1 course summary cs 202 aaron bloomfield. 2 outline a review of the proof methods gone over in class...

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Course SummaryCourse Summary

CS 202CS 202

Aaron BloomfieldAaron Bloomfield

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OutlineOutline

A review of the proof methods gone over A review of the proof methods gone over in classin class

Interspersed with the most popular Interspersed with the most popular demotivators from CS 101demotivators from CS 101

Course summaryCourse summary

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Demotivator winners!Demotivator winners!

MethodologyMethodology– 11stst place vote counted for 3 points place vote counted for 3 points– 22ndnd place vote counted for 2 points place vote counted for 2 points– 33rdrd place vote counted for 1 point place vote counted for 1 point

Have not tallied the Toolkit ones yetHave not tallied the Toolkit ones yet– I won’t get those for a few weeksI won’t get those for a few weeks

Will then buy two demotivators and Will then buy two demotivators and hang them in my office…hang them in my office…

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Proof methods learned so farProof methods learned so far

Logical equivalencesLogical equivalences via truth tablesvia truth tables via logical equivalencesvia logical equivalences

Set equivalencesSet equivalences via membership tablesvia membership tables via set identitiesvia set identities via mutual subset proofvia mutual subset proof via set builder notation and via set builder notation and

logical equivalenceslogical equivalencesRules of inferenceRules of inference

for propositionsfor propositions for quantified statementsfor quantified statements

Pigeonhole principlePigeonhole principleCombinatorial proofsCombinatorial proofs

Ten proof methods in section 1.5:Ten proof methods in section 1.5: Direct proofsDirect proofs Indirect proofsIndirect proofs Vacuous proofsVacuous proofs Trivial proofsTrivial proofs Proof by contradictionProof by contradiction Proof by casesProof by cases Proofs of equivalenceProofs of equivalence Existence proofsExistence proofs

ConstructiveConstructiveNon-constructiveNon-constructive

Uniqueness proofsUniqueness proofs CounterexamplesCounterexamples

InductionInduction Weak mathematical inductionWeak mathematical induction Strong mathematical inductionStrong mathematical induction Structural inductionStructural induction

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Idempotent Law

Associativity of Or

rqprqrp )()()( Definition of implication

Using Logical EquivalencesUsing Logical Equivalences

rqprqrp )()()(

rqprqp rqprrqp

rqprqrp )()()(

rqprqrp

Re-arranging

Original statement

DeMorgan’s Law

qpqp

qpqp )(

rqrprqrp )()(

rrr

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In 12In 12thth place (11 votes) place (11 votes)

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Definition of differenceDefinition of difference

Definition of differenceDefinition of difference

DeMorgan’s lawDeMorgan’s law

Complementation lawComplementation law

Distributive lawDistributive law

Complement lawComplement law

Identity lawIdentity law

Commutative lawCommutative law

Set equivalencesSet equivalences (§1.7) (§1.7)Prove that A∩B=B-(B-A)Prove that A∩B=B-(B-A)

)AB-(BBA

)A(BB

)AB(B

A)B(B

A)(B)B(B

A)(BA)(BBA

88

Proof by set builder notation Proof by set builder notation and logical equivalences 2and logical equivalences 2

Original statement

Definition of difference

Negating “element of”

Definition of difference

DeMorgan’s Law

Distributive Law

Negating “element of”

Negation Law

Identity Law

Definition of intersection

)}(|{ AxBxBxx

)( ABB )}(|{ ABxBxx

))}((|{ ABxBxx

}|{ AxBxx

)}(|{ AxBxBxx

}|{ AxBxBxBxx })(|{ AxBxBxBxx

}|{ AxBxFx

BA

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In 11In 11thth place (1/3) (12 votes) place (1/3) (12 votes)

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Rules of inference (Rules of inference (§1.5)§1.5)

Example 6 of Rosen, section 1.5Example 6 of Rosen, section 1.5We have the hypotheses:We have the hypotheses: ““It is not sunny this afternoon and it It is not sunny this afternoon and it

is colder than yesterday”is colder than yesterday” ““We will go swimming only if it is We will go swimming only if it is

sunny”sunny” ““If we do not go swimming, then we If we do not go swimming, then we

will take a canoe trip”will take a canoe trip” ““If we take a canoe trip, then we will If we take a canoe trip, then we will

be home by sunset”be home by sunset”

Does this imply that “we will be Does this imply that “we will be home by sunset”?home by sunset”?

““It isIt is not not sunny this afternoonsunny this afternoon and it and it

is colder than yesterday”is colder than yesterday” ““We will go swimming only if We will go swimming only if it is it is

sunnysunny”” ““If we do not go swimming, then we If we do not go swimming, then we




““It isIt is not not sunny this afternoonsunny this afternoon and and it it

is colder than yesterdayis colder than yesterday”” ““We will go swimming only if We will go swimming only if it is it is

sunnysunny”” ““If we do not go swimming, then we If we do not go swimming, then we





is colder than yesterdayis colder than yesterday”” ““We will go swimmingWe will go swimming only if only if it is it is

sunnysunny”” ““If If we dowe do not not go swimminggo swimming, then we , then we






sunnysunny”” ““If If we dowe do not not go swimminggo swimming, then , then we we

will take a canoe tripwill take a canoe trip”” ““If If we take a canoe tripwe take a canoe trip, then we will , then we will





sunnysunny”” ““If If we dowe do not not go swimminggo swimming, then , then we we

will take a canoe tripwill take a canoe trip”” ““If If we take a canoe tripwe take a canoe trip, then , then we will we will

be home by sunsetbe home by sunset””

Does this imply that “Does this imply that “we will be we will be home by sunsethome by sunset”?”?

p

q

r

s

t

¬p q

r → p

¬r → s

s → t

t

1111


1.1. ¬p ¬p q q 11stst hypothesis hypothesis

2.2. ¬p¬p Simplification using step 1Simplification using step 1

3.3. r → pr → p 22ndnd hypothesis hypothesis

4.4. ¬r¬r Modus tollens using steps 2 & 3Modus tollens using steps 2 & 3

5.5. ¬r → s¬r → s 33rdrd hypothesis hypothesis

6.6. ss Modus ponens using steps 4 & 5Modus ponens using steps 4 & 5

7.7. s → t s → t 44thth hypothesis hypothesis

8.8. tt Modus ponens using steps 6 & 7Modus ponens using steps 6 & 7

p

qp

q

qp

p

p

qp

q

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Rosen, section 1.5, question 10aRosen, section 1.5, question 10a

Given the hypotheses:Given the hypotheses: ““Linda, a student in this class, owns a Linda, a student in this class, owns a

red convertible.”red convertible.” ““Everybody who owns a red convertible Everybody who owns a red convertible

has gotten at least one speeding ticket”has gotten at least one speeding ticket”

Can you conclude: “Somebody in Can you conclude: “Somebody in this class has gotten a speeding this class has gotten a speeding ticket”?ticket”?

C(Linda)R(Linda)

x (R(x)→T(x))

x (C(x)T(x))

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1.1. x (R(x)→T(x))x (R(x)→T(x)) 33rdrd hypothesis hypothesis

2.2. R(Linda) → T(Linda)R(Linda) → T(Linda) Universal instantiation using step Universal instantiation using step

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3.3. R(Linda)R(Linda) 22ndnd hypothesis hypothesis

4.4. T(Linda)T(Linda) Modes ponens using steps 2 & 3Modes ponens using steps 2 & 3

5.5. C(Linda)C(Linda) 11stst hypothesis hypothesis

6.6. C(Linda) C(Linda) T(Linda) T(Linda) Conjunction using steps 4 & 5Conjunction using steps 4 & 5

7.7. x (C(x)x (C(x)T(x))T(x)) Existential generalization using Existential generalization using

step 6step 6Thus, we have shown that “Somebody in this class has gotten a speeding ticket”

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Proof methods (Proof methods (§1.5)§1.5)

Ten proof methods:Ten proof methods:1.1. Direct proofsDirect proofs2.2. Indirect proofsIndirect proofs3.3. Vacuous proofsVacuous proofs4.4. Trivial proofsTrivial proofs5.5. Proof by contradictionProof by contradiction6.6. Proof by casesProof by cases7.7. Proofs of equivalenceProofs of equivalence8.8. Existence proofsExistence proofs9.9. Uniqueness proofsUniqueness proofs10.10. CounterexamplesCounterexamples

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Direct proof exampleDirect proof example

Rosen, section 1.5, question 20Rosen, section 1.5, question 20 Show that the square of an even number is an Show that the square of an even number is an

even numbereven number Rephrased: if Rephrased: if nn is even, then is even, then nn22 is even is even

Assume Assume nn is even is even Thus, Thus, nn = 2 = 2kk, for some , for some kk (definition of even (definition of even

numbers)numbers) nn22 = (2 = (2kk))22 = 4 = 4kk22 = 2(2 = 2(2kk22)) As As nn22 is 2 times an integer, is 2 times an integer, nn22 is thus even is thus even

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Indirect proof exampleIndirect proof example

If If nn22 is an odd integer then is an odd integer then nn is an odd integer is an odd integer

Prove the contrapositive: If Prove the contrapositive: If nn is an even integer, is an even integer, then then nn22 is an even integer is an even integer

Proof: Proof: nn=2=2kk for some integer for some integer kk (definition of even (definition of even numbers)numbers)

nn22 = (2 = (2kk))22 = 4 = 4kk22 = 2(2 = 2(2kk22))

Since Since nn22 is 2 times an integer, it is even is 2 times an integer, it is even

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Proof by contradictionProof by contradiction

Given a statement p, assume it is falseGiven a statement p, assume it is false Assume ¬pAssume ¬p

Prove that ¬p cannot occurProve that ¬p cannot occur A contradiction existsA contradiction exists

Given a statement of the form p→qGiven a statement of the form p→q To assume it’s false, you only have to consider the To assume it’s false, you only have to consider the

case where p is true and q is falsecase where p is true and q is false

2222

Proof by contradictionProof by contradiction

Theorem (by Euclid): There are infinitely many Theorem (by Euclid): There are infinitely many prime numbers. prime numbers.

Proof. Assume there are a finite number of primesProof. Assume there are a finite number of primes

List them as follows: List them as follows: pp11, , pp22 …, …, ppnn..

Consider the number Consider the number qq = = pp11pp22 … … ppnn + 1 + 1 This number is not divisible by any of the listed primesThis number is not divisible by any of the listed primes

If we divided If we divided ppii into into qq, there would result a remainder of 1, there would result a remainder of 1 We must conclude that We must conclude that qq is a prime number, not among is a prime number, not among

the primes listed abovethe primes listed aboveThis contradicts our assumption that all primes are in the list This contradicts our assumption that all primes are in the list pp11, , pp22 …, …, ppnn..

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Proof by contradiction example 2Proof by contradiction example 2

Rosen, section 1.5, question 21 (b)Rosen, section 1.5, question 21 (b) Prove that if n is an integer and nProve that if n is an integer and n33+5 is odd, then n is even+5 is odd, then n is even Rephrased: If nRephrased: If n33+5 is odd, then n is even+5 is odd, then n is even

Thus, p is “nThus, p is “n33+5” is odd, q is “n is even”+5” is odd, q is “n is even”

Assume p and Assume p and qq Assume that nAssume that n33+5 is odd, and n is odd+5 is odd, and n is odd

Since n is odd:Since n is odd: n=2k+1 for some integer k (definition of odd numbers)n=2k+1 for some integer k (definition of odd numbers) nn33+5 = (2k+1)+5 = (2k+1)33+5 = 8k+5 = 8k33+12k+12k22+6k+6 = 2(4k+6k+6 = 2(4k33+6k+6k22+3k+3)+3k+3) As nAs n33+5 = 2(4k+5 = 2(4k33+6k+6k22+3k+3) is 2 times an integer, n must be even+3k+3) is 2 times an integer, n must be even Thus, we have concluded qThus, we have concluded q

Contradiction!Contradiction! We assumed q was false, and showed that this assumption implies that q must We assumed q was false, and showed that this assumption implies that q must

be truebe true As q cannot be both true and false, we have reached our contradictionAs q cannot be both true and false, we have reached our contradiction

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Vacuous proof exampleVacuous proof example

When the antecedent is falseWhen the antecedent is false

Consider the statement:Consider the statement: All criminology majors in CS 202 are femaleAll criminology majors in CS 202 are female Rephrased: If you are a criminology major and you Rephrased: If you are a criminology major and you

are in CS 202, then you are femaleare in CS 202, then you are femaleCould also use quantifiers!Could also use quantifiers!

Since there are no criminology majors in this Since there are no criminology majors in this class, the antecedent is false, and the class, the antecedent is false, and the implication is trueimplication is true

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Trivial proof exampleTrivial proof example

When the consequence is trueWhen the consequence is true

Consider the statement:Consider the statement: If you are tall and are in CS 202 then you are If you are tall and are in CS 202 then you are

a studenta student

Since all people in CS 202 are students, Since all people in CS 202 are students, the implication is true regardlessthe implication is true regardless

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Proof by cases exampleProof by cases example

Prove that Prove that Note that b ≠ 0Note that b ≠ 0

Cases:Cases: Case 1: a ≥ 0 and b > 0Case 1: a ≥ 0 and b > 0

Then |a| = a, |b| = b, andThen |a| = a, |b| = b, and Case 2: a ≥ 0 and b < 0Case 2: a ≥ 0 and b < 0

Then |a| = a, |b| = -b, andThen |a| = a, |b| = -b, and Case 3: a < 0 and b > 0Case 3: a < 0 and b > 0

Then |a| = -a, |b| = b, andThen |a| = -a, |b| = b, and Case 4: a < 0 and b < 0Case 4: a < 0 and b < 0

Then |a| = -a, |b| = -b, andThen |a| = -a, |b| = -b, and

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

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Proofs of equivalence exampleProofs of equivalence example

Rosen, section 1.5, question 40Rosen, section 1.5, question 40 Show that mShow that m22=n=n22 if and only if m=n or m=-n if and only if m=n or m=-n Rephrased: (mRephrased: (m22=n=n22) ↔ [(m=n)) ↔ [(m=n)(m=-n)](m=-n)]

Need to prove two parts:Need to prove two parts: [(m=n)[(m=n)(m=-n)] → (m(m=-n)] → (m22=n=n22))

Proof by cases!Proof by cases!Case 1: (m=n) → (mCase 1: (m=n) → (m22=n=n22))

(m)(m)22 = m = m22, and (n), and (n)22 = n = n22, so this case is proven, so this case is proven

Case 2: (m=-n) → (mCase 2: (m=-n) → (m22=n=n22)) (m)(m)22 = m = m22, and (-n), and (-n)22 = n = n22, so this case is proven, so this case is proven

(m(m22=n=n22) → [(m=n)) → [(m=n)(m=-n)](m=-n)]Subtract nSubtract n22 from both sides to get m from both sides to get m22-n-n22=0=0Factor to get (m+n)(m-n) = 0Factor to get (m+n)(m-n) = 0Since that equals zero, one of the factors must be zeroSince that equals zero, one of the factors must be zeroThus, either m+n=0 (which means m=n)Thus, either m+n=0 (which means m=n)Or m-n=0 (which means m=-n)Or m-n=0 (which means m=-n)

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Constructive existence proof Constructive existence proof exampleexample

Show that a square exists that is the sum Show that a square exists that is the sum of two other squaresof two other squares Proof: 3Proof: 322 + 4 + 422 = 5 = 522

Show that a cube exists that is the sum of Show that a cube exists that is the sum of three other cubesthree other cubes Proof: 3Proof: 333 + 4 + 433 + 5 + 533 = 6 = 633

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Non-constructive existence proof Non-constructive existence proof exampleexample

Rosen, section 1.5, question 50Rosen, section 1.5, question 50Prove that either 2*10Prove that either 2*10500500+15 or 2*10+15 or 2*10500500+16 is not a +16 is not a perfect squareperfect square

A perfect square is a square of an integerA perfect square is a square of an integer Rephrased: Show that a non-perfect square exists in the set Rephrased: Show that a non-perfect square exists in the set

{2*10{2*10500500+15, 2*10+15, 2*10500500+16}+16}

Proof: The only two perfect squares that differ by 1 are 0 Proof: The only two perfect squares that differ by 1 are 0 and 1and 1

Thus, any other numbers that differ by 1 cannot both be perfect Thus, any other numbers that differ by 1 cannot both be perfect squaressquares

Thus, a non-perfect square must exist in any set that contains Thus, a non-perfect square must exist in any set that contains two numbers that differ by 1two numbers that differ by 1

Note that we didn’t specify which one it was!Note that we didn’t specify which one it was!

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Uniqueness proof exampleUniqueness proof example

If the real number equation 5x+3=a has a If the real number equation 5x+3=a has a solution then it is uniquesolution then it is unique

ExistenceExistence We can manipulate 5x+3=a to equal (a-3)/5We can manipulate 5x+3=a to equal (a-3)/5 Is this constructive or nonconstructive?Is this constructive or nonconstructive?

UniquenessUniqueness If there are two such numbers, then they would fulfill If there are two such numbers, then they would fulfill

the following: a = 5x+3 = 5y+3the following: a = 5x+3 = 5y+3 We can manipulate this to yield that x = yWe can manipulate this to yield that x = y

Thus, the one solution is unique!Thus, the one solution is unique!

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CounterexamplesCounterexamples

Given a universally quantified statement, find a single Given a universally quantified statement, find a single example which it is not trueexample which it is not true

Note that this is DISPROVING a UNIVERSAL statement Note that this is DISPROVING a UNIVERSAL statement by a counterexampleby a counterexample

x ¬R(x), where R(x) means “x has red hair”x ¬R(x), where R(x) means “x has red hair” Find one person (in the domain) who has red hairFind one person (in the domain) who has red hair

Every positive integer is the square of another integerEvery positive integer is the square of another integer The square root of 5 is 2.236, which is not an integerThe square root of 5 is 2.236, which is not an integer

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Weak mathematical induction (§ 3.3)Weak mathematical induction (§ 3.3)

Rosen, question 7: ShowRosen, question 7: Show

Base case: Base case: nn = 1 = 1

Inductive hypothesis: assume Inductive hypothesis: assume

6

)12)(1(

1

2

nnni

n

i

116

61

6

)12)(11(1

2

1

1

2

i

i

6

)12)(1(

1

2

kkki

k

i

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Weak mathematical induction (§ 3.3)Weak mathematical induction (§ 3.3)

Inductive step: show Inductive step: show 6

)1)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)32)(2)(1()1(

1

22

kkkik

k

i

6139261392 2323 kkkkkk

)32)(2)(1()12)(1()1(6 2 kkkkkkk

6

)32)(2)(1(

6

)12)(1()1( 2

kkkkkk

k

6

)1)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)12)(1(

1

2

kkki

k

i

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Strong mathematical induction (§ 3.3)Strong mathematical induction (§ 3.3)

Show that any number > 1 can be written Show that any number > 1 can be written as the product of primesas the product of primes

Base case: P(2)Base case: P(2) 2 is the product of 2 (remember that 1 is not 2 is the product of 2 (remember that 1 is not

prime!)prime!)

Inductive hypothesis: P(1), P(2), P(3), …, Inductive hypothesis: P(1), P(2), P(3), …, P(P(kk) are all true) are all trueInductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true

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Strong mathematical induction (§ 3.3)Strong mathematical induction (§ 3.3)

Inductive step: Show that P(Inductive step: Show that P(kk+1) is true+1) is true

There are two cases:There are two cases: kk+1 is prime+1 is prime

It can then be written as the product of It can then be written as the product of kk+1+1 kk+1 is composite+1 is composite

It can be written as the product of two composites, It can be written as the product of two composites, a and b, where 2 ≤ a and b, where 2 ≤ aa ≤ ≤ bb < < kk+1+1

By the inductive hypothesis, both P(By the inductive hypothesis, both P(aa) and P() and P(bb) is ) is truetrue

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Structural induction Structural induction (§ 3.4)(§ 3.4)

Show that Show that nn((TT) ) ≥≥ 2 2hh((TT) + 1) + 1Inductive hypothesisInductive hypothesis:: Let Let TT11 and and TT22 be full binary trees be full binary trees

Assume that Assume that nn((TT11) ) ≥≥ 2 2hh((TT11) + 1 for some tree ) + 1 for some tree TT11

Assume that Assume that nn((TT22) ) ≥≥ 2 2hh((TT22) + 1 for some tree ) + 1 for some tree TT22

Recursive stepRecursive step:: Let Let TT = = TT11 ∙ ∙ TT22

Here the ∙ operator means creating a new tree with a root note Here the ∙ operator means creating a new tree with a root note rr and subtrees and subtrees TT11 and and TT22

New element is New element is TT By the definition of height and size, we know:By the definition of height and size, we know:

nn((TT) = 1 + ) = 1 + nn((TT11) + ) + nn((TT22))hh((TT) = 1 + max ( ) = 1 + max ( hh((TT11), ), hh((TT22) )) )

Therefore:Therefore:nn((TT) = 1 + ) = 1 + nn((TT11) + ) + nn((TT22))≥≥ 1 + 21 + 2hh((TT11) + 1 + 2) + 1 + 2hh((TT22) + 1) + 1≥≥ 1 + 2*max ( 1 + 2*max ( hh((TT11), ), hh((TT22) )) ) the sum of two non-neg #’s is at least the sum of two non-neg #’s is at least

as large as the larger of the as large as the larger of the twotwo= 1 + 2*= 1 + 2*hh((TT))

Thus, Thus, nn((TT) ) ≥≥ 2 2hh((TT) + 1) + 1

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Induction methods comparedInduction methods comparedWeak Weak

mathematicalmathematicalStrongStrong

MathematicalMathematical StructuralStructural

Used forUsed forUsually Usually

formulaeformulaeUsually formulae not Usually formulae not

provable via provable via mathematical inductionmathematical induction

Only things defined Only things defined via recursionvia recursion

AssumptionAssumption Assume P(k)Assume P(k) Assume P(1), P(2), …, Assume P(1), P(2), …, P(k)P(k)

Assume statement is Assume statement is true for some "old" true for some "old"

elementselements

What to What to proveprove

True for True for P(k+1)P(k+1) True for P(k+1)True for P(k+1)

Statement is true for Statement is true for some "new" elements some "new" elements

created with "old" created with "old" elementselements

Step 1 Step 1 calledcalled

Base caseBase case Base caseBase case Basis stepBasis step

Step 3 Step 3 calledcalled

Inductive stepInductive step Inductive stepInductive step Recursive stepRecursive step

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Pigeonhole principle (Pigeonhole principle (§4.2)§4.2)

Consider 5 distinct points (Consider 5 distinct points (xxii, , yyii) with integer values, where ) with integer values, where ii = 1, = 1, 2, 3, 4, 52, 3, 4, 5Show that the midpoint of at least one pair of these five points Show that the midpoint of at least one pair of these five points also has integer coordinatesalso has integer coordinates

Thus, we are looking for the midpoint of a segment from (Thus, we are looking for the midpoint of a segment from (aa,,bb) to ) to ((cc,,dd))

The midpoint is ( (The midpoint is ( (aa++cc)/2, ()/2, (bb++dd)/2 ))/2 )

Note that the midpoint will be integers if Note that the midpoint will be integers if aa and and cc have the same have the same parity: are either both even or both oddparity: are either both even or both odd

Same for Same for bb and and dd

There are four parity possibilitiesThere are four parity possibilities (even, even), (even, odd), (odd, even), (odd, odd)(even, even), (even, odd), (odd, even), (odd, odd)

Since we have 5 points, by the pigeonhole principle, there must Since we have 5 points, by the pigeonhole principle, there must be two points that have the same parity possibilitybe two points that have the same parity possibility

Thus, the midpoint of those two points will have integer coordinatesThus, the midpoint of those two points will have integer coordinates

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In 3In 3rdrd place (1/2) (27 votes) place (1/2) (27 votes)

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Combinatorial proof (Combinatorial proof (§§4.3)4.3)

Let Let nn be a non-negative integer. Then be a non-negative integer. Then

Combinatorial proofCombinatorial proof A set with A set with nn elements has 2 elements has 2nn subsets subsets

By definition of power setBy definition of power set

Each subset has either 0 or 1 or 2 or … or Each subset has either 0 or 1 or 2 or … or nn elements elementsThere are subsets with 0 elements, subsets with 1 There are subsets with 0 elements, subsets with 1 element, … and subsets with element, … and subsets with nn elements elements

Thus, the total number of subsets isThus, the total number of subsets is

Thus,Thus,

n

k

n

k

n

0

2

nn

k k

n2

0

0

n

n

k k

n

0

1

n

n

n

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In 3In 3rdrd place (2/2) (27 votes) place (2/2) (27 votes)

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Official course goalsOfficial course goals

1.1. Logic: Introduce a formal system (propositional and predicate Logic: Introduce a formal system (propositional and predicate logic) which mathematical reasoning is based on. (sections 1.1-logic) which mathematical reasoning is based on. (sections 1.1-1.4) 1.4)

2.2. Proofs: Develop an understanding of how to read and construct Proofs: Develop an understanding of how to read and construct valid mathematical arguments (proofs) and understand valid mathematical arguments (proofs) and understand mathematical statements (theorems), including inductive proofs. mathematical statements (theorems), including inductive proofs. Also, introduce and work with various problem solving strategies Also, introduce and work with various problem solving strategies and techniques. (sections 1.5, 3.1, 3.3, 3.4) and techniques. (sections 1.5, 3.1, 3.3, 3.4)

3.3. Counting: Introduce the basics of integer theory, combinatorics, Counting: Introduce the basics of integer theory, combinatorics, and counting principles, including a brief introduction to discrete and counting principles, including a brief introduction to discrete probability. (sections 2.4, 4.1-4.4, 5.1) probability. (sections 2.4, 4.1-4.4, 5.1)

4.4. Structures: Introduce and work with important discrete data Structures: Introduce and work with important discrete data structures such as sets, relations, sequences, and discrete structures such as sets, relations, sequences, and discrete functions. (sections 1.6-1.8, 2.7, 3.2, 7.1, 7.3-7.6) functions. (sections 1.6-1.8, 2.7, 3.2, 7.1, 7.3-7.6)

5.5. Applications: Gain an understanding of some application areas of Applications: Gain an understanding of some application areas of the material covered in the course. (sections 2.6, 3.6, 10.3) the material covered in the course. (sections 2.6, 3.6, 10.3)

The written final will be based on these goals!The written final will be based on these goals!

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In 2In 2ndnd place (30 votes) place (30 votes)

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Unofficial course goalsUnofficial course goals

Ensure that all students have a “base” level of discrete Ensure that all students have a “base” level of discrete mathematical proficiencymathematical proficiency

In particular, ensure that they know a number of covered topics, In particular, ensure that they know a number of covered topics, such as sets, logic, etc.such as sets, logic, etc.

Used in CS 216, in particularUsed in CS 216, in particular

To prepare for more advanced computer theory classes To prepare for more advanced computer theory classes (CS 302, CS 432)(CS 302, CS 432)

To give the students more experience in mathTo give the students more experience in math The best way to get better at anything is through practiceThe best way to get better at anything is through practice

To help create a To help create a computer scientistcomputer scientist, not a computer , not a computer programmerprogrammer

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And the winner is…And the winner is…with 45 votes…with 45 votes…

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Have a summer!Have a summer!

Final is 9 a.m. this Saturday!Final is 9 a.m. this Saturday!

1 course summary cs 202 aaron bloomfield. 2 outline a review of the proof methods gone over in class...

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methodology1st place

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