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 such that . Let , where and are integers. ThenNote that .
If , then . Hence we have , which is impossible.
If , then . We have . Similarly, we have , 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 as follows,
Take as the set of numbers in -th column. Then it is clear that for and .
Exercise 1.2.5
(a) If , hence . Therefore, we have or which implies that or . Thus and hence .
(b) If , then or . Hence or , namely . Thus . Therefore, we have .
(c) If , then . Hence and , namely and . Thus . We conclude that .
If , Then and , namely and . Therefore and thus . We conclude that .
As a consequence, .
Exercise 1.2.6
(a) If and are both non-negative, then , , and . HenceIf and are both non-positive, then , , and . Hence
(b) Note that and Since , we have Since implies , we have
(c) We applies the triangle inequality several times,
(d) Note that and Since , we have . HenceTherefore
Exercise 1.2.7
See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.6
Exercise 1.2.8
(a) for .
(b) and for .
(c) , and for all .
Exercise 1.2.9
See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.7
Exercise 1.2.10
(a) & (b) False. Take as a counterexample.
(c) True. Let us prove it. Consider the “if” part. Suppose , then we are done. Suppose , thenBy assumption, for every . By Theorem 1.2.6, we have . Hence we always have .
“Only if” part. If , then it is clear thatfor every .
Exercise 1.2.11
(a) True. When is very large, then 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 . Now we suppose that . Now we show that . Since , we haveBy induction, we have for all .
(b)It is equivalent to show that for all . Note that , hence we have Now we suppose that , namely , we would like to show that . Since , we haveHence . By induction, we have for all , which implies that the sequence is decreasing.
Exercise 1.2.13
See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.12