proofs refresher for math 335 { fall 2020 an overview of ......9 solutions to all exercises45 this...

Proofs Refresher for Math 335 – Fall 2020

An Overview of Logic, Proofs,

Set Theory, and Functions

for a Point Set Topology Course

aBa Mbirika∗

Contents

1 Numerical Sets and Other Preliminary Symbols 3

2 Statements and Truth Tables 5

3 Implications 9

4 Predicates and Quantifiers 13

5 Writing Formal Proofs 20

6 Mathematical Induction 27

7 Quick Review of Set Theory & Set Theory Proofs 32

8 Functions, Bijections, Compositions, Etc. 39

Index 44

9 Solutions to all exercises 45

∗This document is based on an earlier version originally written jointly by aBa Mbirika and Shanise Walker.This current version is geared specifically for those about to take Math 335 – Point Set Topology.

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Section 1 NUMERICAL SETS AND OTHER PRELIMINARY SYMBOLS

1 Numerical Sets and Other Preliminary Symbols

The following are numerical sets that you should familiarize yourself with:

natural numbers 1 N “ t1, 2, 3, . . .u

integers Z “ t. . . ,´3,´2,´1, 0, 1, 2, 3, . . .u

even integers 2Z “ t2k | k P Zu

odd integers 2Z` 1 “ t2k ` 1 | k P Zu

rational numbers Q “

ab | a, b P Z and b ‰ 0

(

irrational numbers RzQ (or equivalently, R´Q)

real numbers R “ trational and irrational numbersu

complex numbers C “ ta` bi | a, b P R and i “?´1u

Gaussian integers Zris “ ta` bi | a, b P Z and i “?´1u

Eisenstein integers Zrρs “ ta` bρ | a, b P Z and ρ “ e2πi3 u

CULTURAL QUESTION 1: Why are the integers denoted Z?

ANSWER: The German word for “numbers” is Zahlen. Germans con-tributed much to Zahlentheorie (number theory).

CULTURAL QUESTION 2: Why are the rationals denoted Q?

ANSWER: It was first denoted Q in 1895 by Giuseppe Peano afterquoziente, Italian for “quotient”.

1We do not consider the number 0 to be a natural number, though it is common in fields such as computerscience where loops and array elements start with the 0th counter.

Section 1 NUMERICAL SETS AND OTHER PRELIMINARY SYMBOLS

Below is a list of symbols with their associated meanings that we will comeacross in this proofs refresher.

SymbolNZQRCH

UY

Xš

P

A Ă B

A Ą BA Ć B

A Č B

A´B or AzBAˆBa | b

gcdpa, bqlcmpa, bq

Meaningset of natural numbers (we exclude 0)set of integersset of rational numbersset of real numbersset of complex numbersthe nullset or emptysetthe universal setunionintersectiondisjoint unionis an element ofA is a subset of BB is a subset of AA is not a subset of BB is not a subset of Aset difference “A without B”Cartesian product of sets A and Ba divides bgreatest common divisor of a and b

least common multiple of a and b

We also have some common abbreviations used in proofs for the most part asfollows.

Abbreviations@

D

BWOCWLOGTFAE

s.t.: or equivalently |

ùñ

WWTSQ.E.D.

Meaningfor allthere existsby way of contradictionwithout loss of generalitythe following are equivalentsuch thatsuch that (used in set-theory notation)implieswe want to showquod erat demonstrandum [end of proof]

Section 2 STATEMENTS AND TRUTH TABLES

2 Statements and Truth Tables

Definition 2.1. A statement (or proposition) is a declarative sentence thatis true or false but not both.

Exercise 2.2. Which of the following are statements? Write NS if it is not astatement and S if it is. Also give the truth value if it is a statement. [You Do!]NOTE: To answer these you may need to first peruse Chapter 0 of our textbook.

(a) A set A Ă Rn is called bounded if there exists a b P R such that |x| ě b forall x P A.

(b) Covid-19 totally sucks.

(c) The torus and the sphere are not topologically equivalent.

(d) Topology is popularly considered to have begun with Leonard Euler’s (1707–1783) solution to the famous Konigsberg bridges problem. [NOTE: Be carefulthinking about this one!]

(e) The set of points in a filled-in triangle Ĳ in R2 is a convex subset of R2.

(f) The set of points in a filled-in star ‹ in R2 is a convex subset of R2.

(g) This statement is false.

(h) Every relation between a set X and a set Y is a function from X to Y .

(i) Every function from a set X to a set Y is a relation between X and Y .

Definition 2.3. A compound statement is a statement which results fromthe application of one or more logical connectives (for example, “not” „, “and”^, “or” _, or “if-then” ùñ ) to a collection of simple statements.

Definition 2.4. A truth table is a mechanism for determining the truthvalues of compound statements.


Example 2.5. Below are the truth tables for negations, conjunctions, disjunc-tions, and implications.

(i) NEGATION: “not” symbol „

A „„„ A

T FF T

The rule is: “Not” reverses the truth value.

(ii) CONJUNCTION: “and” symbol ^

A B A ^ B

T T TT F FF T FF F F

The rule is: An “and” is true ONLY WHEN both sides are true.

(iii) DISJUNCTION: “or” symbol _

A B A___B

T T TT F TF T TF F F

The rule is: An “or” is false ONLY WHEN both side are false.

(iv) IMPLICATION: “if-then” symbol ñ

A B A ùñ B

T T TT F FF T TF F T

The rule is: An “if-then” is false ONLY WHEN the hypothesis holdsbut the conclusion fails.


Definition 2.6. A vacuously true statement is an ‘if-then” statement inwhich the “if”-part (i.e., hypothesis) is false. This statement is true regardlessof the truth value of the “then”-part (i.e., the conclusion). For example,

• If 1 “ 0, then Dr. Trump wins the 2020 Presidential Election.

Equivalently, a vacuous truth is a statement that asserts that all members ofan empty set possess a certain property. For example,

• Every green topology student in our clas is named Dr. Fauci.

Definition 2.7. (Tautology versus Contradiction)

• A tautology is a statement that is always true.

• A contradiction is a statement that is always false.

Exercise 2.8. Is the statement pB ^ „ Aq _ A a tautology, contradiction, orneither? [You Do!]

A B „„„ A B ^ „„„ A pppB ^ „„„ Aqqq ___A

T T F F T

T F F F T

F T T T T

F F T F F

We conclude that pB ^ „ Aq _ A is: neither a tautology nor a contradiction.

Definition 2.9. Two statements are logically equivalent if they have thesame truth tables. We use the symbol ” to denote this.


Exercise 2.10. Is „ pA^Bq ” „ A _ „ B? [You Do!]

A B A ^ B „„„ pppA ^ Bqqq „„„ A „„„ B „„„ A ___ „„„ B

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

We conclude that: „ pA^Bq is logically equivalent to „ A _ „ B.

Definition 2.11. De Morgan’s Laws assert the following logical equivalen-cies:

• „ pA_Bq is logically equivalent to „ A ^ „ B.

• „ pA^Bq is logically equivalent to „ A _ „ B.

Exercise 2.12. Use De Morgan’s Law to negate the following statement:

“I tested positive for Covid-19, but I have no symptoms.” [You Do!]

The negation is: I tested negative for Covid-19, or I have symptoms.

CAREFUL! Sometimes the word “but” is used in place of “and” when the partof the independent clause that follows the “but” is unexpected.

Exercise 2.13. Use De Morgan’s Law to negate the following statement:

´5 ď x ă 4 [You Do!]

The negation is: This is secretly an AND-statement: ´5 ď x and x ă 4. So thenegation is x ă ´5 or x ě 4.

Section 3 IMPLICATIONS

3 Implications

First recall the “If-Then” truth table below:

A B A ùñ B

T T T

T F F

F T T

F F T

Exercise 3.1. Compute the truth table for „ B ùñ „ A. [You Do!]

A B „„„ A „„„ B „„„ B ùñ „„„ A

T T F F T

T F F T F

F T T F T

F F T T T

QUESTION: Is A ùñ B logically equivalent to „ B ùñ „ A? Hell Yah!

Definition 3.2. The contrapositive of the statement A ùñ B is the state-ment „ B ùñ „ A.

Exercise 3.3. Compute the truth table for B ùñ A. [You Do!]

A B B ùñ A

T T T

T F T

F T F

F F T

QUESTION: Is A ùñ B logically equivalent to B ùñ A? Nope!

Definition 3.4. The converse of the statement A ùñ B is the statementB ùñ A.


Exercise 3.5. Compute the truth table for „ A ùñ „ B. [You Do!]

A B „„„ A „„„ B „„„ A ùñ „„„ B

T T F F T

T F F T T

F T T F F

F F T T T

QUESTION: Is A ùñ B logically equivalent to „ A ùñ „ B? Nope!

Definition 3.6. The inverse of the statement A ùñ B is the statement„ A ùñ „ B.

Exercise 3.7. Consider the statement “If the determinant of A is 1, then A is theidentity matrix.”

State the contrapositive and give its truth value?: [You Do!] If A is notthe identity matrix, then the determinant of A is not 1. - FALSE.

State the converse and give its truth value?: [You Do!] If A is the identitymatrix, then the determinant of A is 1. - TRUE

State the inverse and give its truth value?: [You Do!] If the determinant ofA is not 1, then A is not the identity matrix. - TRUE

If implication P ùñ Q is false, is it necessarily true that the inverse istrue? (Yes or No): [You Do!] No.


The Or-Form of an Implication

Exercise 3.8. Compute the truth table for „ A_B and compare it to A ùñ B.[You Do!]

A B „„„ A „„„ A_B A ùñ B

T T F T T

T F F F F

F T T T T

F F T T T

QUESTION: Is A ùñ B logically equivalent to „ A_B? Hell Yah!

Exercise 3.9. Consider the statement “If I don’t wear a mask, then I am helpingspread coronavirus.”

Write the Or-Form of the implication statement: [You Do!] Either I weara mask or I am helping spread coronavirus.

NOTE: This exercise above is sponsored by Dr. Fauci and the C.D.C.

The Negation of an Implication

Since we know that A ùñ B logically equivalent to „ A_B, then it is simpleto find the negation of A ùñ B. We simply take the negation of its equivalentform „ A_B.

Exercise 3.10. What is the negation of „ A _ B? A ^ „ B Verify this isindeed „ pA ùñ Bq via the truth table. [You Do!]

A B „„„ B A ^ „„„ B A ùñ B „„„ pA ùñ Bq

T T F F T F

T F T T F T

F T F F T F

F F T F T F


Exercise 3.11. Consider the statement “If I practice social distancing, thenCovid-19 won’t find me.”

Write the negation: [You Do!] I did practice social distancing, but Covid-19did find me.

Exercise 3.12. Write a sequence of logical equivalencies to prove again (but with-out truth tables that the negation of A ùñ B is indeed A ^ „ B. [You Do!]

„ pA ùñ Bq ” „ p„ A_Bq ” „„ A ^ „ B ” A ^ „ B

Example 3.13. Consider the statement

“Today is not Thanksgiving, or tomorrow is Friday.”

What is the truth value of this statement?NOTE: Do not assume today is this very day.

HINT: Consider all possible cases of the day of the week it can be, and whethertoday is Thanksgiving or tomorrow is Friday holds in each case.

If today is Friday, then it is DEFINITELY not Thanksgiving. The samegoes for the cases of Saturday, Sunday, Monday, Tuesday, and Wednesday.Now on the final case when it is Thursday, then we don’t know if itis Thanksgiving; however, it is certain that “tomorrow is Friday” willdefinitely hold. So the OR-STATEMENT is true.

Exercise 3.14. Write down the If-Then form of the Or-Statement in the exampleabove. [You Do!]

If today is Thanksgiving, then tomorrow is Friday.

Section 4 PREDICATES AND QUANTIFIERS

4 Predicates and Quantifiers

Definition 4.1. A predicate (or propositional function) is a sentence thatcontains a finite number of variables and becomes a proposition when specificvalues are substituted for the variables.

The domain of a predicate is the set of all values that may be substitutedin place of the variable.

Example 4.2. Let D be the set of all UWEC math students this Fall 2020. LetP pxq be the predicate

“x is taking topology”.

Exercise 4.3. For which x in the subset tTyler,Rayna,Collinu Ă D is the predi-cate P pxq from the example above true? [You Do!]Rayna and Collin are, but Tyler is not.

Definition 4.4. A quantifier is a logical symbol which makes an assertionabout the set of values which make one or more predicates true.

QUESTION: What are two VERY WELL-KNOWN quantifiers that youuse a lot in mathematics?

ANSWER: “for all” @ and “there exists” DNote: “there exists” is usually followed by a “such that”. We some-times abbreviate “such that” by “s.t.”.

Exercise 4.5. Let P px, yq be the predicate x´ y P Z. [Answer “Yes” or “No”!] If“No”, then provide x, y-values that give a counterexample.

• If D “ R, is P px, yq true @x, y P D? No, set x “ π and y “ e.

• If D “ Q, is P px, yq true @x, y P D? No, set x “ 12 and y “ 3.

• If D “ Z, is P px, yq true @x, y P D? Yes.

• If D “ N, is P px, yq true @x, y P D? No, set x “ 3 and y “ 4.


Universal and Existential StatementsREMARK: In the following, let Qpxq be a predicate with domain D.

Definition 4.6. A universal statement is a statement using a universalquantifier @ as follows:

@x P D, Qpxq.

For example, let D “ tmammalsu and Qpxq “ “x has mammary glands”.

• “@x P D, Qpxq” in words means “Every mammal has mammary glands.”

Definition 4.7. An existential statement is a statement using an existen-tial quantifier D as follows:

Dx P D s.t. Qpxq.

For example, let D “ tmammalsu and Qpxq “ “x is a primate”.

• “Dx P D s.t. Qpxq” in words means “There is a mammal that is a pri-mate.”

How to (Dis)Prove a Universal and Existential Statements:

• To Prove: @x P D, Qpxq

Let x P D. Then show that Qpxq holds.

• To Disprove: @x P D, Qpxq

Find a counterexample (i.e., show Dx P D s.t. Qpxq fails).

• To Prove: Dx P D s.t. Qpxq

Find at least one x P D such that Qpxq holds.

• To Disprove: Dx P D s.t. Qpxq

Show that for all x P D, the statement Qpxq fails.


Exercise 4.8. Let D “ Z and consider the universal statement:

@x P D, we havex

x2 ` 7ą ´0.187

Is the predicate P pxq is “ xx2`7 ą ´0.187 ” true for all x P D? If false, give

counterexample.

ANSWER: [You Do!] Setting x “ ´3, we have xx2`7 equals ´ 3

16 , and ´ 316 ă

´0.187, so the predicate is not true for all x P D.

Exercise 4.9. Let D “ tNatalie, Dan, Quotes, Zoe, Katieu be a subset of the setof UWEC math majors’ names (in real life and/or according to aBa), and considerthe universal statement:

@x P D, we have x is taking topology.

Is the predicate P pxq is “x is taking topology” true for all x P D?

ANSWER: [You Do!] True, since Quotes is the same as Katie.

Remark 4.10. To prove the above you can use a technique called a proof byexhaustion for obvious reasons. ,

Exercise 4.11. Let D “ Z and consider the universal conditional2

x ă y and w ă z ùñ xw ă yz.

Is the predicate P px, y, w, zq is “x ă y and w ă z ùñ xw ă yz ” true for allx P D?

ANSWER: [You Do!] Setting x “ ´2 and y “ 0 and w “ 1 and z “ 2, we have´2 ă 0 and 1 ă 2, but xw ă yz does not hold since

xw “ p´2q1 “ ´2 ć 1 “ 02“ yz.

So the predicate is not true for all x P D.

2This term “universal conditional” will be explained on the next page in Remark 4.12.


Remark 4.12. The statement in the previous example is called a universal con-ditional statement since it is understood to mean

@x, y, w, z P R, P1px, y, w, zq ùñ P2px, y, w, zq

where P1px, y, w, zq is x ă y and w ă z, and P2px, y, w, zq is xw ă yz.

Example 4.13. Universal conditional statements in informal English are quiteoften implied rather than specific. Let us write the following statements formally(using set-theoretic terms for the numerical sets at the start of Section 1).

(i) All integers are either odd or even.

For all x P Z, we have x is either odd or even.

(ii) No irrational numbers are integers.

For all x P RzQ, we have x R Z.

A Helpful Proof Tip (using negations)

To prove a statement is true, it is often helpful (or easier) to insteadconsider proving the NEGATION of the statement is false.

Exercise 4.14. Read the next page where we show how to formally negate auniversal conditional statement. Then write the negation for each of the twouniversal conditional statements above in Example 4.13. [You Do!]

(i) There exists an x P Z such that x is neither even nor odd.

(ii) There exists an x P RzQ such that x P Z.


Negations of Universal and ExistentialStatements

How to Negate Universal and Existential Statements:

• To negate a universal statement:

„ p@x P D, Qpxq q ” Dx P D s.t. „ Qpxq

• To negate an existential statement:

„ p D x P D s.t. Qpxq q ” @x P D, „ Qpxq

Exercise 4.15. Negate “there is a natural number that is even and prime”.

ANSWER: [You Do!] Every natural number is odd or composite.

Exercise 4.16. Negate “all dogs go to heaven”.

ANSWER: [You Do!] There exists a dog that does not go to heaven.

NOTE: This negation is false because it is well known that it is indeed truethat ALL dogs go to heaven as can be seen in the image below3. And hence theuniversal statement is true.

3This painting was done by artist Jim Warren and appears with his permission. You can see more of his workat https://jimwarren.com/.

https://jimwarren.com/


Statements with Multiple Quantifiers

Many statements mix @ and D, It important to note the following;

• Order of the symbols @ and D matters!!

• Order does not matter if there are two @’s or two D’s.

Example 4.17. Let the domain D be the set of all people. And let P px, yq bethe predicate “x loves y”.

• Interpret the following: @x P D, D y P D s.t. P px, yq

Everyone loves at least one person.

• Interpret the following: Dx P D s.t. @ y P D, P px, yq

There exists a person who loves everyone.

How to Negate Multiply-Quantified Statements:

• To negate a universal-existential statement:

„ p@x P D, D y P E s.t. P px, yq q is logically equivalent to

Dx P D s.t. @ y P E, „ P px, yq

• To negate an existential-universal statement:

„ p D x P D s.t. @ y P E, P px, yq, q is logically equivalent to

@x P D, D y P E s.t. „ P px, yq


How to Prove a Universal-Existential Statement:

• To Prove: @x P D, D y P E s.t. P px, yq

Let x P D. And find a y P E such that P px, yq holds.

How to Prove an Existential-Universal Statement:

• To Prove: Dx P D s.t. @ y P E, P px, yq

Find an x P D for which no matter what choice of y P E,we have P px, yq holds.

The Calculus II definition of sequence convergence is a multiply-quantified statement!

A sequence panq in R converges to a point a P R if

@ ε ą 0, DN0 P N such that n ě N0 ùñ |an ´ a| ă ε.

Exercise 4.18. Negate the continuity definition above (using English and not @-Dsymbols).

ANSWER: [You Do!] There exists an ε ą 0 such that for any N P N, we havefor all n ě N the an are outside the ε-tunnel of a (i.e., |an ´ a| ě ε).

Section 5 WRITING FORMAL PROOFS

5 Writing Formal Proofs

The Four Types of Proofs:

• (I) Direct Proofs

• (II) Proof by Cases

• (III) Proof by Contradiction

• (IV) Proof by Contrapositive

Remark 5.1. Tips and best starting points for a proof:

1. Start by writing what hypotheses (i.e., assumptions) are given (e.g., x P Q).

2. Write down your WWTS in the proof RIGHT AFTER stating the assump-tions, and perhaps put it in a box or a bubble like so (to remind you that isNOT part of your proof):

WWTS: Blah Blah Blah.

NOTE: In our class, you are REQUIRED to use the WWTS bubble (exceptin the “Proof by Contradiction” method which never has a WWTS).

3. Mathematically/Symbolically say what that means. (e.g., x “ ab for some

a, b P Z with b ‰ 0). This may be done at times right before the WWTS.

4. Some things NOT to do!!!!!!!!:

(a) NEVER replace English words with math blackboard slang.

(b) Do not put mathmode symbols both before [AND] after a comma.

Do not write garbage like: “Choose a R x s.t. @y, y ą x.”

EXAMPLE: Rewriting the garbage in the box above so it looks less “trashy”:

Choose a real number x such that for all y, we have y ą x. 4

4Of course, this statement though written very pretty, is not true. [WHY?]


How to Prove the Four Types of Proofs

(I) DIRECT PROOF:

How to Prove the Universal Conditional Statement:

For all x P D, we have P pxq implies Qpxq.

• If the statement is not already in the form above, then write it in this form.

• *FIRST SENTENCE OF PROOF*

Assume x P D and suppose P pxq holds.

• *SECOND SENTENCE OF PROOF*

WWTS: Qpxq holds.

NOTE: The exact words of Qpxq need not be in the bubble. Sometimes youput there only what it suffices to show.

• *BODY OF PROOF*

• *LAST LINE OF PROOF* We conclude that Qpxq holds.

• Put a Q.E.D. or just a .

(II) PROOF BY CASES:

How to Prove the Conditional Statement:

If A1 or A2 or A3 or . . . or An, then B.

• (CASE 1) Assume A1 holds.WWTS: B holds.

Then prove this.

• (CASE 2) Assume A2 holds.WWTS: B holds.

Then prove this.

• ... etc...

• (CASE n) Assume An holds.WWTS: B holds.

Then prove this.

• *LAST LINE OF PROOF* We conclude that B holds.



(III) PROOF BY CONTRADICTION:


If A, then B.

• Assume A holds.

• Suppose by way of contradiction (BWOC) that B is false.

• THERE IS NO WWTS IN THIS TYPE OF PROOF!!!

• Derive any contradiction.

• Hence it cannot be that B is false.

• *LAST LINE OF PROOF* We conclude that B holds.


(IV) PROOF BY CONTRAPOSITIVE:


If A, then B.

• Restated it in its contrapositive form, we have

If „ B, then „ A.

NOTE: You don’t write this contrapositive form in your formal proof.

• *FIRST SENTENCE OF PROOF*

Assume „ B holds.

• *SECOND SENTENCE OF PROOF*

WWTS: „ A holds.

NOTE: The exact words of „ A need not be in the bubble. Sometimes youput there only what it suffices to show.

• *BODY OF PROOF*

• *LAST LINE OF PROOF* We conclude that „ A holds.



An Example of a Direct Proof

Example 5.2. Prove the following multiply-quantified statement:

@x P R, D y P R such that x` y “ 0.

Proof. Let x P R.

WWTS: D y P R s.t. x` y “ 0.

Set y :“ ´x. Observe that y P R and it follows that

x` y “ x` p´xq “ 0

as desired. Thus the claim holds.

Q.E.D.

Exercise 5.3. Prove that the following statement is false:

Dx P R such that @ y P R, we have x2` y2

“ 8. (1)

RECALL: To prove a statement is false either find a counterexample, OR provethat the statement’s negation is true. Statement (1) does not lend itself to findinga counterexample, so let’s prove the negation is true. WHAT IS THE NEGATIONOF STATEMENT (1)? [You Do!]

@x P R, D y P R such that x2` y2

‰ 8.

Now prove the negation. [You Do!]

Proof. Let x P R.

WWTS: D y P R s.t. x2 ` y2 ‰ 8.

Either x “ ˘2 or it is not. If x “ ˘2, then choose y :“ 0. Observe thatx2 ` y2 “ 22 ` 02 “ 4 ‰ 8. But if x ‰ ˘2, then choose y :“ x. Observe thatx2 ` y2 “ 2x2 ‰ 2p˘2q2 “ 8. Thus the negation of Statement (1) is true, so weconclude that Statement (1) is false.

Q.E.D.


An Example of a Proof by Contrapositive

This lemma below literally CANNOT be proven directly!!

Think about it. Give it a shot! Your proof would start as follows:

Proof. Assume n2 is even.

WWTS: n is even.

Since n2 is even, then n2 “ 2k for some k P Z.

THERE IS LITERALLY NOWHERE TO GO FROM HERE! THINKABOUT IT!

Lemma 5.4. If n2 is even, then n is even.

Proof. Let n be an odd integer.

WWTS: n2 is odd.

Since n is odd, then n “ 2k ` 1 for some k P Z. Squaring n we get the

following sequence of equalities

n2“ p2k ` 1qp2k ` 1q “ 4k2

` 4k ` 1

“ 2p2k2` 2kq ` 1.

Set M :“ 2k2 ` 2k. Observe that M P Z by closure of multiplication

and addition in Z. Hence n2 “ 2M ` 1 is odd. Therefore we conclude

If n2 is even, then n is even.

Q.E.D.

NOTE: It is nice to end a proof by contrapositive by stating the statement asoriginally written. But it is not necessary too; ending with “n2 is odd” is fine too.


An Example of a Proof by Contradiction

The so-called FIRST CRISIS IN MATH!!

Long ago, in the 6th century BC, the followers of the school of Pythagoras,the Pythagoreans, came to a crisis in math. It was a long held view that allnumbers are the ratio of two integers. However, the Pythagorean memberHippasus is thought to be the one to discover that there is a length thatcannot be the ratio of two integers – a Greek Crisis! This length was the

?2.

It is rumored that the drowning of Hippasus was the punishment from thegods for divulging this secret5.

Theorem 5.5. The value?

2 is irrational.

Proof. Assume by way of contradiction that?

2 P Q. Then?

2 “ ab

where a, b P Z with b ‰ 0, and we may assert that the greatest common

divisor of a and b is 1. Then we have the following sequence of implications?

2 “a

bùñ

?2b “ a

ùñ 2b2“ a2 by squaring both sides

ùñ a2 is even

ùñ a is even by Lemma 5.4.

Hence a “ 2k for some k P Z. Substituting this value a in the equality?

2b “ a, we have?

2b “ 2k and the following sequence of implications?

2b “ 2k ùñ 2b2“ 4k2 by squaring both sides

ùñ b2“ 2k2

ùñ b2 is even

ùñ b is even by Lemma 5.4.

However, both a and b being even contradicts the fact that 1 is theirgreatest common divisor. Therefore we conclude that

?2 is irrational.

Q.E.D.

5We would provide a reference for this, but there is a lot of documented debate with historians about the firstGreek discoverer of the irrationality of

?2. But for sure, the drowning of Hippasus is certain.


An Example of a Proof by Cases

An Important Consequence of the Division Algorithm

Theorem 5.6 (Division Algorithm). Given integers a and b, with b ą 0,there exist unique integers q and r satisfying

a “ qb` r such that 0 ď r ă b.

A consequence of this theorem is one super well-known result that everyinteger is even (i.e., in 2Z) or odd (i.e. in 2Z` 1). The 2 is replaceable here.For instance, every integer lies in ONE AND ONLY one of the sets 3Z or3Z` 1 or 3Z` 2.

Theorem 5.7. If n is an integer not divisible by 3, then n2 P 3Z` 1.

Proof. Assume n is an integer not divisible by 3. Then either n lies in

3Z` 1 or 3Z` 2.

(CASE 1:) Assume n P 3Z` 1.

WWTS: n2 P 3Z` 1.

Since n P 3Z` 1 then n “ 3m` 1 for some m P Z. Thus we have

n2“ p3m` 1q2 “ 9m2

` 6m` 1 “ 3p3m2` 2mq ` 1.

Set M :“ 3m2` 2m. Clearly M P Z, and hence n2 “ 3M ` 1 P 3Z` 1.

(CASE 2:) Assume n P 3Z` 2.

WWTS: n2 P 3Z` 1.

Since n P 3Z` 2 then n “ 3m` 2 for some m P Z. Thus we have

n2“ p3m` 2q2 “ 9m2

` 12m` 4 “ 3p3m2` 4m` 1q ` 1.

SetM :“ 3m2`4m`1. ClearlyM P Z, and hence n2 “ 3M`1 P 3Z`1.

Q.E.D.

Section 6 MATHEMATICAL INDUCTION

6 Mathematical Induction

The Falling Dominoes Analogy

Imagine that we want to prove that an infinite number of statements S1, S2,S3, etc. are all true. The dominoes analogy of math induction is the following:

• Visualize each statement as a domino. And visualize the act of the nth dominofalling to mean that statement Sn is proven true.

• Suppose you can knock over the first domino (i.e., you can prove S1).

• Suppose you can show that if any kth domino falling will definitely force thepk ` 1qth domino to fall.

• We can conclude that S1 falls and knocks down S2. Next S2 falls and knocksdown S3. Then S3 knocks down S4, etc.

• Hence all dominoes fall!!! Q.E.D.


The Principle of Mathematical Induction

Let P pnq be a property defined on the integers n. Let a P N be a fixedinteger. Suppose the following two statements are true:

1. P paq is true

2. For every k ě a, we have P pkq implies P pk ` 1q.

Then P pnq is true for all n ě a.

The Method of Mathematical Induction

PROVE: For all n ě a, it follow that P pnq is true.

(Proof)• Step 1: (Base Case) Show that P paq is true.

• Step 2: (Induction Hypothesis) Assume P pkq holds for some k ě a.WARNING!: The word “some” is VERY IMPORTANT! This exis-tential quantifying word asserts that the k is a PARTICULAR (i.e.,fixed) but arbitrarily chosen integer.

• Step 3: Show that P pk ` 1q holds. WARNING!: If you manage toprove P pk ` 1q without ever using the fact that P pkq holds, then yourproof is DEFINITELY wrong.

• Step 4: Conclude that P pnq is true for all n ě a.

Exercise 6.1. Suppose that you were to prove that?n ă 1?

1` 1?

2` ¨ ¨ ¨ ` 1?

nfor

all n ě 2. State the induction hypothesis and the WWTS. [You Do!]

Induction Hypothesis: Assume that?k ă 1?

1` 1?

2` ¨ ¨ ¨ ` 1?

kfor some k ě 2.

WWTS:?k ` 1 ă 1?

1` 1?

2` ¨ ¨ ¨ ` 1?

k`1.


Example 6.2. Prove that 2n` 1 ă 2n for all n ě 3.

Base Case: (n “ 3) Observe that 2p3q ` 1 “ 6 ă 8 “ 23, and hence

the base case holds.

Induction Hypothesis: Assume that 2k ` 1 ă 2k for some k ě 3.

WWTS: 2pk ` 1q ` 1 ă 2k`1.

It suffices to show that 2k ` 3 ă 2k`1. Consider the following sequence

of equalities and inequalities:

2k ` 3 “ p2k ` 1q ` 2

ă 2k ` 2 by Induction Hypothesis

ă 2k ` 2k since 2 ă 2k for all k ě 2

“ 2 ¨ 2k

“ 2k`1,

as desired. Hence 2n` 1 ă 2n for all n ě 3.

Q.E.D.


Exercise 6.3. Prove by induction that n2 ă 2n for all n ě 5. [You Do!]

HINT: In your attempt to prove the k`1 case holds, you mayfind it useful at some point to use the result from Example 6.2.

Base Case: (n “ 5) Observe that 52 “ 25 ă 32 “ 25, and hence the base caseholds.

Induction Hypothesis: Assume that k2 ă 2k for some k ě 5.

WWTS: pk ` 1q2 ă 2k`1.

Consider the following sequence of equalities and inequalities:

pk ` 1q2 “ k2` 2k ` 1

ă 2k ` 2k ` 1 by Induction Hypothesis

ă 2k ` 2k since 2k ` 1 ă 2k for all k ě 3 by Example 6.2

“ 2 ¨ 2k

“ 2k`1,

as desired. Hence n2 ă 2n for all n ě 5.

Q.E.D.


Exercise 6.4. For all n ě 3, prove by induction that the angle sum of a convexn-gon is 180pn´ 2q degrees. [You Do!]

NOTE: You may find it helpful somewhere in your proof to draw a schematic orpictorial representation to help illustrate your thought.

aBa to do still

Q.E.D.

Section 7 QUICK REVIEW OF SET THEORY & SET THEORY PROOFS

7 Quick Review of Set Theory & Set Theory Proofs

Definition 7.1. A set is any collection of objects.

• Repetition of objects is ignored. For instance tx, x, yu “ tx, yu.

• Order does not matter. For instance tx, yu “ ty, xu.

A member of a set is called an element. It is customary to use a capital letterto denote a set.

Definition 7.2. The cardinality of a set is the number of elements in the set.We use the symbols NpAq, npAq, or |A| to denote this “size” of a set A.

Two types of infinity 8 that will arise in this class are ℵ0 (aleph-naught) andℵ1 (aleph-one). The former is the cardinality of the natural numbers, while thelatter is the cardinality of the real numbers.

Definition 7.3. The nullset, or empty set, is the set that contains no ele-ments. It is denoted H or equivalently tu.

Exercise 7.4. Compute the following cardinalities. [You Do!]

A “ t H , tHu , tH, tH, tHuu u ùñ |A| “ 3

A “ ta, b, cu ùñ |A| “ 3

A “ ta, a, bu ùñ |A| “ 2

A “ R ùñ |A| “ ℵ1

A “ Q ùñ |A| “ ℵ0

A “

"ˆ

a11 a12

a21 a22

˙

|aij P Z and 0 ď aij ď i` j

*

ùñ |A| “ ℵ1

A “ tHu ùñ |A| “ 1


WARNING! Two Easily Confused Symbols: P versus Ă

Definition 7.5. The notation x P A means “x is an element of A”.

Definition 7.6. The notation A Ă B means “A is a subset of B”.

A Ă B if and only if pFor every x P A, we have x P Bq.

Exercise 7.7. Let A “ tπ, e, tπ, eu, tπuu. Answer the following. [True or False?]

π P A True

tπu R A False

e Ă A True

teu P A False

teu Ă A True

tteu, te, πuu Ă A False

ttπu, te, πuu Ă A True

QUESTION: Why is the nullset a subset of any set A?

ANSWER: Consider the statement H Ă A. Since there are no ele-ments in H, it is clear that ALL of the elements in H are contained inA, regardless of what set A is. This is called a vacuously true statement.

Definition 7.8. The power set of a given set A is the set of all subsets of A.We denote the power set of A by PpAq.

For example if A “ ta, b, cu, then the power set of A is

PpAq “!

H, tau, tbu, tcu, ta, bu, ta, cu, tb, cu, ta, b, cu)

.

Exercise 7.9. Let |A| “ 10. What is the cardinality of the power set of A? Ingeneral if |B| “ n, then what is |PpBq| equal to? [You Do!]


Definition 7.10. The universal set is a larger known set wherein a givenset lives. We denote it U . This set is generally understood in context ordeliberately stated.

EXAMPLE: Find the universal sets.

A “ tdogsu ùñ U “ animals

A “ teven integersu ùñ U “ integers

Definition 7.11. Let A and B be sets. Then the intersection of A and B(denoted A X B) is the set of elements in U that lie in both A and B. Set-theoretically,

AXB :“ tx P U | x P A and x P Bu.

Definition 7.12. Let A and B be sets. Then the union of A and B (denotedAYB) is the set of elements in U that lie in A or B (or both). Set-theoretically,

AYB :“ tx P U | x P A or x P Bu.

Definition 7.13. Let A and B be sets. The difference A´ B of A and B isdefined as tx P A | x R Bu. This is also called the complement of B in A.

NOTE: This set difference is sometimes written as AzB.

Definition 7.14. Let A be a subset of the universal set U . The complementof the set A (denoted Ac or A1) is the set of elements in U that do not lie in A.Set-theoretically,

Ac :“ tx P U | x R Au.


(ATTENTION!!!!) Notation for the complement

Remark 7.15. In topology, we will never be writing Ac or A1. We use set-difference notation. It is VITAL to know the ambient space for which the setA lives in. So if A is a subset of a topological space X. Then the complementof A is written ONLY as X ´ A.

Just as we have De Morgan’s Laws for logic symbols _ and ^ (recall Defini-tion 2.11), we have De Morgan’s Law for Y and X, respectively, as follows:

Definition 7.16. Let A and B be subsets of a set X. Then De Morgan’sLaws assert the following set equalities:

• X ´ pAYBq is equal to pX ´ Aq X pX ´Bq.

• X ´ pAXBq is equal to pX ´ Aq Y pX ´Bq.

Exercise 7.17. Prove that the following containment of De Morgan’s law: [YouDo!]

X ´ pAXBq Ă pX ´ Aq Y pX ´Bq

.

Proof. Let x P X ´ pAXBq.

WWTS: x P pX ´ Aq Y pX ´Bq.

Since x P X ´ pAXBq, then x P X and x R AXB. It suffices to show that eitherx P X ´ A or x P X ´ B. Since x R A X B, then x R A. Hence x P X and x R A,so thus x P X ´A. We conclude that x P pX ´Aq Y pX ´Bq as desired, and theresult follows.

Q.E.D.


Set Theory on Collections of Sets

Let J be any indexing set of any size (finite or infinite). Suppose for each j P J ,we have a set Aj. Consider the collection of sets tAj | j P Ju.

Definition 7.18. We have the following:

• “Big Union”:Ť

jPJ

Aj “ tx P U | x P Aj for some j P Ju

• “Big Intersection”:Ş

jPJ

Aj “ tx P U | x P Aj for every j P Ju

• Let A1, A2, . . . , An be a finite collection of sets. The Cartesian productof the collection is denoted A1 ˆ A2 ˆ ¨ ¨ ¨An and is defined as

tpa1, a2, . . . , anq | ai P Ai for i “ 1, 2, . . . , nu.

• Two sets A and B are disjoint if A X B “ H. Moreover the collectionof sets tAj | j P Ju is pairwise disjoint if Ai X Aj “ H for i, j P J suchthat i ‰ j.

• A collection of sets tAj | j P Ju is a partition of a set X if and only if

1. X “Ť

jPJ

Aj, and

2. the collection is pairwise disjoint.

We use the notation X “š

jPJ

Aj to denote this pairwise disjoint union.


Example 7.19. Prove that the sets 3Z, 3Z` 1, and 3Z` 2 form a partition of Z.

Proof. Consider the sets 3Z, 3Z` 1, and 3Z` 2.

WWTS: Z “š

j“0,1,2

3Z` j.

By the division algorithm (see Theorem 5.6), every integer n can be

represented in exactly one of three forms:

n “ 3k or n “ 3k ` 1 or n “ 3k ` 2

for some integer k, and hence Z “ 3Z Y 3Z ` 1 Y 3Z ` 2. This alsoimplies that the three sets are pairwise disjoint and hence Z “

š

j“0,1,2

3Z`

j.

Q.E.D.

How to prove two sets are equal?

Claim 7.20. Let A and B be sets. Then

A “ B if and only if A Ă B and B Ă A.

Exercise 7.21. Let A,B1, B2, . . . , Bn be sets. Prove that

AY

˜

nč

i“1

Bi

¸

Ą

nč

i“1

pAYBiq [You Do!]

Proof. We first show the Ă containment holds. Let x P 10Z` 7.

WWTS: x P 5Z` 2.

Since x P 10Z` 7 then x “ 10k ` 7 for some k P Z. Thus

x “ 10k ` 5` 2 “ 5p2k ` 1q ` 2.

Set M :“ 2k ` 1. Observe that M P Z and so x “ 5M ` 2 P 5Z ` 2. Thus10Z ` 7 Ă 5Z ` 2 as desired. We now show Ą containment does not hold. Sety “ 12 “ 5 ¨ 2` 2 P 5Z` 2. If y P 10Z` 7, then

12 “ 10k ` 7 for some k P Z ùñ 10k “ 12´ 7 ùñ k “5

10P Z

which gives a contradiction so 10Z` 7 Č 5Z` 2.

Q.E.D.


Group Work in Class and/or Homework

Exercise 7.22. Let A,B,C and D be sets. Determine which of the statementsbelow are true. If a double implication fails, determine whether one or the otherof the possible implications holds. If an equality fails, determine whether thestatement becomes true if the “equals” symbol is replaced by one or the other ofthe inclusion symbols Ă or Ą.[We’ll do some in class and some in a later HW assignment!]

(a) A Ă B and A Ă C ðñ A Ă pB Y Cq.

(b) A Ă B or A Ă C ðñ A Ă pB Y Cq.

(c) A Ă B and A Ă C ðñ A Ă pB X Cq.

(d) A Ă B or A Ă C ðñ A Ă pB X Cq.

(e) A´ pA´Bq “ B.

(f) A´ pB ´ Aq “ A´B.

(g) AX pB ´ Cq “ pAXBq ´ pAX Cq.

(h) AY pB ´ Cq “ pAYBq ´ pAY Cq.

(i) pAXBq Y pA´Bq “ A. [Let’s draw some Venn diagrams in class!]

(j) A Ă C and B Ă D ùñ AˆB Ă C ˆD.

(k) The converse of (j).

(l) The converse of (j), assuming that A and B are nonempty.

(m) pAˆBq Y pC ˆDq “ pAY Cq ˆ pB YDq.

(n) pAˆBq X pC ˆDq “ pAX Cq ˆ pB XDq.

(o) Aˆ pB ´ Cq “ pAˆBq ´ pAˆ Cq.

(p) pA´Bq ˆ pC ´Dq “ pAˆ C ´B ˆ Cq ´ AˆD.

(q) pAˆBq ´ pC ˆDq “ pA´ Cq ˆ pB ´Dq.

Section 8 FUNCTIONS, BIJECTIONS, COMPOSITIONS, ETC.

8 Functions, Bijections, Compositions, Etc.

The following is a variety of important definitions related to functions defined onarbitrary sets. For the following definitions, let X and Y be sets and consider afunction f : X Ñ Y .

Definition 8.1. The domain of f is the set X, and the codomain of f is theset Y .

Definition 8.2. The range of f is the set

ty P Y | fpxq “ y for some x P Xu.

Equivalently, this set is called the image of f .

Definition 8.3. Let A Ă X and B Ă Y . Then we have the following sets:

• The image of A under f is denoted fpAq and defined as

fpAq “ ty P Y | fpaq “ y for some a P Au.

• The preimage of B under f is denoted f´1pBq and defined as

f´1pBq “ tx P X | fpxq “ b for some b P Bu.

WARNING!!!!!: By using the symbol f´1pBq, we do not mean that thefunction f has an inverse. The symbol f´1pBq simply denotes a set. Forevery B Ă Y , the set f´1pBq (which may possibly be empty) is well-definedREGARDLESS if the function f has an inverse.

Exercise 8.4. Let X “ ta, b, c, d, eu and Y “ t1, 2, 3u. Define f : X Ñ Y suchthat fpaq “ 1, fpbq “ 2, and fpcq “ fpdq “ fpeq “ 3. Then compute the followingthree sets: image fpta, buq, preimage f´1pt3uq, and range of f . [You Do!]


Injective, Surjective, and Bijective Functions

Definition 8.5. Consider a function f : X Ñ Y . The function f is calledinjective if the following implication holds for all elements x1, x2 P X:

If fpx1q “ fpx2q, then x1 “ x2.

We sometimes refer to such a function as being one-to-one.

Definition 8.6. Consider a function f : X Ñ Y . The function f is calledsurjective if the following universal-existential statement holds:

For each y P Y , there exists an x P X such that fpxq “ y.

We sometimes refer to such a function as being onto.

Exercise 8.7. For each function denote in the corresponding box (Yes/No) if thefunction is injective and/or surjective. The first is done for you. [You Do!]

the function injective? surjective?

f : RÑ R by fpxq “ x2 No No

f : RÑ R´ t0u by fpxq “ x3 Yes No

f : RÑ R by fpxq “ x´1x2`3 No Yes

f : Ppta, b, cuq Ñ Z by fpXq “ |X| for each

subset X P Ppta, b, cuq. (See footnote6.) No No

Definition 8.8. Consider a function f : X Ñ Y . The function f is calledbijective if it is both injective and surjective. If so, we say that f gives aone-to-one correspondence between A and B.

6Recall the definition of “power set” given in Definition 7.8.


Exercise 8.9. Consider the following four sets

A “ tf : t1, 2, 3, 4u Ñ ta, b, cu | f is injectiveu

B “ tg : ta, b, cu Ñ t1, 2, 3, 4u | g is injectiveu

C “ th : t1, 2, 3, 4u Ñ ta, b, cu | h is surjectiveu

D “ ti : ta, b, cu Ñ t1, 2, 3, 4u | i is surjectiveu.

Compute the cardinalities of each of the sets, and provide a justification for each.[You Do!]


Composition of Functions

Definition 8.10. Consider functions f and g with f : X Ñ Y and g : Y Ñ Z.We define the composition of functions f and g by g ˝ f : X Ñ Z by

pg ˝ fqpxq :“ gpfpxqq for all x P X.

Exercise 8.11. Let f and g be functions with f : X Ñ Y and g : Y Ñ Z givenby the following arrow diagram:

Answer the following questions: [You Do!]

• What is`

g ˝ f˘

px1q,`

g ˝ f˘

px2q,`

g ˝ f˘

px3q, and`

g ˝ f˘

px4q?`

g ˝ f˘

px1q “ z2,`

g ˝ f˘

px2q “ z3,`

g ˝ f˘

px3q “ z4, and`

g ˝ f˘

px4q “ z1

• Fill in the table below with answers “Yes” or “No”.


g ˝ f Yes Yes

f Yes No

g No Yes


Theorem 8.12. Suppose that f : X Ñ Y and g : Y Ñ Z. Then

(i) If g ˝ f is injective, then f is injective.

(ii) If g ˝ f is surjective, then g is surjective.

NOTE: We will prove part (i) and leave it to you to prove part (ii).

Proof of (i). Assume g ˝ f is injective.

WWTS: f is injective.

Suppose we have fpaq “ fpbq for a, b P X . It suffices to show thata “ b. Since fpaq “ fpbq, then gpfpaqq “ gpfpbqq. It follows thatpg ˝ fqpaq “ pg ˝ fqpbq. Since g ˝ f is injective, then a “ b as desired.Hence f is injective.

Q.E.D.

Exercise 8.13. Prove part (ii) of Theorem 8.12. [You Do!]

Proof of (ii). Assume g ˝ f is surjective.

WWTS: g is surjective.

Let z P Z. It suffices to show that there exists a y P Y such that gpyq “ z.Since g ˝ f is surjective, there exists an x P X such that pg ˝ fqpxq “ z. Hencegpfpxqq “ z. Set y “ fpxq. Then gpyq “ z as desired. Thus g is surjective.

Q.E.D.

Index

Bbijective, 40

Ccardinality, 32Cartesian product, 36codomain, 39complement of a set, 34composition of functions, 42compound statement, 5conjunction, 6contradiction, 7contrapositive statement, 9converse statement, 9

DDe Morgan’s laws for logic, 8De Morgan’s laws for sets, 35disjoint sets, 36disjoint union, 36disjunction, 6division algorithm, 26, 37domain, 39domain of a predicate, 13

Eelement of a set, 32empty set, 32existential quantifier, 14existential statement, 14

HHippasus (fl. 500 B.C.), 25

Iimage, 39image of A under f , 39implication, 6induction, 28injective, 40intersection, 34inverse statement, 10

Llogically equivalent, 7

Nnegation, 6nullset, 32

Oone-to-one, 40one-to-one correspondence, 40onto, 40

Ppartition of a set, 36power set, 33predicate, 13preimage of B under f , 39proof by cases, 21proof by contradiction, 22proof by contrapositive, 22proof by exhaustion, 15proof, direct method, 21Pythagoras (c. 570–495 B.C.), 25

Qquantifier, 13

Rrange, 39

Sset, 32statement, 5surjective, 40

Ttautology, 7truth table, 5

Uunion, 34universal conditional statement, 16universal quantifier, 14universal set, 34universal statement, 14

Vvacuously true statement, 7

44

Section 9 SOLUTIONS TO ALL EXERCISES

9 Solutions to all exercises

Exercise 2.2: (a) S - False, (b) NS - “Even though the virus does indeed suck,this sentence is an opinion.”, (c) S - True, (d) NS - Even though this is written inour book, it is the authors’ opinion., (e) S - True, (f) S - False, (g) NS - Google“Liar’s Paradox”, (h) S - False, and (i) S - True.

Exercise 2.8: Is the statement pB ^ „ Aq _ A a tautology, contradiction, orneither?

A B „„„ A B ^ „„„ A pppB ^ „„„ Aqqq ___A

T T F F T

T F F F T

F T T T T

F F T F F

We conclude that pB ^ „ Aq _ A is: neither a tautology nor a contradiction.

Exercise 2.10: Is „ pA^Bq ” „ A _ „ B?

A B A ^ B „„„ pppA ^ Bqqq „„„ A „„„ B „„„ A ___ „„„ B

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

We conclude that: „ pA^Bq is logically equivalent to „ A _ „ B.

Exercise 2.12: I tested negative for Covid-19, or I have symptoms.

Exercise 2.13: x ă ´5 or x ě 4.

Exercise 3.1: Compute the truth table for „ B ùñ „ A.

A B „„„ A „„„ B „„„ B ùñ „„„ A

T T F F T

T F F T F

F T T F T

F F T T T

QUESTION: Is A ùñ B logically equivalent to „ B ùñ „ A? Hell Yah!


Exercise 3.3: Compute the truth table for B ùñ A.

A B B ùñ A

T T T

T F T

F T F

F F T

QUESTION: Is A ùñ B logically equivalent to B ùñ A? Nope!

Exercise 3.5: Compute the truth table for „ A ùñ „ B.

A B „„„ A „„„ B „„„ A ùñ „„„ B

T T F F T

T F F T T

F T T F F

F F T T T

QUESTION: Is A ùñ B logically equivalent to „ A ùñ „ B? Nope!

Exercise 3.7: Consider the statement “If the determinant of A is 1, then A isthe identity matrix.”State the contrapositive and give its truth value?: If A is not the identitymatrix, then the determinant of A is not 1. - FALSE.State the converse and give its truth value?: If A is the identity matrix,then the determinant of A is 1. - TRUEState the inverse and give its truth value?: If the determinant of A is not1, then A is not the identity matrix. - TRUEIf implication P ùñ Q is false, is it necessarily true that the inverse istrue? (Yes or No): No.

Exercise 3.8: Compute the truth table for „ A_B and compare it to A ùñ B.

A B „„„ A „„„ A_B A ùñ B

T T F T T

T F F F F

F T T T T

F F T T T

QUESTION: Is A ùñ B logically equivalent to „ A_B? Hell Yah!


Exercise 3.9: Consider the statement “If I don’t wear a mask, then I am helpingspread coronavirus.”

Write the Or-Form of the implication statement: Either I wear a mask orI am helping spread coronavirus.

Exercise 3.10: What is the negation of „ A _ B? A ^ „ B Verify this isindeed „ pA ùñ Bq via the truth table.

A B „„„ B A ^ „„„ B A ùñ B „„„ pA ùñ Bq

T T F F T F

T F T T F T

F T F F T F

F F T F T F

Exercise 3.11: The negation of the statement “If I practice social distancing,then Covid-19 won’t find me.” is the following

I did practice social distancing, but Covid-19 did find me.

Exercise 3.12: Write a sequence of logical equivalencies to prove again (butwithout truth tables that the negation of A ùñ B is indeed A ^ „ B.

„ pA ùñ Bq ” „ p„ A_Bq ” „„ A ^ „ B ” A ^ „ B

Exercise 3.14: Write down the If-Then form of the Or-Statement “Today is notThanksgiving, or tomorrow is Friday.”

If today is Thanksgiving, then tomorrow is Friday.

Exercise 4.3: Rayna and Collin are, but Tyler is not.

Exercise 4.5: Let P px, yq be the predicate x´ y P Z.

• If D “ R, is P px, yq true @x, y P D? No, set x “ π and y “ e.

• If D “ Q, is P px, yq true @x, y P D? No, set x “ 12 and y “ 3.

• If D “ Z, is P px, yq true @x, y P D? Yes.

• If D “ N, is P px, yq true @x, y P D? No, set x “ 3 and y “ 4.


Exercise 4.8: Setting x “ ´3, we have xx2`7 equals ´ 3

16 , and ´ 316 ă ´0.187, so

the predicate is not true for all x P D.

Exercise 4.9: Setting x “ 2, we get 22 ą 2` 1. Setting x “ 5, we get 52 ą 5` 1.Setting x “ 9, we get 92 ą 9 ` 1. Setting x “ 10, we get 102 ą 10 ` 1. So thepredicate is true for all x P D.

Exercise 4.11: Setting x “ ´2 and y “ 0 and w “ 1 and z “ 2, we have ´2 ă 0and 1 ă 2, but xw ă yz does not hold since

xw “ p´2q1 “ ´2 ć 1 “ 02“ yz.

So the predicate is not true for all x P D.

Exercise 4.14: ‘An ‘’informal” negation for each of the two universal conditionalstatements from Example 4.13 are:

(i) There exists an x P Z such that x is neither even nor odd.

(ii) There exists an x P RzQ such that x P Z.

Exercise 4.15: Negation of “there is a natural number that is even and prime”is: Every natural number is odd or composite.

Exercise 4.16: Negation of “all dogs go to heaven” is: There exists a dog thatdoes not go to heaven.

Exercise 4.18: A negation of the continuity definition (using English and not @-Dsymbols is the following

There exists an ε ą 0 such that for any N P N, we have for all n ě N the anare outside the ε-tunnel of a (i.e., |an ´ a| ě ε).

Exercise 5.3 (Part 1 of 2) This question had two parts. First give the negationof the Statement (1):

@x P R, D y P R such that x2` y2

‰ 8.


Exercise 5.3 (Part 2 of 2) And here is a proof for the negation of Statement (1):Proof. Let x P R.

WWTS: D y P R s.t. x2 ` y2 ‰ 8.

Either x “ ˘2 or it is not. If x “ ˘2, then choose y :“ 0. Observe thatx2 ` y2 “ 22 ` 02 “ 4 ‰ 8. But if x ‰ ˘2, then choose y :“ x. Observe thatx2 ` y2 “ 2x2 ‰ 2p˘2q2 “ 8. Thus the negation of Statement (1) is true, so weconclude that Statement (1) is false.

Q.E.D.

Exercise 6.1: Place the induction hypothesis and WWTS below.Induction Hypothesis: Assume that

?k ă 1?

1` 1?

2` ¨ ¨ ¨ ` 1?

kfor some k ě 2.

WWTS:?k ` 1 ă 1?

1` 1?

2` ¨ ¨ ¨ ` 1?

k`1.

Exercise 6.3: Prove by induction that n2 ă 2n for all n ě 5.

Base Case: (n “ 5) Observe that 52 “ 25 ă 32 “ 25, and hence the base caseholds.

Induction Hypothesis: Assume that k2 ă 2k for some k ě 5.

WWTS: pk ` 1q2 ă 2k`1.

Consider the following sequence of equalities and inequalities:

pk ` 1q2 “ k2` 2k ` 1

ă 2k ` 2k ` 1 by Induction Hypothesis

ă 2k ` 2k since 2k ` 1 ă 2k for all k ě 3 by Example 6.2

“ 2 ¨ 2k

“ 2k`1,

as desired. Hence n2 ă 2n for all n ě 5.


Q.E.D.

Exercise 6.4: For all n ě 3, prove by induction that the angle sum of a convexn-gon is 180pn´ 2q degrees. aBa to do still!!!!!

Q.E.D.


Exercise 7.4:

A “ t H , tHu , tH, tH, tHuu u ùñ |A| “ 3

A “ ta, b, cu ùñ |A| “ 3

A “ ta, a, bu ùñ |A| “ 2

A “ R ùñ |A| “ ℵ1

A “ Q ùñ |A| “ ℵ0

A “

"ˆ

a11 a12

a21 a22

˙

|aij P Z and 0 ď aij ď i` j

*

ùñ |A| “ ℵ1

A “ tHu ùñ |A| “ 1

Exercise 7.7: Let A “ tπ, e, tπ, eu, tπuu. Answer the following. [True or False?]

π P A True

tπu R A False

e Ă A True

teu P A False

teu Ă A True

tteu, te, πuu Ă A False

ttπu, te, πuu Ă A True

Exercise 7.9: The corresponding sizes of the power sets of A and B are respec-tively:

• |PpAq| “ 210 “ 1024, and

• |PpBq| “ 2n.


Exercise 7.17: Prove that the following containment of De Morgan’s law:

X ´ pAXBq Ă pX ´ Aq Y pX ´Bq

.Proof. Let x P X ´ pAXBq.

WWTS: x P pX ´ Aq Y pX ´Bq.

Since x P X ´ pAXBq, then x P X and x R AXB. It suffices to show that eitherx P X ´ A or x P X ´ B. Since x R A X B, then x R A. Hence x P X and x R A,so thus x P X ´A. We conclude that x P pX ´Aq Y pX ´Bq as desired, and theresult follows.

Q.E.D.

Exercise 7.21: Let A,B1, B2, . . . , Bn be sets. Prove that

AY

˜

nč

i“1

Bi

¸

Ą

nč

i“1

pAYBiq [You Do!]

Proof. We first show the Ă containment holds. Let x P 10Z` 7.

WWTS: x P 5Z` 2.

Since x P 10Z` 7 then x “ 10k ` 7 for some k P Z. Thus

x “ 10k ` 5` 2 “ 5p2k ` 1q ` 2.

Set M :“ 2k ` 1. Observe that M P Z and so x “ 5M ` 2 P 5Z ` 2. Thus10Z ` 7 Ă 5Z ` 2 as desired. We now show Ą containment does not hold. Sety “ 12 “ 5 ¨ 2` 2 P 5Z` 2. If y P 10Z` 7, then

12 “ 10k ` 7 for some k P Z ùñ 10k “ 12´ 7 ùñ k “5

10P Z

which gives a contradiction so 10Z` 7 Č 5Z` 2.

Q.E.D.


Exercise 8.4: The image fpta, buq, preimage f´1pt3uq, and range of f are:

• fpta, buq “ t1, 2u

• f´1pt3uq “ tc, d, eu

• range of f is Y

Exercise 8.7:


f : RÑ R by fpxq “ x2 No No

f : RÑ R´ t0u by fpxq “ x3 Yes No

f : RÑ R by fpxq “ x´1x2`3 No Yes

f : Ppta, b, cuq Ñ Z by fpXq “ |X| for each

subset X P Ppta, b, cuq. No No

Exercise 8.9:aBa to do still


Exercise 8.11: Refer to the diagram given in this exercise.

• What is`

g ˝ f˘

px1q,`

g ˝ f˘

px2q,`

g ˝ f˘

px3q, and`

g ˝ f˘

px4q?`

g ˝ f˘

px1q “ z2`

g ˝ f˘

px2q “ z3`

g ˝ f˘

px3q “ z4`

g ˝ f˘

px4q “ z1

• Fill in the table below with answers “Yes” or “No”.


g ˝ f Yes Yes

f Yes No

g No Yes

Exercise 8.13: Prove part (ii) of Theorem 8.12.

Proof. Assume g ˝ f is surjective.

WWTS: g is surjective.

Let z P Z. It suffices to show that there exists a y P Y such that gpyq “ z.Since g ˝ f is surjective, there exists an x P X such that pg ˝ fqpxq “ z. Hencegpfpxqq “ z. Set y “ fpxq. Then gpyq “ z as desired. Thus g is surjective.

Q.E.D.

proofs refresher for math 335 { fall 2020 an overview of ......9 solutions to all exercises45 this...

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