This establishes the desired result Exercise 4.5 Suppose that limn→∞an= 0 and {bn}∞ n=1is bounded.. Show that M − cannot be an upper bound ofthe sequence... d By the previous exercise,
Trang 1If a ≥ 0 then −a ≤ 0 so that |a| = max{−a, a} = a If a < 0 then −a > 0 sothat |a| = max{−a, a} = −a The graph is shown in Figure 1
Figure 1Exercise 1.3
Show that |a| ≥ 0 with |a| = 0 if and only if a = 0
Suppose first that a ≥ 0 Then |a| = a
• If b ≥ 0 then |b| = b In this case, |a| = |b| implies that a = b
Trang 2• If b < 0 then |b| = −b In this case, |a| = |b| implies that a = −b.
Suppose now that a < 0 Then |a| = −a
• If b ≥ 0 then |b| = b In this case, |a| = |b| implies that −a = b which isthe same as a = −b
• If b < 0 then |b| = −b In this case, |a| = |b| implies that −a = −b which
Show that | − a| = |a|
Solution
If a ≥ 0 then −a < 0 Thus, |a| = a and | − a| = −(−a) = a Hence,
| − a| = |a| If a < 0 then −a > 0 In this case, |a| = −a and | − a| = −a.That is, | − a| = |a|
Exercise 1.7
Show that |ab| = |a| · |b|
Solution
Suppose first that a ≥ 0 Then |a| = a
• If b ≥ 0 then |b| = b Moreover, ab ≥ 0 In this case, |ab| = ab = |a| · |b|
• If b < 0 then |b| = −b Moreover, ab ≤ 0 In this case, |ab| = −ab =a(−b) = |a| · |b|
Suppose now that a < 0 Then |a| = −a
• If b ≥ 0 then |b| = b Moreover, ab ≤ 0 In this case, |ab| = −ab = (−a)b =
Trang 3If a > 0 then 1a > 0 Thus, 1
a
= 1a = |a|1 If a < 0 then 1a < 0 Thus,
Solution
Using both Exercise 1.7 and Exercise 1.8 we can write the following a
b
= ...
n
∞ n=1 converges to 1, where C 6= is a con-stant
Solution
Let > We want to find a positive integer N such that 1... through by an+1 to obtain
an+2
an+1 = +
an
an+1.Take the limit as n → ∞ to obtain
L =...
n=1 is bounded from above with an upperbound equals to
(d) By the previous exercise, the sequence is convergent say to A Using thearithmetic operations of sequences, we can write