Rudin 【Principle of Mathematical Analysis】Notes & Exercise Solutions

10042010, 05:41 PM
(This post was last modified: 12152014 02:52 PM by elim.)
Post: #1




Rudin 【Principle of Mathematical Analysis】Notes & Exercise 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 $p$. $\quad\quad$ If there were such a $p$, we could write $p = m/n$ where $m$ and $n$ are integers that are not both even. $\quad\quad$ Let us assume this is done. Then (1) implies (2) $m^2 = 2n^2$. This shows that $m^2$ is even. Hence $m$ is even (if m were odd, m2 would be odd), $\quad\quad$ and so $m^2$ is divisible by $4$. It follows that the right side of (2) is divisible by $4$, so that $n^2$ is even, $\quad\quad$ which implies that n is even. The assumption that (1) holds thus leads to the conclusion that $\quad\quad$ both $m$ and $n$ are even, contrary to our choice of $m$ and $n$. Hence (1) is impossible for rational p. We now examine this 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 a 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) $\displaystyle{q = p\frac{p^22}{p+2} = \frac{2p+2}{p+2}}$ Then (4) $\displaystyle{q^2 2 =2(p^22)/(p+2)^2}$ . If $p$ is in $A$ then $p^2 2 < 0$, (3) shows that $q > p$, and (4) shows that $q^2 < 2$. Thus $q$ is in $A$. If $p$ is in $B$ then $p^2 2> 0$, (3) shows that $0 < q < p$, and (4) shows that $q^2 > 2$. Thus $q$ is in $B$. There is no problem to understand the text, but how the expression \[q = p  \frac{p^22}{p+2}=\frac{2p+2}{p+2} \] been constructed? What's the thoughts behind? 

12152010, 02:32 PM
Post: #3




1.1 Example The idea of $q = p  (p^22)/(p+2) = (2p+2)/(p+2)$
We'd like to define, for each rational $p>0$, a rational $q > 0$ such that $\displaystyle{0<\frac{q\sqrt{2}}{p\sqrt{2}}<1}$
This ensures that $p^2>2 \Leftrightarrow q^2>2$ and $q\sqrt{2}<p\sqrt{2}$ Let $q = p R(p)$, then $\displaystyle{\frac{q\sqrt{2}}{p\sqrt{2}}=1\frac{R(p)}{p\sqrt{2}}}$ and so $\displaystyle{0<\frac{R(p)}{p\sqrt{2}}<1}$. $R(p)$ must be rational with the same sign as $p^22$, and so $\displaystyle{R(p)=\frac{p^22}{Q(p)}}$, $\displaystyle{\frac{R(p)}{p\sqrt{2}}=\frac{p+\sqrt{2}}{Q(p)}}$ with $Q(p)$ a rational function of $p$ and to meet the condition $\displaystyle{0<\frac{p+\sqrt{2}}{Q(p)}<1}$, the simplest $Q(p)$ is $p+2$. Finally, this also ensures $\displaystyle{q=p\frac{p^22}{p+2}=\frac{2p+2}{p+2}>0, \quad (p>0)}$ 

12152010, 02:32 PM
(This post was last modified: 11302012 01:54 PM by elim.)
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 p then q= p + r for some r>0. 

12182010, 11:41 PM
(This post was last modified: 11302012 01:55 PM by elim.)
Post: #5




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


12182010, 11:41 PM
(This post was last modified: 11302012 01:55 PM by elim.)
Post: #6




1.3~1.4 Basic Set and Notations Rudin [Principle of Mathematical Analysis] Notes
1.3 Definitions
$\quad\quad$ Use $x \in A$ to indicate that $x$ is a member(an element) of $A$; Otherwise $x \not\in A$ $\quad\quad$ The set $\varnothing$ that contains no element is called the empty set; $A \ne \varnothing$ is called nonempty. $\quad\quad$ Write $A \subset B$ or $B \supset A$ when $\forall x\in A \; (x\in B)$; We call $A$ is a subset of $B$ when $A \subset B$ $\quad\quad$ Write $A=B$ if $(A\subset B)\wedge (B\subset A)$. Otherwise $A \ne B$ $\quad\quad$ Write $A \subsetneq B$ to indecate that $(A \subset B) \wedge (A \ne B)$ and call $A$ a proper subset of $B$ 1.4 Definition Throughout Chap. I, the set of all rational numbers is denoted by $\mathbb{Q}$ 

12272010, 01:38 PM
(This post was last modified: 12122012 04:38 AM by elim.)
Post: #7




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

12282010, 04:47 PM
Post: #8




1.11 Theorem [Symmetric Properties between $\inf$ and $\sup$]
1.11 Theorem Suppose ordered set $S$ has leastupperbound property,
$\quad \varnothing \ne B\subset S,\quad \varnothing \ne L=\{x\in S \mid \forall y \in B, x \le y\}$ Then $\exists \;\alpha \in S, \quad \inf B =\alpha = \sup L$ $\quad$ Proof. Since $L$ is not empty and bounded above, $\exists \;\alpha \in S\quad \alpha = \sup L$ $\quad$ If $\gamma < \alpha$ then $\exists \lambda \in L,\;\forall x \in B \quad \gamma < \lambda \le x$ and so $\gamma \not\in B$, therefore $\forall x \in B, \quad \alpha \le x$ $\quad$ i.e. $\alpha \in L$ or $\alpha$ is a lower bound of $B$. $\quad$ If $\alpha < \beta$, then $\beta \not\in L$ since $\alpha = \sup L$. So $\alpha$ is 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 Mathematical Analysis] Notes
1.12 Definition A field is a set $F$ with two operations addition and multiplication, satisfying axioms (A),(M) and (D):
(A) Axioms for addition $\quad\quad$ (A1) Addition is closed in $F$: $(x, y \in F)\Rightarrow (x+y\in F)$ $\quad\quad$ (A2) Addition is commutative: $x + y = y + x\quad (\forall x,y\in F)$ $\quad\quad$ (A3) Addition is associative: $(x+y)+z = x+(y+z)\quad (\forall x,y,z\in F)$ $\quad\quad$ (A4) $F$ has additive unit: $\exists 0\in F\ \forall x\in F\quad 0 + x = x$ $\quad\quad$ (A5) $F$ has additive inverse:$\forall x\in F\ \exists x\in F \quad(x+(x) = 0)$ (M) Axioms for multiplication $\quad\quad$ (M1) multiplicationis closed in $F$: $(x, y \in F)\Rightarrow (xy\in F)$ $\quad\quad$ (M2) multiplication is commutative: $xy = yx\quad (\forall x,y\in F)$ $\quad\quad$ (M3) multiplication is associative: $(xy)z = x(yz)\quad (\forall x,y,z\in F)$ $\quad\quad$ (M4) $F$ has multiplicative unit: $\exists 1\in F\ \forall x\in F\quad 1x = x$ $\quad\quad$ (M5) $F$ has multiplicative inverse:$\forall x\in F\ \exists 1/x\in F \quad(x(1/x) = 1)$ (D) The distributive law: $x(y+z) = xy+xz\quad (\forall x,y,z\in F)$ 1.13 Remarks (a) In any field one usually write $\quad\quad \displaystyle{xy, \frac{x}{y}, x+y+z, xyz, x^2, x^3, 2x, 3x,\cdots}$ $\ \quad$ In place of $\quad\quad \displaystyle{x+(y), x\cdot \left(\frac{1}{y} \right),(x+y)+z,(xy)z, xx, xxx,x+x, x+x+x,\cdots}$ (b) Rational numbers $\mathbb{Q}$ with its customary addition and multiplication is a field. (c) We shall see that some familiar properties of $\mathbb{Q}$ are actually the $\quad\quad$consequences of field axioms thus hold in any field (complex numbers, etc) 

04232012, 11:04 AM
Post: #10




1.14  1.16 Propositions from Field Axioms. Rudin 【Principle of Mathematical Analysis】Notes
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) $y\overset{A_4}{=} 0+y \overset{A_5}{=} (x + x) + y \overset{A_3}{=} x + (x+y) \overset{former}{=} x + (x+z) \overset{A_3}{=} (x+x) + z \overset{A_5}= 0+z \overset{A_4}{=}z$ (b) This is the case $z = 0$ in (a) (c) By $A_5$, this is the case $z = x$ in (a) (d) $(x) = (x+(x))(x) = x + (x  (x)) = x + 0 = x$ 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 $1 = (1/x)x(1/y)y = xy(1/x)(1/y) = 0(1/x)(1/y)\overset{(a)}{=}0$ Let's prove (c') $(x = (1)x)$: $\bigg((1)x + x = (1)x + 1x = (1+1) x = 0 \bigg)\overset{1.14c}{\implies} ((1)x = x)$ Now (c) $(x)y =((1)x)y = x(1)y = x(y) = (1)(xy) = (xy)$, Also (d) $(x)(y) =  (x(y)) = ((xy)) \overset{1.14(d)}{=} xy$ 

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