Rudin 【Principle of Mathematical Analysis】Notes & Solutions

10042010, 05:41 PM
Post: #1




Rudin 【Principle of Mathematical Analysis】Notes & Solutions
Index


10042010, 05:41 PM
Post: #2




Rudin [Principle of Mathematical Analysis] Notes
1.1 Example We now show that the equation
(1) $p^2 = 2$ is not satisfied by any rational $p$. $\quad$If there were such a $p$, we could write $p = m/n$ where $m,$ $\quad n(\in\mathbb{N}^+)$ are not both even. Assuming this is done, then $\quad$(1) implies (2) $m^2 = 2n^2$. This shows that $m^2$ is even. Hence $m$ is even $\quad$(if $m$ were odd, $m^2$ would be odd), and so $m^2$ is divisible $\quad$by $4$. It follows that the right side of (2) is divisible by $4$, so $\quad$that $n^2$ is even, which implies that n is even. $\quad$The assumption that (1) holds thus leads to the conclusion $\quad$that both $m$ and $n$ are even, contrary to our choice of $m$ $\quad$and $n$. Hence (1) is impossible for rational p. We now examine the situation a little more closely. Let $A$ be the set of all positive rationals $p$ such that $p^2 < 2$ and let $B$ consist of all positive rationals $p$ such that $p^2 > 2$. We shall show that A contains no largest number and $B$ contains no smallest. More explicitly, for every $p$ in $A$ we can find some rational $q$ in $A$ such that $p < q$, and for every $p$ in $B$ we can find a rational $q$ in $B$ such that $q < p$. Then (4) To do this, we associate with each rational p > 0 the number (3) $q = p\small\dfrac{p^22}{p+2} = \dfrac{2p+2}{p+2}$ Then (4) $q^2 2 =2\scriptsize\dfrac{p^22}{(p+2)^2}$ . If $p$ is in $A$ then $p^2 2 < 0$, (3) shows that $q > p$, and (4) shows that $q^2 < 2$. Thus $q\in A$. If $p\in B$ then $p^2 2> 0$, (3) implies $0 < q < p$, and (4) implies $q^2 > 2$. Thus $q\in B$. There is no problem to understand the text, but how the expression $q = p  \small\dfrac{p^22}{p+2}=\dfrac{2p+2}{p+2}$ been constructed? What's the thoughts behind? 

12152010, 02:32 PM
Post: #3




The idea of $q = p  \frac{p^22}{p+2} = \frac{2p+2}{p+2}$ in Example1.1
For each rational $p> 0\underset{\tiny\,}{,}$ we'd like to define a rational
$q > 0$ such that $0< \frac{q\sqrt{2}}{p\sqrt{2}}<1.\;$This implies that $(p^2>2 \underset{\tiny\,}{\iff} q^2>2)\wedge(q\sqrt{2}<p\sqrt{2})$ Let $q = p R(p)$, then $\frac{q\sqrt{2}}{p\sqrt{2}}=1\frac{R(p)}{p\sqrt{2}},\,$so we've $0<\frac{R(p)}{p\sqrt{2}}<1.\quad R(p)$ must be rational with the same sign as $p^22$, thus $R(p)=\frac{p^22}{Q(p)}\;$with some rational $Q(p).\,$Now $\frac{R(p)}{p\sqrt{2}}=\frac{p+\sqrt{2}}{Q(p)}\,$so $0<\frac{p+\sqrt{2}}{Q(p)}<1.\;$ Clearly the simplest $Q(p)$ meets the condition is $p+2$. and we then have $q=p\frac{p^22}{p+2}=\frac{2p+2}{p+2}>0\;\; (p>0).$ Here is a better try: We are trying to find a rational function $q = R(p)$ such that $R^2(x) 2 = \frac{k(x^2 2)}{g^2(x)}$ with rational $g(x)$ satisfies $g^2 > k > 0\;(x > 0).$ For simplicity let $g(x) \underset{\tiny\,}{=}x + a$ and so $R^2(x) = \frac{k(x^22)+2(x+a)^2}{(x+a)^2} = \frac{(k+2)x^2+4ax +2(a^2k)}{(x+a)^2}$. Since $k = a = 2$ makes the numerator a perfect square $4(x+1)^2$. We've found the simplest $R(x) = \small\dfrac{2x+2}{x+2}.$ In general, if $m\in\mathbb{N}^+$ is not a perfect square, then $({\small R}(x))^2 m = \frac{k(x^2m)}{(x+b)^2}$ has solution $k = b^2  m > 0,\;$ $b=\min\{j\in\mathbb{N}:\;j^2 > m\}$ thus ${\small R}(x) = \small\dfrac{bx +m}{x + b}$ i.e. the corresponding recursion $c_{n+1} = \frac{bc_n+m}{c_n + b}\to \sqrt{m}.$ Let $c_n \underset{\tiny\,}{=} \small\beta_n+\sqrt{2}$, by the recursion formula, $\,\beta_{n+1}^{1}+\frac{1}{2\sqrt{2}}=(3+2\sqrt{2})(\beta_n^{1}+\frac{1}{2\sqrt{2}}).$ $\,\beta_n^{1}+\frac{1}{2\sqrt{2}}={\small(3+2\sqrt{2})}^{n1}(\beta_1^{1}+\frac{1}{2\sqrt{2}}).$ Let $\{c_n^{\pm}\}\;\,$be the resulting sequences with initial value $\beta_1= 1\sqrt{2},\;2\sqrt{2}$ respectively. This yields $\beta_{n+1}^{1}=\frac{1\pm(3+2\sqrt{2})^n}{2\sqrt{2}}\;$and$\;{\large c}_n^{\pm}=\sqrt{2}\pm\frac{2\sqrt{2}}{(3+2\sqrt{2})^n\mp 1}.\;\;$ Thus $\;c_1^{}< c_2^{}< \cdots< c_n^{}< \sqrt{2}< c_n^{+}< \cdots< c_2^{+}< c_1^{+}$ or $\;\;1< \frac{4}{3}< \frac{7}{5}< \frac{24}{17}< \frac{41}{29}{\small <\cdots \underset{\tiny\,}{\;}}$ $\qquad \qquad\cdots< c_{n+1}^{}< c_n^{}< \sqrt{2}< c_n^{+}< c_{n+1}^{+}< \cdots\underset{\tiny\,}{\,}$ $\qquad\qquad\qquad\qquad\qquad {\small \cdots<} \frac{58}{41}< \frac{17}{12}< \frac{10}{7}< \frac{3}{2}< 2.$ 

12152010, 02:32 PM
Post: #4




Insert: Rudin [Principle of Mathematical Analysis] Notes
(10042010 05:41 PM)elim Wrote: 1.1 Example.... Hi, i was wondering the same thing so i construct my solution, its not that elegant but there it is: we want a rational q such that p< q for every p on A and q, if q > p then q= p + r for some r>0. if $q^2$ < 2 then $ p^2 + 2pr + r^2 \underset{\,}{<} 2,$ then $ r< \small\dfrac{2p^2}{2p + r}$ we know that r < 2 so we take $r = \small\dfrac{2p^2}{2p+2}$ and then the previously inequality is true for every p so $q = p +\small \dfrac{2p^2}{2p+2}$ is on A and is bigger than p. 

12182010, 11:41 PM
(This post was last modified: 01292019 03:56 PM by elim.)
Post: #5




1.2 Remark on Rationals [Principle of Math Analysis]
1.2 Remark 1.1 Example Shows that $\mathbb{Q}$ has certain gaps
$\quad$in sprite off the fact that it's dense. The real number $\quad$system fills these gaps. 

12182010, 11:41 PM
Post: #6




1.3~1.4 Basic Set and Notations Rudin [P.M.A] Notes
1.3 Definitions
$\quad$We write $x \in A$ when $x$ is a member(an element) of $A$ $\qquad x\not\in A$ means that $x$ is not a member of $A$. $\quad$The set $\varnothing$ contains no element is called Empty set. $\quad$A set $A$ is called nonempty if $A\ne\varnothing$. $\quad A$ is a sebset of $B$(Write $\small A\subset B$ or $\small B\supset A$) if $\qquad\qquad\forall x\in A \; (x\in B)$; $\quad$Write $A=B$ if $(A\subset B)\wedge (B\subset A)$. Otherwise $A \ne B$ $\quad A$ is a proper sebset of $B$(Write $\small A \subsetneq B$) if $\qquad\qquad(A \subset B) \wedge (A \ne B).$ 1.4 DefinitionWe use $\mathbb{Q}$ for the set of all rational numbers. 

12272010, 01:38 PM
Post: #7




1.5  1.10 Ordered set: Definitions and Examples. Rudin [P.M.A]
ORDERED SETS
1.5 Definition An order on a set $S$ is a relation $<$ such that $\quad\quad$ (i) $\forall x, y\in S$, one and only one of the following is true: $\qquad\qquad\qquad x < y,\;\; x = y,\;\; y < x$ $\quad\quad$ (ii) $\forall x, y, z \in S, \quad (x < y) \wedge (y < z) \Rightarrow (x < z)$ $\quad$We read $x < y$ as "$x$ is less (smaller) than $y$" or "$y > x$ $\quad$($y$ is bigger (greater) than $x$)" or "$x$ precedes $y$" $\quad$Write $x \le y$ for $ (x < y) \vee (x = y)$ $\qquad\qquad\qquad$ i.e. $\small \lnot (y < x)\,$ i.e.$\,\small\lnot (x > y)$ 1.6 Definition. An ordered set is a set with an order defined. $\quad\quad \mathbb{Q}$ is an ordered set where $r < s$ means $s  r >0.$ 1.7 Definition. If $S$ is an ordered set, $E\subset S$ $\qquad$and $\exists \beta \in S, \forall x \in E (x \le \beta)$, then $E$ is called bounded $\qquad$above, $\beta$ is called an upper bound of $E$. $\qquad$Lower bounds are defined similarly with $\ge$ in place of $\le$. 1.8 Definition. If $E \subset S$ is bounded above in ordered set $S$, $\qquad$and $\alpha \in S$ is an upper bound of $E$ such that $\qquad(\gamma < \alpha) \implies (\gamma$ is not an upper bound of $E)$, $\qquad$then $\alpha$ is called the least upper bound of $E$, or the $\qquad$supremum of $E$, and we write $\alpha = \sup E$ $\qquad$Similarly we define the greatest lower bound, or infimum $\qquad$of a set. Thus $\alpha = \inf E$ means that $\alpha$ is a lower bound $\qquad$of $E$ and $E$ has no bigger lower bound than $\alpha$. 1.9 Examples. $\quad$(a) In Example 1.1, $\small B(A)$ is exactly the set of upper(lower) $\qquad$bounds of $\small A$($B$) and $\small A$($\small B$) has no supremum(infimum) $\qquad$in $\mathbb{Q}$. $\quad$(b)Even if $\alpha = \sup E$ exists, $\alpha$ may or may not be a $\qquad$member of $E$. $\quad$(c) Let $\small E = \{1/n \mid n \in \mathbb{N}\}$, then $\sup E (\small = 1 \in E\not\ni 0=)\inf E$ 1.10 Definition. An ordered set $S$ is said to have the $\qquad$leastupperbound property If $\qquad\quad\forall E\subset S \; ((E \ne \varnothing) \wedge (E$ is bounded above$))$ $\qquad\qquad\implies(\exists \alpha \in S, \; \alpha = \sup E )$ 

12282010, 04:47 PM
Post: #8




1.11 Theorem [Symmetric Properties between $\inf$ and $\sup$]
1.11 Theorem Suppose $S$ has leastupperbound property,
$\quad \varnothing \ne B\subset S,\quad \varnothing \ne L=\{x\in S \mid x \le y(\forall y \in B)\}$ $\quad$Then $\exists \;\alpha \in S\, (\inf B =\alpha = \sup L).$ Proof. Since $L(\ne\varnothing)$ is bounded above, $\alpha:=\sup L\in S.$ $\quad$If $\gamma< \alpha$ then $\exists\lambda\in L\,\forall x \in B\,(\gamma < \lambda \le x)$ and so $\quad\gamma \not\in B$, therefore $\forall x \in B\,(\alpha \le x).$ i.e. $\alpha \in L$ or $\alpha$ is a $\quad$lower bound of $B$. $\quad$If $\alpha < \beta$, then $\beta \not\in L$ since $\alpha = \sup L$. Therefore $\alpha$ is $\quad$the greatest lower bound of $B$, i.e. $\alpha = \inf B \quad\quad \square$ 

01102011, 10:16 AM
Post: #9




1.12 Fields, 1.13 Remarks Rudin [Principle of Math. Analysis]
1.12 Definition A field is a set $F$ with two operations
$\quad$addition and multiplication, satisfying axioms (A),(M) $\quad$and (D) below: (A) Axioms for addition $\quad$ (A1) Addition is closed in $F$: $(x, y \in F)\Rightarrow (x+y\in F)$ $\quad$ (A2) Addition is commutative: $x + y = y + x\;(\forall x,y\in F)$ $\quad$ (A3) Addition is associative: $\qquad\qquad\quad (x+y)+z = x+(y+z)\;(\forall x,y,z\in F)$ $\quad$ (A4) $F$ has additive unit: $\exists 0\in F\ \forall x\in F\;(0 + x = x)$ $\quad$ (A5) $F$ has additive inverse: $\qquad\qquad\quad \forall x\in F\ \exists x\in F \,(x+(x) = 0)$ (M) Axioms for multiplication $\quad$ (M1) multiplicationis closed in $F$: $(x, y \in F)\Rightarrow (xy\in F)$ $\quad$ (M2) multiplication is commutative: $xy = yx\,(\forall x,y\in F)$ $\quad$ (M3) multiplication is associative: $\qquad\qquad\quad (xy)z = x(yz)\,(\forall x,y,z\in F)$ $\quad$ (M4) $F$ has multiplicative unit: $\exists 1\in F\ \forall x\in F\,(1x = x)$ $\quad$ (M5) $F$ has multiplicative inverse: $\qquad\qquad\quad\forall x\in F\ \exists 1/x\in F \;(x(1/x) = 1)$ (D) The distributive law: $x(y+z) = xy+xz\,(\forall x,y,z\in F)$ 1.13 Remarks (a) In any field one usually write $\qquad\displaystyle{xy, \frac{x}{y}, x+y+z, xyz, x^2, x^3, 2x, 3x,\cdots}$ $\qquad$In place of $\quad\quad \displaystyle{x+(y), x\cdot \left(\frac{1}{y} \right),(x+y)+z,(xy)z, }$ $\qquad xx, xxx,x+x, x+x+x,\cdots$ (b) Rational numbers $\mathbb{Q}$ with its customary addition and $\qquad$multiplication is a field. (c) We shall see that some $\qquad$familiar properties of $\mathbb{Q}$ are actually the $\quad\quad$consequences of field axioms thus hold in any field $\qquad$(complex numbers, etc) 

04232012, 11:04 AM
Post: #10




1.14  1.16 Propositions from Field Axioms. Rudin 【P.M.A】
1.14 Proposition From axioms for addition
(a) $(x+y=x+z) \implies (y = z)$ (b) $(x+y = x) \implies (y = 0)$ (c) $(x+y=0) \implies (y=x)$ (d) $(x) = x$ Proof: (a) $y\overset{A_4}{=} 0+y \overset{A_5}{=} (x + x) + y \overset{A_3}{=} x + (x+y)$ $\qquad\overset{\small\text{asm}}{=} x + (x+z) \overset{A_3}{=} (x+x) + z \overset{A_5}= 0+z \overset{A_4}{=}z$ (b) This is the case $z = 0$ in (a) (c) By $A_5$, this is the case $z = x$ in (a) (d) $(x) = (x+(x))(x)$ $\qquad\qquad\;\;= x + (x  (x)) = x + 0 = x$ 1.15 Proposition From axioms for multiplication (a) $(x \ne 0)\wedge (xy = xz) \implies (y = z)$ (b) $(x \ne 0) \wedge (x y = x) \implies (y = 1)$ (c) $(x)y =  (xy) = x(y)$ (d) $(x)(y) = xy$ Proof: (a) $0x + 0x \overset{Distri.}= (0+0)x \overset{A_4}{=} 0 x \overset{1.14(b)}{\implies} 0x = 0$ (c) $(x\ne 0)\wedge (xy = 1)\implies (y=1/x)$ (d) $(x\ne 0)\implies 1/(1/x) = x$ The proof is parallel to that of Proposition 1.14. 1.16 Proposition $0$ and additive inverse with multiplication (a) $0x = 0$ (b) $(x\ne 0)\wedge (y\ne 0)\implies (xy \ne 0)$ (c) $(x)y =(xy) = x(y)$ (d) $(x)(y) = xy$ Proof: (a) $\left(0x + 0x \overset{Distri}{=} (0+0) x \overset{A_5}{=} 0x \right ) \overset{1.14(b)}{\implies} (0x = 0)$ (b) If $x\ne 0 \ne y$ but $xy = 0$, then $\qquad\small 1 = (1/x)x(1/y)y = xy(1/x)(1/y) = 0(1/x)(1/y)\overset{(a)}{=}0$ Let's prove $(c')\quad (x = (1)x)$: $\qquad\small((1)x + x = (1)x + 1x = (1+1) x = 0)\overset{1.14c}{\implies} (c')$ Now (c) $(x)y =((1)x)y = (1)(xy) = (xy)$, Also (d) $(x)(y) =  (x(y)) = ((xy)) \overset{1.14(d)}{=} xy$ 

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