Real Analysis (Math 452) Chapter 1 Assignments Dr. Friedman

Real Analysis (Math 452) Chapter 1 Assignments
Dr. Friedman
Section 1
Assignment
Ross: #2, 4, 5, 8, 12
E1: Let
be the sequence defined by
is bounded above by 2.
. Prove that
Supplemental
Ross: #3, 6, 9, 11
Section 2
Assignment
E1: Prove that
is irrational. Does a similar argument work to show
is irrational?
E2: Where does the proof done in class that
is irrational break down if we try to use it to
prove that
is irrational?
Section 3
Assignment
Ross: #5, 6
Supplemental
Ross: #3, 4
Review/Preliminaries
Assignment
E1: Decide which of the following represent true statements about the nature of sets. If you
believe any to be false, you must, of course, provide a specific counterexample.
a)
b)
c)
E2: Given a function f and a subset A of its domain, let
a) Let
. If
.
, find
. Does
in this case?
b) Find two sets A and B for which
c) Show that, for an arbitrary function
.
.
, it is always true that
E3: Show that, for an arbitrary function
, and for all sets
, it is always true
that
. (Note: A similar relationship holds for unions.)
E4: Form the logical negation of each claim. Try to embed the word “not” as deeply into the
resulting sentence as possible (or avoid using it altogether).
a) For all real numbers satisfying
, there exists an
such that
.
b) Between every two distinct real numbers, there is a rational number.
c) For all
is either a natural number or an irrational number.
d) Given any real number
satisfying
.
Section 4
Assignment
Ross #1-4 (selected parts TBA), 6, 8, 14
Supplemental
Ross #1-4, 9, 10
Countable/Uncountable Sets
Assignment
E1: Prove that the interval (0,1) can be put into one-to-one correspondence with the set of real
numbers.
E2: Supply a rebuttal to the following concern about the proof that the interval (0,1) is
uncountable:
Every rational number has a decimal expansion so we could apply this same argument to
show that the set of rational numbers between 0 and 1 is uncountable. But we know that
subsets of the rational numbers are countable so there is a flaw in this proof!