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--- math104-f21:hw3 [2021/09/11 01:59]
pzhou
+++ math104-f21:hw3 [2022/01/11 08:36] (current)
pzhou ↷ Page moved from math104:hw3 to math104-f21:hw3
@@ Line 1: / Line 1: @@
-====== HW 3 ======
+====== HW 3 with solution ======
   - [Tao] Ex 5.4.3,
   - [Tao] Ex 5.4.5 (you may assume result in 5.4.4)
   - [Tao] Ex 5.4.7,
   - [Tao] Ex 5.4.8
-  - Let  $A$  be the subset of  $\Q$  consisting of rational numbers with denominators of the form  $2^m$ . Prove that for any  $x \in \R$ , there is a Cauchy sequence  $(a_n)$  in  $A$ , such that $x = \LIM a_n$.
+  - Let  $A$  be the subset of  $\Q$  consisting of rational numbers with denominators of the form  $2^m$ . Prove that for any  $x \in \R$ , there is a Cauchy sequence  $(a_n)$  in  $A$ , such that  $x = LIM a_n$ .
+====== Solution ======
+==== 5.4.3 ====
+Problem: Show that for any real number  $x$ , there is exactly one integer  $N$ , such that  $N \leq x \leq N+1$ .
+Sol: If  $x$  is an integer, then let  $N=x$ . Assume  $x$  is not an integer now. Then  $|x| > 0$ , hence by Prop 5.4.12,  there exists a positive integer  $N_1$  such that  $|x| < N_1$ . Consider the set  $S = \{-N_1, -N_1 + 1, \cdots, N_1 \}$  and the subsets  $S_- = \{ n \in S \mid n < x \}$ ,  $S_+ = \{n \in S \mid n > x\}$ . Since  $x$  is not an integer, then  $S = S_- \sqcup S_+$  (disjoint union, meaning  $S_- \cap S_+ = \emptyset$ ). Let  $N = \max (S_-)$ , since  $S_-$  is a finite set, the max element exists in  $S_-$ .
+Remark: we will call such  $N$  the 'floor' of  $x$ , and write  $N = \lfloor x \rfloor$ .
+==== 5.4.5 ====
+Given any real numbers  $x < y$ , show that there is a rational number  $q$  such that  $x < q < y$ .
+Sol: Since  $0 < y-x$ , we can find a large enough positive integer  $N$ , such that  $N (y-x) > 1$  (by Archimedian property). Let  $m$  be the unique integer, such that  $m \leq N x < m+1$  (proven in the problem 1), then  $m + 1 \leq N x + 1 < N y$ , hence we have  $N x < m+1 < N y$ . Let $q = (m+1)/N$.
+==== 5.4.7 ====
+Let  $x,y$  be real nubmers. Show that  $x \leq y+\epsilon$  for all real $\epsilon>0 $if and only if$ x \leq y $. Show that$ |x-y| < \epsilon $for all real$ \epsilon > 0 $if and only if$ x=y$.
+Solution: If  $x \leq y$ , then for any  $\epsilon > 0$ , we have  $x \leq y < y + \epsilon$ , hence  $x < y + \epsilon$ . In the other direction, if  $x \leq y +\epsilon$  for all  $\epsilon > 0$ , we claim that  $x - y \leq 0$ . If not, say  $c = x-y > 0$ , then, we can let  $\epsilon = c / 2$ , then we get
+ c = (x-y) \leq  \epsilon  = c / 2 \Rightarrow c - c/2 \leq 0 \Rightarrow c \leq 0
+which contradict with assumption  $c > 0$ , hence the claim is true.
+For the second statement, if  $|x-y| < \epsilon$  for all real  $\epsilon > 0$ , then  $|x-y| \leq 0$  by the same argument. Since we also have  $|x-y| \geq 0$  by definition of absolute value, hence  $|x-y|=0$ . Thus  $x = y$ .
+==== 5.4.8 ====
+Let  $(a_n)$  be a Cauchy sequence of rationals, and let  $x$  be a real number. Show that, if  $a_n \leq x$  for all  $n$ , then  $LIM a_n \leq x$ . Similarly, if  $a_n \geq x$  for all  $n$ , then  $LIM a_n \geq x$ .
+Solution: We only prove the first half of the statement, since the other half is similar. We prove by contradiction, if  $LIM a_n  > x$ , then there is a rational number  $q$ , such that  $LIM a_n > q  > x$ . Since we have  $a_n < x < q$ , we have  $LIM a_n \leq LIM q = q$  (Corollary 5.4.10), contradict with the condition of $LIM a_n > q.
+==== Problem 5 ====
+Let  $A$  be the subset of  $\Q$  consisting of rational numbers with denominators of the form  $2^m$ . Prove that for any  $x \in \R$ , there is a Cauchy sequence  $(a_n)$  in  $A$ , such that  $x = LIM a_n$ .
+Solution: Let  $x \in \R$ , and let  $(b_n)$  be a Cauchy sequence in  $\Q$ , such that  $x = LIM b_n$ , we are going to construct  $(a_n)$  in  $A$ , such that  $(a_n)$  is equivalent to  $(b_n)$ , hence  $x = LIM a_n$ . For  $n \geq 1$ , we let  $a_n = \frac{\lfloor 2^n b_n \rfloor}{2^n}$ , where  $\lfloor .. \rfloor$  is the floor function (see Problem 1). Then
+ $|a_n - b_n| = | \frac{\lfloor 2^n b_n \rfloor}{2^n} - \frac{2^n b_n }{2^n} | = | \frac{\lfloor 2^n b_n \rfloor - b_n}{2^n}| \leq 1 / 2^n$
+hence  $(a_n)$  is equivalent to  $(b_n)$ .

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