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Solution to Understanding Analysis Exercise 1.2


Exercise 1.2.1

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.1


Exercise 1.2.2

We prove it by contradiction.

Suppose there exists a rational number $r$ such that $2^r=3$. Let $r=p/q$, where $p$ and $q$ are integers. Then\[2^{p/q}=3\Longrightarrow 2^p=3^q.\]Note that $q\ne 0$.

If $q>0$, then $p>0$. Hence we have $3|2^p$, which is impossible.

If $q<0$, then $p<0$. We have $2^{-p}=3^{-q}$. Similarly, we have $3|2^{-p}$, which is impossible. Thus completing the proof.


Exercise 1.2.3

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.2


Exercise 1.2.4

Rearrange $\mathbf N$ as follows,

Take $A_i$ as the set of numbers in $i$-th column. Then it is clear that $A_i\cap A_j=\emptyset$ for $i\ne j$ and $\cup_{i=1}^\infty A_i=\mathbf N$.


Exercise 1.2.5

(a) If $x\in (A\cap B)^c$, hence $x\notin A\cap B$. Therefore, we have $x\notin A$ or $x\notin B$ which implies that $x\in A^c$ or $x\in B^c$. Thus $x\in A^c\cup B^c$ and hence $(A\cap B)^c\subseteq A^c\cup B^c$.

(b) If $x\in A^c \cup B^c$, then $x\in A^c$ or $x\in B^c$. Hence $x\notin A$ or $x\notin B$, namely $x\notin A\cap B$. Thus $x\in (A\cap B)^c$. Therefore, we have $A^c\cup B^c\subseteq (A\cap B)^c$.

(c) If $x\in (A\cup B)^c$, then $x\notin A\cup B$. Hence $x\notin A$ and $x\notin B$, namely $x\in A^c$ and $x\in B^2$. Thus $x\in A^c\cap B^c$. We conclude that $(A\cup B)^c\subseteq A^c\cap B^c$.

If $x\in A^c\cap B^c$, Then $x\in A^c$ and $x\in B^c$, namely $x\notin A$ and $x\notin B$. Therefore $x\notin A\cup B$ and thus $x\in (A\cup B)^c$. We conclude that $A^c\cap B^c\subseteq (A\cup B)^c$.

As a consequence, $(A\cup B)^c=A^c \cap B^c$.


Exercise 1.2.6

(a) If $a$ and $b$ are both non-negative, then $|a+b|=a+b$, $|a|=a$, and $|b|=b$. Hence\[|a+b|=a+b=|a|+|b|.\]If $a$ and $b$ are both non-positive, then $|a+b|=-(a+b)$, $|a|=-a$, and $|b|=-b$. Hence\[|a+b|=-(a+b)=-a-b=|a|+|b|.\]

(b) Note that $$(a+b)^2=a^2+2ab+b^2$$ and $$(|a|+|b|)^2=|a|^2+2|a||b|+|b|^2=a^2+2|a||b|+b^2.$$ Since $ab\leqslant |a b|$, we have $$(a+b)^2\leqslant (|a|+|b|)^2.$$Since $x^2\leqslant y^2$ implies $|x|\leqslant |y|$, we have $$|a+b|\leqslant |a|+|b|.$$

(c) We applies the triangle inequality several times, \begin{align*}&~|a-c|+|c-d|+|d-b|\\ \geqslant &~|(a-c)+(c-d)|+|d-b|\\ =&~|a-d|+|d-b|\geqslant |(a-d)+(d-b)|\\ =&~|a-b|.\end{align*}

(d) Note that $$||a|-|b||^2=(|a|-|b|)^2=a^2-2|a||b|+b^2$$ and $$|a-b|^2=(a-b)^2=a^2-2ab+b^2.$$Since $ab\leqslant |a||b|$, we have $-2|a||b|\leqslant 2ab$. Hence\begin{align*}||a|-|b||^2=&~(|a|-|b|)^2=a^2-2|a||b|+b^2\\ \leqslant & ~a^2-2ab+b^2=|a-b|^2.\end{align*}Therefore $$||a|-|b||\leqslant |a-b|.$$


Exercise 1.2.7

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.6


Exercise 1.2.8

(a) $f(n)=n+1$ for $n\in\mathbf N$.

(b) $f(1)=1$ and $f(n)=n-1$ for $n\geqslant 2$.

(c) $f(1)=0$, $f(2n)=n$ and $f(2n+1)=-n$ for all $n\mathbf N$.


Exercise 1.2.9

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.7


Exercise 1.2.10

(a) & (b) False. Take $a=b=0$ as a counterexample.

(c) True. Let us prove it. Consider the “if” part. Suppose $a < b$, then we are done. Suppose $a\geqslant b$, then\[|a-b|=a-b.\]By assumption, \[|a-b|=a-b < \varepsilon\] for every $\varepsilon>0$. By Theorem 1.2.6, we have $a=b$. Hence we always have $a\leqslant b$.

“Only if” part. If $a\leqslant b$, then it is clear that\[a< b+\varepsilon\]for every $\varepsilon>0$.


Exercise 1.2.11

(a) True. When $n$ is very large, then $\dfrac{1}{n}$ is very small and close to zero.

(b) False. See above. We cannot find a positive number which is small than any positive number.

(c) True. See Theorem 1.4.3.


Exercise 1.2.12

(a) It is clear $y_1=6> -6$. Now we suppose that $y_n> -6$. Now we show that $y_{n+1}> -6$. Since $y_{n+1}=\dfrac{2y_n-6}{3}$, we have\begin{align*}y_{n+1}=&~\dfrac{2y_n-6}{3}\\ >&~\frac{2\times (-6)-6}{3}=\frac{-18}{3}=-6.\end{align*}By induction, we have $y_n> -6$ for all $n\in\mathbf N$.

(b)It is equivalent to show that $y_n> y_{n+1}$ for all $n\in\mathbf N$. Note that $y_2=\dfrac{2\cdot 6-6}{3}=2$, hence we have \[y_1 > y_2.\]Now we suppose that $y_n > y_{n+1}$, namely $y_n-y_{n+1}< 0$, we would like to show that $y_{n+1}> y_{n+2}$. Since $y_{n+1}=\dfrac{2y_n-6}{3}$, we have\begin{align*}y_{n+2}-y_{n+1}=&~\frac{2y_{n+1}-6}{3}-\frac{2y_n-6}{3}\\ =&~\frac{2(y_{n+1}-y_n)}{3}< 0.\end{align*}Hence $y_{n+2}< y_{n+1}$. By induction, we have $y_n> y_{n+1}$ for all $n\in\mathbf N$, which implies that the sequence $(y_1,y_2,\cdots)$ is decreasing.


Exercise 1.2.13

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.12

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

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