Math 323: Homework 1 Solutions
David Glickenstein
January 26, 2010
1.3b) The interval [0; 3] is nite. Its negation is The interval [0; 3] is innite.
1.3e) If x > 3 then f (x) > 7: Its negation is x > 3 and f (x)
1.4d) x < 5 or x > 7: Its negation is 5
x
7
Math 323
Exam 2B Solutions 0-3
Soluution to Problem 1(a) corrected Jun 13-14
Slight arithmetic error in Problem 3 corrected Jun 15, 11:05 AM
21st century
0. (a) Give the standard way discussed in this course for starting a proof of a statement of the form
Lesson 3a
1.
If a disjunction is true, then only one part of the disjunction is true.
This sentence is false. And it is ambiguous because the word only excludes the
possibility that if two parts are both true, then a disjunction is true.
2.
If only one pa
Lesson 8
1.
If a is an element of S and b is an integer, then ab is an element of S.
Proof: S is a sticky subset of the integers. Assume a is an element of S and b is an integer.
By the definition of a sticky set (Whenever a is an element of S and b is an
Lesson 9
6. (a) A sequence is convergent if it is Cauchy.
Antecedent: it is Cauchy. (Assume this statement is p)
Consequent: a sequence is convergent. (Assume this statement is q)
Converse: if a sequence is convergent, then it is Cauchy. (q=>p)
Contraposi
1.
Suppose that n is an element of S. Then n and n2 is are elements of S.
Proof: S is a sticky subset of the integers. Assume n is an element of S.
In this case, n is an element of S and S is a sticky subset of the integers.
So, n is an integer. (-1) n=-n
Lesson 29
Sect.1.4:
14. Prove: If x/(x-2)
3, then x < 2 or x
3. ( x Rx 2 )
Proof: By the results of Sect.1.3.1,
The statement is equivalent to:
x R , x 2 , [x/(x-2) 3 and x < 2] x 3
Suppose x R , x 2 , [x/(x-2) 3 and x < 2] is true,
Then x -2 > 0,
So, x/
Lesson 29
Sect.1.3
1. p
(q r)
p
T
T
T
T
F
F
F
F
(p
q)
p
q
T
F
T
F
T
F
T
F
qr
r
T
T
F
F
T
T
F
F
p
T
T
T
F
T
T
T
F
(q r)
T
T
T
F
T
T
T
T
r
q
p
r
q
T
T
T
F
T
T
T
F
F
T
F
F
F
T
F
F
Because the truth tables of [p
T
F
T
T
F
F
F
T
T
F
T
T
F
F
F
T
(q r)] and [(p
Lesson 30
(b) If R and S are reflexive, then R S is reflexive.
The statement is true.
Proof: Suppose (x, y) R S,
then (x, y) R or (x, y) S.
Assume R and S are reflexive, then by the definition of reflexive,
(x, x), (y, y) R or (x, x), (y, y) S.
So, we pro
Lesson 34
(g) If R and S are equivalence relations, then R S is an equivalence relation.
The statement is true.
Proof: Suppose R and S are equivalence relations, then
Reflexive: Assume R and S are reflexive,
then by the result of 31(a), which is if R and
Lesson 6
1.
Why the statement (A) Chris is tall or Robin is tall does not mean the same as the
statement (C) Chris is tall and Robin is not tall.
Explanation: Suppose statement p is Chris is tall and statement p is Robin is
tall, so the statement (A) is p
Lesson 1
1.
All integers are not even.
In this sentence, the placement of not can be in are not or not even. That means
this sentence could be considered as all integers are not even, or all integers are not
even. The meaning of these two sentences are di
Math 323
Exam 2C Solutions 6-7
21st century
6. For the purposes of this exam, we will say that an integer p is regular iff
there exists an integer k such that p = 4 k + 1.
For the purposes of this exam, we will say that an integer p is constipated iff it
Math 323
Exam 3, Problem 4, Solution and Comment
21st century
Problem 4 on the Third Exams was some version of the following:
4. For each real number x, define f (x) = x2. Then f : R R .
[This is obvious; you dont have to prove it.]
Let J be the open inte
Math 323
Exam 2D Solutions 8-9
21st century
8. (a) Explain briefly why R = cfw_ ( -1, -1), (2, 2), is a relation.
SOLUTION. R is a relation because it is a set of ordered pairs i.e., it is a set all of whose
elements are ordered pairs.
(b) Let A = cfw_ -1
Math 323
Final Exam, COMMENTS on Easy Problems, no Solutions
21st century
Minor typographical corrections, 5 July 2013, 10:10 AM
In numerical order, quick and easy problems only.
Notice the recurring theme: How do you start the proof? and DEFINITIONS.
1.
Math 323
Exam 3 , Solutions, 1-6
21st century
0. Suppose you are given a set S and a set T, and you want to prove that S T. Describe the standard
way of starting the proof and explain what you have to do to complete the proof.
SOLUTION. Consider an (arbit
Math 323
Exam 3 , Solutions, 7-9
21st century
7. Define f : cfw_1, 2, 3 R by f (1) = 10, f (2) = 11, f (3) = 11. Let T = cfw_11, 12.
Find f -1 (T) and explain your answer briefly.
SOLUTION. f -1 (T) = cfw_2, 3. Explanation:
The only possible elements of f
Review Session (Tuesday 6 July) Math 323 10/20/14 8:48 AM
7 questions
1) definition/theorem: 5 questions, nothing to prove, only
statements
o Need to be able to prove using the definition
n) prove inf/sup/max/min
m) limits
o If general, then exactly th
Study Guide Math 323 Exam 2
10/20/14 8:49 AM
Definitions
Set: A set is a collection of objects called members or elements
with some defining property.
A set A is a subset of set B if every element of A is an element of B.
If a is an element of A and b
Lesson 4
Sect. 1.1: Exercises
4. (b) The relation R is reflexive or symmetric.
Negation: The relation R is not reflexive and it is not symmetric.
(d) x is in A or x is not in B.
Negation: x is not in A and x is in B.
(e) If x <7, then f(x) is not in C.
Ne
Lesson 32
(c) if x is rational and y is irrational, then x + y is irrational.
Contrapositive: if x + y is rational, then x is irrational or y is rational.
Rewrite: (x + y is rational)
Proof: Since [p
(q r)]
(x is irrational y is rational)
q)
[(p
To do th
Lesson 35
Part I: Determine whether the following relations are transitive, and prove your answer
clearly and completely.
1. The empty set, .
The empty set is transitive.
Proof: Suppose (a, b)
and (b, c)
Since there is no elements in
(a, b)
,
,
and (b,
Lesson 31a
24. Define a relation R on the set of all integers
by x R y iff x y = 3k for some integer k.
Verify that R is an equivalence relation and describe the equivalence class E 5. How many distinct
equivalence classes are there?
Assume R is an equiva
Math 323 Formal Mathematical Reasoning and Writing
Problem Set 6 Due Wednesday, October 8
Ryleigh Fitzpatrick
p
0(PS5.7). Prove that the order (l) in the eld Q( 2) = [a + b |a, bQ] given by
s + t l u + v if f |s
u|(s
u) < 2|v
t|(v
t)
satises trichotomy.
p
Math 323 Formal Mathematical Reasoning and Writing
Problem Set 7 Due Friday, October 17
Ryleigh Fitzpatrick
1. Let a, b R with a < b. Prove that there exists s Q such that a < s < b.
/
By the Density Theorem we know that there exists r Q such that a < r <
Math 323 Formal Mathematical Reasoning and Writing
Problem Set 10 Due Friday, November 14
Ryleigh Fitzpatrick
1. Give an example of two functions f : R R and g : R R that do not commute under
composition.
For two functions to not commute under composition
Math 323 Formal Mathematical Reasoning and Writing
Problem Set 12 Due Wednesday, November 26
Ryleigh Fitzpatrick
1. Let f : R R be dened by
f (x) =
2x 3
2x 7
if x < 1
if x 1
Compute the following images and pre images, and write your answers in as simple
Math 323 Formal Mathematical Reasoning and Writing
Problem Set 13 Due Friday, December 5
Ryleigh Fitzpatrick
0.5. Prove if f : A B is a function with domain A and Ti with i I is a family of sets where
i I, Ti B, then
f 1
f 1 (Ti )
Ti =
iI
iI
Assume f : A
Math 323 Formal Mathematical Reasoning and Writing
Problem Set 14Extra Credit Due Friday, December 12
Ryleigh Fitzpatrick
1. Find the interior, boundary, closure, isolated points and accumulation points of the following
sets. Also say if they are open or
Lesson 35
Part II: A student makes the following claim:
Suppose R is a symmetric relation on a set A.
Suppose x and y are in A.
Then (x, y) is in R and (y, x) is in R.
Explanation:
Let A = cfw_1, 2, 3, R = cfw_(1, 2), (2, 1).
By the definition of symmetri
Lesson 34
(f) If R and S are transitive, then R S is transitive.
Pretend that the statement is true.
I start the proof by assuming R and S are transitive.
Suppose (x, y) R S, (y, z) S R.
By the definition of union,
(x, y) and (y, z) R S.
Then, I get stuck
Lesson 33
(e) If R and S are transitive, then R S is transitive.
The statement is true.
Proof: Assume R and S are transitive.
Suppose (x, y) (y, z) R S,
then (x, y) (y, z) R and (x, y) (y, z) S.
Since R is transitive, (x, z) R.
Similarly, (x, z) S.
By the
Lesson 30
(a) If R and S are reflexive, then R S is reflexive.
The statement is true.
Proof: Suppose (x, y) R S,
then (x, y) R and (x, y) S.
Assume R and S are reflexive, then by the definition of reflexive,
(x, x), (y, y) R and (x, x), (y, y) S.
So, (x,
Lesson 34
(h) If R and S are equivalence relations, then R S is an equivalence relation.
The statement is false.
Explanation: Assume R and S are transitive,
then by the result of 31(f),
R S is transitive is false.
Therefore, the original statement is fals