Math 295 (Fall 2011). Solutions for Problem Set 2
2.1. (a) Suppose rst that x y and dene a := x and b := y . Then a and b are both
elements of the set S = cfw_x, y . Let us verify that a = min S and b = max S . Choose an
arbitrary z S . We must show that
Math 295 (Fall 2011). First midterm
Monday, September 26
You have 50 minutes
Please do not use any written or printed materials other than notes that were written
by you either in class or at home. Remember to justify all steps in your solutions.
Feel fre
Math 295 (Fall 2011). Final exam solutions
These solutions include all 10 of the problems that were originally handed out
F.1. (1) False. The sequence (an ) dened by an = n for all n N has no convergent
n=1
subsequence. (Note: if you stated that every seq
Math 295 (Fall 2011). Second midterm solutions
1
1
M.2.1. Let x0 R be given. For every n N, we have x0 n < x0 + n . By homework
problem 2.7(d), we can choose a rational number qn (possibly depending on n) such that
1
1
x0 n < qn < x0 + n . Thus we obtain
Math 295 (Fall 2011). Second midterm
Tuesday, November 8
You have 3 hours.
Please read the instructions below carefully.
Please do not use any written or printed materials other than notes that were written
by you either in class or at home. Remember to j
Math 295 (Fall 2011). First midterm solutions
M.1.1. (a) The word smallest means: suppose Y is any subset of R such that n Y
and if k Y , then k + 1 Y . Then Sn Y .
Fix n N and let Sn denote the collection of all subsets X R such that n X and
if k X , the
Math 295 (Fall 2011). Final exam problems
The in-class nal exam will consist of 8 out of the 10 problems that appear below. All
problems have equal weight. You will be asked to choose 6 of the 8 problems. Please do
not discuss the problems or your solutio
Math 295 (Fall 2011). Solutions for Problem Set 0
0.1. Throughout the solution we x some a R.
(a) First we deal with the only if direction, which is easier.
Assume that a has a real square root. By denition, this means that there exists b R
such that b2 =
Math 295 (Fall 2011). Solutions for Problem Set 5
5.1. If x = 0, it is obvious that both series converge absolutely. Indeed, when x = 0,
all terms of the rst series are equal to 0, and all terms except for the rst one of the
second series are also equal t
Math 295 (Fall 2011). Solutions for Problem Set 4
4.1. (a) Let > 0 be given. Since an 0 as n by assumption, we can nd N N
such that |0 an | < for all n N . But |0 an | = |an | and |0 bn | = |bn |. Since
|bn | |an | for all n by assumption, we conclude tha
Math 295 (Fall 2011). Solutions for Problem Set 3
3.1. (a) Let x R be given. We consider two cases. If x 0, then |x| = x, so |x| 0.
Moreover, in this case |x| = 0 if and only if x = 0.
Next suppose that x < 0. Then |x| = x, which implies that |x| > 0. In
Math 295 (Fall 2011). Solutions for Problem Set 1
1.1. First consider n = 1. Here dene 1! := 1. Next suppose that for some n N, we
already know what n! means. Then dene (n + 1)! := n! (n + 1). By the principle of
induction, this means that we have dened n
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