MATH 301 SOLUTIONS 8, KOGAN SPRING 2013
MAMIKON GULIAN
3.4
12. We define the sebsequence xnk as follows. There exists n1 such
that |xn1 | > 1. This is clear from the definition of unboundedness. Now suppose xn1 , ., xnk1 have been chosen. Then
there exist
MATH 301 SOLUTIONS 7, KOGAN SPRING 2013
MAMIKON GULIAN
Comments. On # 14 in Section 3.4, some of you had the right
idea, but did not develop your proof fully. You have to define the
subsequence before you can use it. Many proofs simply began by using
a su
MATH 301 SOLUTIONS 3, KOGAN SPRING 2013
MAMIKON GULIAN
Graded problems were Section 2.1 #7, Section 2.2 #4 and #17, and
Section 2.3 #5, #7.
1. Section 2.1
(7) We prove this by contradiction. Suppose that there exists a
rational s so that t2 = 3. Write t =
MATH 301 SOLUTIONS 1, KOGAN SPRING 2013
MAMIKON GULIAN
1. Section 1.1, pp. 10
(1) A B C consists of elements of B C which are less than or
equal to 20. Now
B = cfw_2, 5, 8, 11, 14, 17, 20, .
C = cfw_3, 5, 7, 9, 11, 13, 15, 17, 19, .,
so B C = cfw_5, 11, 1
MATH 301 SOLUTIONS 5, KOGAN SPRING 2013
MAMIKON GULIAN
Graded problems were Section 3.1 5a, 5d, 8, and 10. Some additional
problems are included here.
1. Section 2.1
For parts of number (5), some of you tried a squeeze theorem arguement. This is an exampl
MATH 301 SOLUTIONS 10, KOGAN SPRING 2013
MAMIKON GULIAN
Notes
Note that continuity is defined using an statement, but many
statements are easier to prove using the sequential criterion. The problems here serve as examples. My advice is to always check if
MATH 301 SOLUTIONS 4, KOGAN SPRING 2013
MAMIKON GULIAN
Solutions to some problems on HW 4; only three were graded since
the quiz had to be graded that week.
1. Section 2.4
(3) First, we claim u is an upper bound for S. If it were not,
then for some s S, w
MATH 301 SOLUTIONS 2, KOGAN SPRING 2013
MAMIKON GULIAN
Graded problems for this assignment were problems 7 and 20 of
section 1.2, and problems 4, 9, and 20 on Section 2.1.
General comments on writing proofs: Your proofs should be
readable. They should not
MATH 301 SOLUTIONS 6, KOGAN SPRING 2013
MAMIKON GULIAN
1. Section 3.2
(7) Suppose bn < M for all n. We have
0 |bn an 0| M an .
Since an 0, the result follows from the squeeze theorem.
(15) Write
a n
n1
1
1
n
n n
+1
= b(n + 1) n ,
(a + b ) = b
b
a
where
MATH 301 Introduction to Mathematical Analysis I
Section 0201 Fall 2015
Instructor: Dr. Jinglai Shen
E-mail: shenj@umbc.edu
Oce: Math/Psyc 417
Phone: (410) 4552402
Lectures: Tue and Thu, 12:302:15 pm, at SOND 202
Oce hours: Tue and Thu, 2:30 pm3:30