beta =\{t\mid t=r+s, r\in\alpha, s\in\beta\}= \beta+\alpha$
$\quad(A_3)$ This, similar to the above, follows from the associative law in $\mathbb{Q}$
$\quad(A_4)$ If $r\in\alpha,\; s\in 0^*$, then $r+s < r$ thus $\alpha + 0^*\subset \alpha$. On the other hand, if $p\in\alpha$, then $p < r \in\alpha$ for some $r$ so $p = r + (p-r)$ with $p-r\in 0^*$ and so $\alpha \subset \alpha+0^*$. We conclude that $\alpha + 0^* = \alpha$
$\quad(A_5)$ Fix $\alpha\in\mathbb{R}$, let $\beta = \{p\in\mathbb{Q}: \exists r > 0\; (-p -r \in\mathbb{Q}\setminus \alpha)\}$. We'll show $\beta\in\mathbb{R}$ and $\alpha+\beta = 0^*.\quad$ If $s\not\in\alpha$, take $p = -s + 1$ then $-p -1 = s\not\in\alpha$ hence $p\in\beta$; If $q\in\alpha$ then $-q \not\in\beta$ so $\varnothing \ne \beta \ne \mathbb{Q}$.
$\quad\quad$ For $p\in\beta$, pick $r > 0$ such that $-p-r \not\in\alpha$. If $q < p$, then $-q -r > -p -r$ hence $-q -r \not\in\alpha$ and $q\in\beta$. So (II) holds. Take $t = p + (r/2)$, then $-t -(r/2) = -p -r\not\in\alpha$. So $p < t \in\beta$, (III) holds. Therefore $\beta\in\mathbb{R}$.
$\quad\quad$ Let $r\in\alpha,\; s\in\beta$, then $-s\not\in\alpha \implies r < -s \implies r+s < 0$ so $\alpha+\beta \subset 0^*$
$\quad\quad$ Conversely, if $v\in 0^*$, let $w = -v/2$ then $w > 0$ and there is an integer $n$ such that $nw\in\alpha$ but $(n+1)w \not\in\alpha$ (by Archimedean property of \mathbb{Q}). Put $p = -(n+2)w$ then $-p-w = (n+1)w \not\in\alpha \implies p\in\beta$ and $v = -2w = nw + p\in\alpha + \beta$ thus $0^*\subset \alpha+\beta$.
$\quad\quad$ We conclude that $\alpha + \beta = 0^*$ and got the reason to denote $\beta$ as $-\alpha$. (see $(A_4)$)

Step 5 Having proved the addition defined in Step 4 satisfied Axioms (A) in Definition 1.12, Proposition 1.14 is thus valid in $\mathbb{R}$ and we can prove the addition property of ordered field (Definition 1.17):
$\quad\quad\quad \forall \alpha,\beta,\gamma \in\mathbb{R}\; (\beta < \gamma)\implies (\alpha + \beta < \alpha + \gamma)$.
$\quad\quad$ Indeed, it's clear in this case that $\alpha+\beta \subset \alpha+\gamma$ by the definition of $+$ in $\mathbb{R}$. If $\alpha+\beta = \alpha+\gamma$, then the cancellation law (Proposition 1.14) would imply $\beta = \gamma$.
$\quad\quad$ It follows that $\alpha > 0^*$ if and only if $-\alpha < 0^*$.

Step 6 Multiplication is a little more bothersome than addition in the present context. since products of negative rationals are positive. For this reason we confine ourselves first to $\mathbb{R}^+ = \{\alpha\in\mathbb{R}: \alpha > 0^*\}$
$\quad\quad$ For $\alpha,\beta \in \mathbb{R}^+$ define $\alpha\beta = \{p\in\mathbb{Q}: p < rs, (0 $\quad\quad$ The this axioms (M) and (D) of Definition 1.12 hold with $\mathbb{R}^+$ in place of $F$ and $1^*$ in the role of $1$.
$\quad\quad$ The proofs are so similar to the ones given in detail in Step 4 that we omit them.
$\quad\quad$ Note, in particular, the multiplication property of Definition 1.17 holds $\alpha,\beta > 0^* \implies \alpha\beta > 0^*$

Step 7 We complete the definition of multiplication by setting $\alpha 0^* = 0^* \alpha = 0^*$, and \[\alpha\beta =\begin{cases}
(-\alpha)(-\beta)& \text{ if } \alpha < 0^*,\;\beta < 0^* \\
-[(-\alpha)\beta]& \text{ if } \alpha < 0^*,\;\beta > 0^* \\
-[\alpha(-\beta)]& \text{ if } \alpha > 0^*,\;\beta < 0^*
\end{cases}\] The products on the right were defined in Step 6.
$\quad\quad$ Having proved (Step 6) that the axioms (M) hold in $\mathbb{R}^+$, it's perfectly simple to prove them in $\mathbb{R}$ by repeated application of the identity $\gamma = -(-\gamma)$ which is part of Proposition 1.14.
$\quad\quad$ The distributive law $\alpha(\beta+\gamma) = \alpha \beta + \alpha \gamma$ is proved in several cases. For instance, suppose $\alpha > 0^*,\; \beta < 0^*,\; \beta + \gamma > 0^*$, then $\gamma = (\beta+\gamma) + (-\beta)$ and (since we already have the distributive law in $\mathbb{R}^+$) $\alpha\gamma = \alpha(\beta+\gamma)+\alpha(-\beta)$. But $\alpha (-\beta) = -(\alpha\beta)$, we see that $\alpha(\beta+\gamma) = \alpha \beta + \alpha \gamma$ this case.
$\quad\quad$ The other cases are handled in the same way.
$\quad\quad$ We have now completed the proof that $\mathbb{R}$ is an ordered field with the least upper-bound property.

Step 8 For $r\in\mathbb{Q}$, let $r^* = \{p\in\mathbb{Q}: p $\quad\quad$ (a) $r^* + s^* = (r+s)^*$
$\quad\quad$ (b) $r^* s^* = (rs)^*$
$\quad\quad$ (c) $(r^* < s^*) \Longleftrightarrow (r < s)$

$\quad\quad$ To proof (a), assume $p\in (r^*+s^*)$, then $p = u+v$ for some $u < r,\; v < s$. So $p < r+s$ thus $p \in (r+s)^*$.
$\quad\quad$ Conversely, if $p\in (r+s)^*$, then $p < r+s$. Choose $t$ so that $2t = r+s -p$ and put ${r}' = r-t,\; {s}' = s-t$. Then ${r}'\in r^*,\; {s}'\in s^*$ and $p = {r}'+{s}'$, so $p\in (r^* + s^*)$.
$\quad\quad$ This proves (a). The proof of (b) is similar.
$\quad\quad$ (c): $r < s\implies (r\in s^*)\wedge (r\not\in r^*)\implies r^*< s^* \implies$
$\qquad\qquad \exists p\in s^* \; (p\not\in r^*)\implies (r\le p < s)\;\;\square$

Step 9 We saw in Step 8 that the correspondence "$r \to r^*$" preserves sums, products and order. This can be expressed by saying that the ordered field $\mathbb{Q}$ is isomorphic to the ordered field $\mathbb{Q}^*$ whose elements are the rational cuts.
$\quad\quad$ It is this identification of $\mathbb{Q}$ with $\mathbb{Q}^*$ which allows us to regard $\mathbb{Q}$ as a subfield of $\mathbb{R}$.
$\quad\quad$ The 2nd part of Theorem 1.19 is to be understood in terms of this identification.
$\quad\quad$ It is a fact, which we'll not prove here, that any two ordered fields with the least-upper-bound property are isomorphic. The 1st part of Theorem 1.19 therefore characterizes the real field $\mathbb{R}$ completely.

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$\quad\quad$ The book by Landau and Thurston cited inn the Bibliography are entirely devoted to number systems. Chap. 1 of Knopp's book contains a more leisurely description of how $\mathbb{R}$ can be obtained from $\mathbb{Q}$. Another construction defines real number as an equivalence class of Cauchy sequences of rational numbers(see Chap.3), is carried out in Sec. 5 of the book by Hewitt and Stromberg.
$\quad\quad$ The cuts in $\mathbb{Q}$ that we used here were invented by Dedekind. Rational Cauchy sequence equivalence classes' approach of construction is due to Cantor. Both Cantor and Dedekind published their constructions in 1872.

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Very condensed materials.