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Introduction to Abstract Mathematics (MATH 13)
University of California, Irvine
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Math 13 — An Introduction to Abstract Mathematics
- January 21, Neil Donaldson & Alessandra Pantano- 1 Introduction Contents
- 2 Logic and the Language of Proofs
- 2 Propositions
- 2 Methods of Proof
- 2 Quantifiers
- 3 Divisibility and the Euclidean Algorithm
- 3 Remainders and Congruence
- 3 Greatest Common Divisors and the Euclidean Algorithm
- 4 Sets and Functions
- 4 Set Notation and Describing a Set
- 4 Subsets
- 4 Unions, Intersections, and Complements
- 4 Introduction to Functions
- 5 Mathematical Induction and Well-ordering
- 5 Investigating Recursive Processes
- 5 Proof by Induction
- 5 Well-ordering and the Principle of Mathematical Induction
- 5 Strong Induction
- 6 Set Theory, Part II
- 6 Cartesian Products
- 6 Power Sets
- 6 Indexed Collections of Sets
- 7 Relations and Partitions
- 7 Relations
- 7 Functions revisited
- 7 Equivalence Relations
- 7 Partitions
- 7 Well-definition, Rings and Congruence
- 7 Functions and Partitions
8 Cardinalities of Infinite Sets 165 8 Cantor’s Notion of Cardinality................................ 165 8 Uncountable Sets........................................ 173
Useful Texts
Book of Proof, Richard Hammack, 2nd ed 2013. Available free online! Very good on the basics: if you’re having trouble with reading set notation or how to construct a proof, this book’s for you! These notes are deliberately pitched at a high level relative to this textbook to provide contrast.
Mathematical Reasoning, Ted Sundstrom, 2nd ed 2014. Available free online! Excellent resource. If you would like to buy the actual book, you can purchase it on Amazon at a really cheap price.
Mathematical Proofs: A Transition to Advanced Mathematics, Chartrand/Polimeni/Zhang, 3rd Ed 2013, Pearson. The most recent course text. Has many, many exercises; the first half is fairly straightforward while the second half is much more complex and dauntingly detailed.
The Elements of Advanced Mathematics, Steven G. Krantz, 2nd ed 2002, Chapman & Hall and Foundations of Higher Mathematics, Peter Fletcher and C. Wayne Patty, 3th ed 2000, Brooks–Cole are old course textbooks for Math 13. Both are readable and concise with good exercises.
Learning Outcomes
Developing the skills necessary to read and practice abstract mathematics.
Understanding the concept of proof, and becoming acquainted with several proof techniques.
Learning what sort of questions mathematicians ask, what excites them, and what they are looking for.
Introducing upper-division mathematics by giving a taste of what is covered in several areas of the subject.
Along the way you will learn new techniques and concepts. For example:
Number Theory Five people each take the same number of candies from a jar. Then a group of seven does the same. The, now empty, jar originally contained 239 candies. Can you decide how much candy each person took?
Geometry and Topology How can we visualize and compute with objects like the Mobius strip?
Fractals How to use sequences of sets to produce objects that appear the same at all scales.
To Infinity and Beyond! Why are some infinities greater than others?
The mathematics you have learned so far has consisted almost entirely of computations, with the theoretical aspects swept under the rug. At upper-division level, the majority of mathematics is presented in an abstract way. This course will train you in understanding and creating abstract mathematics, and it is our hope that you will develop an appreciation for it.
Proof
The essential concept in higher-level mathematics is that of proof. A basic dictionary entry for the word would cover two meanings:
An argument that establishes the truth of a fact.
A test or trial of an assertion. 2
In mathematics we always mean the former, while in much of science and wider culture the second meaning predominates. Indeed mathematics is one of the very few disciplines in which one can categorically say that something is true or false. In reality we can rarely be so certain. A greasy sales- man in a TV advert may claim that to have proved that a certain cream makes you look younger; a defendant may be proved guilty in court; the gravitational constant is 9− 2. Ask yourself what these statements mean. The advert is just trying to sell you something, but push harder and they might provide some justification: e. 100 people used the product for a month and 75 of them claim to look younger. This is a test, a proof in the second sense of the definition. Is a defendant really guilty of a crime just because a court has found them so; have there never been any miscarriages of justice? Is the gravitational constant precisely 9− 2 , or is this merely a good approximation? This kind of pedantry may seem over the top in everyday life: indeed most of us would agree that if 75% of people think a cream helps, then it probably is doing something beneficial. In mathematics and philosophy, we think very differently: the concepts of true and false and of proof are very precise.
So how do mathematicians reach this blissful state where everything is either right or wrong and, once proved, is forever and unalterably certain? The answer, rather disappointingly, is by cheating. Nothing in mathematics is true except with reference to some assumption. For example, consider the following theorem:
Theorem 1. The sum of any two even integers is even.
We all believe that this is true, but can we prove it? In the sense of the second definition of proof, it might seem like all we need to do is to test the assertion: for example 4 + 6 = 10 is even. In the first sense, the mathematical sense, of proof, this is nowhere near enough. What we need is a definition of even. 3
Definition 1. An integer is even if it may be written in the form 2n where n is an integer.
The proof of the theorem now flows straight from the definition.
2 It is this notion that makes sense of the seemingly oxymoronic phrase The exception proves the rule. It is the exception that tests the validity of the rule. 3 And more fundamentally of sum and integer.
Proof. Let x and y be any two even integers. We want to show that x + y is an even integer. By definition, an integer is even if it can be written in the form 2k for some integer k. Thus there exist integers n, m such that x = 2 m and y = 2 n. We compute:
x + y = 2 m + 2 n = 2 (m + n). (∗)
Because m + n is an integer, this shows that x + y is an even integer.
There are several important observations:
‘Any’ in the statement of the theorem means the proof must work regardless of what even in- tegers you choose. It is not good enough to simply select, for example, 4 and 16, then write 4 + 16 = 20. This is an example, or test, of the theorem, not a mathematical proof.
According to the definition, 2m and 2n together represent all possible pairs of even numbers.
The proof makes direct reference to the definition. The vast majority of the proofs in this course are of this type. If you know the definition of every word in the statement of a theorem, you will often discover a proof simply by writing down the definitions.
The theorem itself did not mention any variables. The proof required a calculation for which these were essential. In this case the variables m and n come for free once you write the definition of evenness! A great mistake is to think that the proof is nothing more than the calculation (∗). This is the easy bit, and it means nothing without the surrounding sentences.
The important link between theorems and definitions is much of what learning higher-level math- ematics is about. We prove theorems (and solve homework problems) because they make us use and understand the subtleties of definitions. One does not know mathematics, one does it. Mathematics is a practice; an art as much as it is a science.
Conjectures
In this course, you will discover that one of the most creative and fun aspects of mathematics is the art of formulating, proving and disproving conjectures. To get a taste, consider the following:
Conjecture 1. If n is any odd integer, then n 2 − 1 is a multiple of 8.
Conjecture 1. For every positive integer n, the integer n 2 + n + 41 is prime.
A conjecture is the mathematician’s equivalent of the experimental scientist’s hypothesis: a state- ment that one would like to be true. The difference lies in what comes next. The mathematician will try to prove that a conjecture is undeniably true by relying on logic, while the scientist will ap- ply the scientific method, conducting experiments attempting, and hopefully failing, to show that a hypothesis is incorrect.
Theorem 1. If n is any odd integer, then n 2 − 1 is a multiple of 8.
Proof. Let n be any odd integer; we want to show that n 2 − 1 is a multiple of 8. By the definition of odd integer, we may write n = 2 k + 1 for some integer k. Then
n 2 − 1 = ( 2 k + 1 ) 2 − 1 = ( 4 k 2 + 1 + 4 k) − 1 = 4 k 2 + 4 k = 4 k(k + 1 ).
We distinguish two cases. If k is even, then k(k + 1 ) is even and so 4k(k + 1 ) is divisible by 8. If k is odd, then k + 1 is even. Therefore k(k + 1 ) is again even and 4k(k + 1 ) divisible by 8. In both cases n 2 − 1 = 4 k(k + 1 ) is divisible by 8. This concludes the proof.
It is now time to explore Conjecture 1. The question here is whether or not n 2 + n + 41 is a prime integer for every positive integer n. We know that when n = 1, 2, 3, 4, 5, 6 or 7 the answer is yes, but examples do not make a proof. At this point, we do not know whether the conjecture is true or false. Let us investigate the question further. Suppose that n is any positive integer; we must ask whether it is possible to factor n 2 + n + 41 as a product of two positive integers, neither of which is one. 4 When n = 41 such a factorization certainly exists, since we can write
412 + 41 + 41 = 41 ( 41 + 1 + 1 ) = 41 · 43.
Our counterexample shows that there exists at least one value of n for which n 2 + n + 41 is not prime. We have therefore disproved the conjecture that ‘for all positive integers n, n 2 + n + 41 is prime,’ and so Conjecture 1 is false!
The moral of the story is this: to show that a conjecture is true you must prove that it holds for all the cases in consideration, but to show that it is false a single counterexample suffices.
Conjectures: True or False?
Do your best to prove or disprove the following conjectures. Then revisit these problems at the end of the course to realize how much your proof skills have improved.
The sum of any three consecutive integers is even.
There exist integers m and n such that 7m + 5 n = 4.
Every common multiple of 6 and 10 is divisible by 60.
There exist integers x and y such that 6x + 9 y = 10.
For every positive real number x, x + 1 x is greater than or equal to 2.
If x is any real number, then x 2 ≥ x. 4 Once again we rely on a definition: a positive integer is prime if it cannot be written as a product of two integers, both greater than one.
If n is any integer, n 2 + 5 n must be even.
If x is any real number, then |x| ≥ −x.
Consider the set R of all real numbers. For all x in R, there exists y in R such that x < y.
Consider the set R of all real numbers. There exists x in R such that, for all y in R, x < y.
The sets A = {n ∈ N : n 2 < 25 } and B = {n 2 : n ∈ N and n < 5 } are equal. Here N denotes the set of natural numbers.
Now we know a little of what mathematics is about, it is time to practice some of it!
Connecting Propositions: Conjunction, Disjunction and Negation
We now define how to combine propositions in natural ways, modeled on the words and, or and not.
Definition 2. Let P and Q be propositions. The conjunction (AND, ∧) of P and Q, the disjunction (OR, ∨) of P and Q, and the negation or denial (NOT, ¬, ∼, ) of P are defined by the truth tables,
P Q P ∧ Q T T T T F F F T F F F F
P Q P ∨ Q
T T T
T F T
F T T
F F F
P ¬P
T F
F T
It is usually better to use and, or and not rather than conjunction, disjunction and negation: the latter may make you sound educated, but at the risk of being misunderstood!
Example. Let P, Q and R be the following propositions:
P. Irvine is a city in California.
Q. Irvine is a town in Ayrshire, Scotland.
R. Irvine has seven letters.
Clearly P is true while R is false. If you happen to know someone from Scotland, you might know that Q is true We can now compute the following (increasingly grotesque) combinations...
P ∧ Q P ∨ Q P ∧ R ¬R (¬R) ∧ P ¬(R ∨ P) (¬P) ∨ [((¬R) ∨ P) ∧ Q] T T F T T F T aThe second syllable is pronounced like the i in bin or win. Indeed the first Californian antecedent of the Irvine family which gave its name to UCI was an Ulster-Scotsman named James Irvine (1827–1886). Probably the family name was originaly pronounced in the Scottish manner.
How did we establish these facts? Some are quick, and can be done in your head. Consider, for instance, the statement (¬R) ∧ P. Because R is false, ¬R is true. Thus (¬R) ∧ P is the conjunction of two true statements, hence it is true. Similarly, we can argue that R ∨ P is true (because R is false and P is true), so the negation ¬(R ∨ P) is false. Establishing the truth value of the final proposition (¬P) ∨ [((¬R) ∨ P) ∧ Q] requires more work. You may want to set up a truth table with several auxiliary columns to help you compute:
P Q R ¬P ¬R (¬R) ∨ P ((¬R) ∨ P) ∧ Q (¬P) ∨ [((¬R) ∨ P) ∧ Q] T T F F T T T T
The importance of parentheses in a logical expressions cannot be stressed enough. For example, try building the truth table for the propositions P ∨ (Q ∧ R) and (P ∨ Q) ∧ R. Are they the same?
Conditional and Biconditional Connectives
In order to logically set up proofs, we need to see how propositions can lead one to another.
Definition 2. The conditional ( =⇒ ) and biconditional ( ⇐⇒ ) connectives have the truth tables
P Q P =⇒ Q T T T T F F F T T F F T
P Q P ⇐⇒ Q
T T T
T F F
F T F
F F T
For the proposition P =⇒ Q, we call P the hypothesis and Q the conclusion.
Observe that the expressions P =⇒ Q and P ⇐⇒ Q are themselves propositions: they are sentences which are either true or false!
Synonyms
=⇒ and ⇐⇒ can be read in many different ways: P =⇒ Q P ⇐⇒ Q P implies Q P if and only if Q Q if P P iff Q P only if Q P and Q are (logically) equivalent P is sufficient for Q P is necessary and sufficient for Q Q is necessary for P
Example. The following propositions all mean exactly the same thing:
If you are born in Rome, then you are Italian.
You are Italian if you are born in Rome.
You are born in Rome only if you are Italian.
Being born in Rome is sufficient to be Italian.
Being Italian is necessary for being born in Rome.
Are you comfortable with what P and Q are here?
The biconditional connective should be easy to remember: P ⇐⇒ Q is true precisely when P and Q have identical truth states. It is harder to make sense of the conditional connective. One way of thinking about it is to consider what it means for an implication to be false. If P =⇒ Q is false, it is impossible to create a logical argument which assumes P and concludes Q. The second row of P =⇒ Q encapsulates the fact that it should be impossible for truth ever to logically imply falsehood.
Proof. Let x and y be any two odd integers. We want to show that product x · y is an odd integer. By definition, an integer is odd if it can be written in the form 2k + 1 for some integer k. Thus there must be integers n, m such that x = 2 n + 1 and y = 2 m + 1. We compute:
x · y = ( 2 n + 1 )( 2 m + 1 ) = 4 mn + 2 n + 2 m + 1 = 2 ( 2 mn + n + m) + 1.
Because 2mn + n + m is an integer, this shows that x · y is an odd integer.
It is common to place a symbol (in this case ) at the end of a proof to tell the reader that your argu- ment is complete. Traditionally the letters Q.E. (from the Latin quod erat demonstrandum, literally ‘which is what had to be demonstrated’) were used, but this has gone out of style may also feel that you want to write more, or less than the above. This is a difficult thing to judge. What do you feel is a convincing argument? Test your argument on your classmates. The appropriate level of de- tail will depend on your readership: a middle school student will need more detail than a graduate student! At the moment, the best guide is to write for someone with the same mathematical sophis- tication as yourself. If, in three weeks’ time, you can return to what you’ve written and understand it, then it’s probably good!
The Converse and Contrapositive
The following constructions are used continually in mathematics: it is vitally important to know the difference between them.
Definition 2. The converse of an implication P =⇒ Q is the reversed implication Q =⇒ P. The contrapositive of P =⇒ Q is ¬Q =⇒ ¬P.
In general, we cannot say anything about the truth value of the converse of a true statement. The contrapositive of a true statement is, however, always true.
Theorem 2. The contrapositive of an implication is logically equivalent the original implication.
Proof. Simply use our definitions of negation and implication to compute the truth table:
P Q P =⇒ Q ¬Q ¬P ¬Q =⇒ ¬P T T T F F T T F F T F F F T T F T T F F T T T T
Since the truth states in the third and sixth columns are identical, we see that P =⇒ Q and its contrapositive ¬Q =⇒ ¬P are logically equivalent.
Example. Let P and Q be the following statements:
P. Claudia is holding a peach. Q. Claudia is holding a piece of fruit.
The implication P =⇒ Q is true, since all peaches are fruit. As a sentence, we have:
If Claudia is holding a peach, then Claudia is holding a piece of fruit.
The converse of P =⇒ Q is the sentence:
If Claudia is holding a piece of fruit, then Claudia is holding a peach.
This is palpably false: Claudia could be holding an apple!
The contrapositive of P =⇒ Q is the following sentence:
If Claudia is not holding any fruit, then she is not holding a peach.
This is clearly true.
Proof by Contrapositive
The fact that P =⇒ Q and ¬Q =⇒ ¬P are logically equivalent allows us, when convenient, to prove P =⇒ Q by instead proving its contrapositive. As an example, consider another basic theorem.
Theorem 2. Let x and y be integers. If x + y is odd, then exactly one of x or y is odd.
The theorem is an implication of the form P =⇒ Q where
P. The sum x + y of integers x and y is odd. Q. Exactly one of x or y is odd.
A direct proof would require that we assume P to be true and logically deduce the truth of Q. For instance we might start our argument with:
Suppose that x + y = 2 n + 1 for some integer n
The problem is that this doesn’t really tell us anything about x and y, which we need to think about in order to demonstrate the truth of Q. Instead we consider the negations of our propositions:
¬Q. x and y are both even or both odd (they have the same parity). ¬P. The sum x + y of integers x and y is even.
Since P =⇒ Q is logically equivalent to the seemingly simpler contrapositive (¬Q) =⇒ (¬P), we choose to prove the latter. This is, by Theorem 2, equivalent to proving the original implication.
Aside. Think about the meaning! In the previous example we saw how negation switches and to or. This is true only when and denotes a conjunction between two propositions. Before applying De Morgan’s laws, think about the meaning of the sentence. For example, the negation of
Mark and Mary have the same height.
is the proposition:
Mark and Mary do not have the same height.
If you blindly appeal to De Morgan’s laws you might end up with the following piece of nonsense:
Mark or Mary do not have the same height.
Logical rules are wonderfully concise, but very easy to misuse. Always think about the meaning of a sentence and you shouldn’t go wrong.
Negating Conditionals
You will often want to understand the negation of a statement. In particular, it is important to under- stand the negation of a conditional P =⇒ Q. Is it enough to say ‘P doesn’t imply Q’? And what could this mean? To answer the question you can use truth tables, or just think. Here is the truth table for P =⇒ Q and its negation: recall that negation simply swaps T and F.
P Q P =⇒ Q ¬(P =⇒ Q) T T T F T F F T F T T F F F T F
The only time there is a T in the final column is when both P is true and Q is false. We have therefore proved the following:
Theorem 2. ¬(P =⇒ Q) is logically equivalent to P ∧ ¬Q (read ‘P and not Q’).
Now think in words rather than calculate. What is the negation of the following implication?
It’s the morning therefore I’ll have coffee.
Hopefully it is clear that the negation is:
It’s the morning and I won’t have coffee.
The implication ‘therefore’ has disappeared and the expression ‘and won’t’ is in its place.
Warning! The negation of P =⇒ Q is not a conditional. In particular it is neither of the following:
The converse, Q =⇒ P. The contrapositive of the converse, ¬P =⇒ ¬Q.
If you are unsure about this, write down the truth tables and compare.
Example. Let x be an integer. What is the negation of the following sentence?
If x is even, then x 2 is even.
Written in terms of propositions, we wish to negate P =⇒ Q , where P and Q are:
P. x is even. Q. x 2 is even.
The negation of P =⇒ Q is P ∧ ¬Q, namely:
x is even and x 2 is odd.
This is very different to ¬P =⇒ ¬Q (if x is odd then x 2 is odd).
Keep yourself straight by thinking about the meaning of these sentences. It should be obvious that ‘x even =⇒ x 2 even’ is true. It negation should therefore be false. The fact that it is false should make reading the negation feel a little uncomfortable.
Tautologies and Contradictions
We finish this section with two related concepts that are helpful for understanding proofs.
Definition 2. A tautology is a logical expression that is always true, regardless of what the compo- nent statements might be. A contradiction is a logical expression that is always false.
The easiest way to detect these is simply to construct a truth table.
Examples. 1. P ∧ (¬P) is a very simple contradiction:
P ¬P P ∧ (¬P) T F F F T F
Whatever the proposition P is, it cannot be true at the same time as its negation.
- (P ∧ (P =⇒ Q)) =⇒ Q is a tautology. This is essentially how we understand a direct proof: if P is true and we have a correct argument P =⇒ Q, then Q must also be true.
P Q P =⇒ Q P ∧ (P =⇒ Q) (P ∧ (P =⇒ Q)) =⇒ Q T T T T T T F F F T F T T F T F F T F T
2.1 Suppose that “Girls smell of roses” and “Boys have dirty hands” are true statements and that “The Teacher is always right” is a false statment. Which of the following are true? Hint: Label each of the given statements, and think about each of the following using connectives. (a) If girls smell of roses, then the Teacher is always right. (b) If the Teacher is always right, then boys have dirty hands. (c) If the Teacher is always right or girls smell of roses, then boys have dirty hands. (d) If boys have dirty hands and girls smell of roses, then the Teacher is always right.
2.1 Write the negation (in words) of the following claim: If Jack and Jill climb up the hill, then they fall down and like pails of water. 2.1 Orange County has two competing transport plans under consideration: widening the 405 freeway and constructing light rail down its median. A local politician is asked, “Would you like to see the 405 widened or would you like to see light rail constructed?” The politician wants to sound positive, but to avoid being tied to one project. What is their response? (Hint: Think about how the word ‘OR’ is used in logic... )
2.1 (a) Rewrite the following sentence using the word ‘necessary.’ If I am to get a new bicycle, I must do my homework. (b) Rewrite the following sentence using the word ‘sufficient.’ The United States must play more soccer if it is to win the World Cup. 2.1 (a) What are the converse and the contrapositive of the statements in the previous question? Write your answers in sentences, like the originals. (b) What are the negations of the statements in the previous question? 2.1 Construct the truth tables for the propositions P ∨ (Q ∧ R) and (P ∨ Q) ∧ R. Are they the same?
2.1 Apply de Morgan’s laws to the result of Theorem 2 to prove that P =⇒ Q is logically equivalent to ¬P ∨ Q. 2.1 Prove that the expressions (P =⇒ Q) ∧ (Q =⇒ P) and P ⇐⇒ Q are logically equivalent (have the same truth table). Why does this make sense?
2.1 (a) Prove that ((P ∨ Q) ∧ ¬P) ∧ ¬Q is a contradiction.
(b) Prove that (¬P ∧ Q) ∨ (P ∧ ¬Q) ⇐⇒ ¬(P ⇐⇒ Q) is a tautology.
2.1 Prove or disprove: (P ∧ ¬Q =⇒ F) ⇐⇒ (P =⇒ Q) is a tautology. Here F represents a contradiction: some proposition which is always false.
2.1 Suppose that “If Colin was early, then no-one was playing pool” is a true statement. (a) What is its contrapositive of this statement? Is it true? (b) What is the converse? Is it true? (c) What can we conclude (if anything?) if we discover each of the following? Treat the two scenarios separately. (i) Someone was playing pool. (ii) Colin was late.
2.1 Suppose that “Ford is tired and Zaphod has two heads” is a false statement. What can we conclude if we discover each of the following? Treat the two scenarios separately. (a) Ford is tired. (b) Ford is tired if and only if Zaphod has two heads.
2.1 Suppose that the following statements are true:
- Every Pig likes mud.
- If a creature cannot fly then it is not an astronaut.
- A creature is an astronaut if it likes mud.
Is it true that ‘Pigs can fly’? Explain your answer. (Hint: try rewriting each of the statements in the form ‘x is/likes =⇒ x is/likes .’)
2.1 (a) Use a truth table to prove the distributive law
(P ∧ Q) ∨ R ⇐⇒ (P ∨ R) ∧ (Q ∨ R)
(b) Use logical algebra (see the aside on page 18) to prove that
((P =⇒ R) ∧ (Q =⇒ R)) ⇐⇒ ((P ∨ Q) =⇒ R)
(Hint: start by using the result of question 8)
2.1 (a) Do there exists propositions P, Q such that both P =⇒ Q and its converse are true?
(b) Do there exist propositions P, Q such that both P =⇒ Q and its converse are false? Justify your answers by giving an example or a proof that no such examples exist.
2.1 Let R be the proposition “The summit of Mount Everest is underwater”. Suppose that S is a proposition such that (R ∨ S) ⇐⇒ (R ∧ S) is false.
(a) What can you say about S? (b) What if, instead, (R ∨ S) ⇐⇒ (R ∧ S) is true?
Hopefully it is obvious to you that R is false...
2.1 (Hard) Suppose that P, Q are propositions. Argue that any of the 16 possible truth tables
P Q? T T T/F T F T/F F T T/F F F T/F
represents an expression? created using only P and Q and the operations ∧, ∨, ¬. Can you extend your argument to show that any truth table with any number of inputs represents some logical expression?
Math13Textbook - Textbook
Course: Introduction to Abstract Mathematics (MATH 13)
University: University of California, Irvine
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