C-N Math 207 Discrete Math

C-N Math 207Discrete Math

Kenneth Massey

September 16, 2011

C-N Math 207 - Massey, 1 / 70

Introduction

Question

What is the smallest positive number?


Introduction

Sets

I a set S is a collection of objects (members,elements)

I S = {VT ,UNC ,GT ,Mia ,UVA,Duke}I the cardinality (size) of S is written as |S |

I S is finite if |S | <∞, and infinite if |S | =∞I empty / null set ∅ = { }, |∅| = 0I “x is an element of S ” is denoted by x ∈ S


Introduction

Numbers

Abstract notions of quantity, position, or size.I natural numbers N = {0, 1, 2, 3, · · · }(use N+ to omit zero)

I integers Z = {· · · ,−2,−1, 0, 1, 2, 3, · · · }I set-builder notation:members “such that” conditions

I rational numbers Q = {nd | n , d ∈ Z, d 6= 0}I real numbers R is the continuumof all points on number lineincluding irrationals e.g.

√2, π


Introduction

Set Examples

1. S = {x ∈ Z | |x − 5| ≤ 2}2. S = {1, 12 ,

14,

18 , · · · }

3. S = {(x , y) ∈ R2 | x 2 + y2 ≤ 1}4. S = {ticker symbols of S&P 500 members}


Introduction

Discrete vs ContinuousI A set is discrete if its members can be listed(enumerated) so that each has a “next”(successor), with nothing in-between

I all finite sets are discrete (loop)I examples:

1. “digital” computers2. money (Office Space rounding)3. siblings (1

2 allowed)4. water discrete, but practically continuous5. “quantum” physics6. movie frames vs real time?7. Z vs R, Q ?


Introduction

Existence and Uniqueness

I “there exists” is denoted by ∃∃x ∈ Z | x 2 = 9

I ∃! means “there exists a unique”∃! x ∈ R | 3x + 4 = 10

I slash through a symbol means “not”@x ∈ Q | x 2 = 2


Introduction

If-Then Statements

“If (condition) then (conclusion)”I The conclusion must be true whenever thecondition is true.

I Any exception (counter-example) means theif-then statement is false.

I If c ∈ Z, then ∃x ∈ R | x 2 + 6x = c


Introduction

Function

f : A→ B is a function that pairs each memberof set A with a unique member of set B .I y = f (x ) where x ∈ A and y ∈ BI the set A is called the domainI the set B is called the codomainI {b ∈ B | ∃a ∈ A with f (a) = b} is called therange or image of f


Introduction

Function Examples

1. f : R→ R defined by y = f (x ) = x 2

2. f : Z→ N defined by y = f (x ) = x 2

3. f : N→ N defined by y = f (n) = b√

nc4. f : N+ → Q defined by f (n) = n+1

n

Which are onto (range equals codomain)?Which are one-to-one (no “y” value hit twice)?


Introduction

Invertible Function

I f is said to be invertible if each y in B ismapped to exactly once

I f −1 : B → A is a function that reverses fI To find the inverse, solve for x .

x = f −1(y)I the composition of a function and itsinverse is the identity function f −1 ◦ f = If −1(f (x )) = x and f (f −1(y) = y


Introduction

Sequence

A sequence is a function whose domain is somesubset of Z (often N).I an ordered indexed list, f (k) = fkI geometric sequence ak = r k for some r ∈ RI harmonic sequence

{ 1k

}∞k=1


Introduction

Sum NotationCapital Greek sigma Σ indicates “the sum of”indexed terms, e.g.

6∑k=3

k 2 = 32 + 42 + 53 + 62 = 86

∞∑k=1

110k = .1+ .01+ .001+ · · · = .11111 =

19

an =

n∑k=1

k =n(n + 1)

2


Introduction

Product Notation

Capital Greek pi Π indicates “the product of”indexed terms, e.g.

4∏k=1

k + 1k

= (2/1)(3/2)(4/3)(5/4) = 5


Introduction

Factorials

n factorial is defined for n ∈ N by

n ! =

{∏nk=1 k n ≥ 1

1 n = 0

I the sequence of factorials is1, 1, 2, 6, 24, 120, · · ·

I n ! = n · (n − 1)!


Integers

IntegersI Z is discrete, but infiniteI Subsets of Z can be used to label otherdiscrete sets.

I Arithmetic operations: addition (+) andmultiplication (·)

I Basic properties: closed, commutative,associative, distributive, identities, inverses,ordered

I Number theory is the branch ofmathematics that studiesthe properties of Z.


Integers

Division Algorithm

Note that 173 = 5+ 2

3 can be written as

17 = 3 · 5+ 2

Division algorithm: let a , b ∈ Z, with b 6= 0,then ∃! q , r ∈ Z such that

a = bq + r , 0 ≤ r < |b|

q is called the quotient, and r the remainder.

I example: a = ±73, b = 11


Integers

Place Value

I Ways to represent natural numbers:tally/tic marks, Egyptian, Roman numeral,place value (brilliant!).

I We are familiar with base 10, but othersocieties have used e.g. base 12 or 60.

I Computers use binary (base 2)octal (base 8), [hexadecimal] (base 16)


http://www.colorpicker.com/

Integers

Place ValueEach x ∈ N can be uniquely represented in baseb as a linear combination of powers of b, witheach coefficient (digit) between 0 and b − 1.

x = (Dd−1 · · ·D2D1D0)b =

d−1∑k=0

Dkbk

I (75)10 = 7 · 101 + 5 · 100I (113)8 = 1 · 82 + 1 · 81 + 3 · 80I (300)5 = 3 · 52I (1001011)2 = 26 + 23 + 21 + 20

I (4B)16 = 4 · 161 + 11 · 160


Integers

Place Value

Write (146)10 in base 3 using the divisionalgorithm.

146 = 3 · 48+ 2= 3 · (3 · 16+ 0) + 2= 3 · (3 · (3 · 5+ 1) + 0) + 2= 3 · (3 · (3 · (3 · 1+ 2) + 1) + 0) + 2

So (146)10 = (12102)3.


Integers

Place Value

There are 10 types of people in the world: thosethat understand binary, and those that don’t.1. Coach Sparks decides to use hexadecimal

for the football team’s jersey numbers. If ajersey reads “AC”, find the decimal andbinary equivalents.

2. How many points did C-N win by if thescoreboard reads 53− 32 in base 7 ?


Integers

Place Value

Arithmetic algorithms work the same way inother bases.1. (234)5 + (312)52. (175)8 · (6)83. (11010)2 · (1011)2


Integers

Divides

Definition: Let a , b ∈ Z.Then b divides a , if and only if (iff) there existsq ∈ Z such that a = bq .

I in the division algorithm, r = 0I Equivalent statements:

I notation: b | aI b divides (evenly into) aI a is divisible by bI b is a factor/divisor of a


Integers

Generally True?

1. n | 02. 0 | n3. n | 14. 1 | n5. n | n6. n | n2

7. (a + b) | (a2 − b2)


Integers

Your First Proof

Theorem: If a | b and b | c, then a | c(divisibility is transitive)

Illustration: 4 | 12 and 12 | 72, therefore 4 | 72.

Proof: Suppose a | b and b | c.By definition, b = aq1 and c = bq2.Then by substitution,c = bq2 = (aq1)q2 = a(q1q2).Therefore a | c. �


Integers

Direct If-Then Proofs1. Understand what you are proving.

Work an example; believe it is true.

2. Scratch work:

I what do you know? (the “if” part)I what do you need to show? (the “then” part)I use definitions to translate symbols/terminologyI deduce the desired result using algebra,

substitution, tricks, previous theorems, etc

3. Rewrite your proof neatly.

I State the theorem.I Prove the theorem, clearly justifying

each step as you move fromassumptions to the conclusion.


Integers

Theorem: Suppose a , b ∈ N.If a | b and b | a , then a = b.

Proof:


Integers

Even and Odd

Definitions:I n ∈ Z is even iff ∃q ∈ Z | n = 2qI n ∈ Z is odd iff ∃q ∈ Z | n = 2q + 1

Prove that1. if n is odd, then n + 5 is even.2. if n is odd, then n2 is odd.


Integers

Modulo

As in the division algorithm, let a = bq + r .Then “a modulo b” is the remainder r .I notation: a (mod b) = rI b | a iff a (mod b) = 0I odd if a (mod 2) = 1I List the possible values of a (mod 5).


Integers

Congruence

a and b are congruent (aliases) mod n

a ≡ b (mod n)

Equivalent definitions:I n | (a − b)I a (mod n) = b (mod n)I (a − b) (mod n) = 0I a − b = qn for some q ∈ Z


Integers

Equivalence

Congruence is an equivalence relation:1. reflexive

a ≡ a (mod n)2. symmetric

if a ≡ b (mod n) then b ≡ a (mod n)3. transitive

if a ≡ b (mod n) and b ≡ c (mod n)then a ≡ c (mod n)


Integers

Integers (mod n)The equivalence class of a (mod n) is the set ofall of a ’s aliases.

[a ]n = {b ∈ Z | a ≡ b (mod n)}

Zn is the set of all equivalence classes (mod n)I Zn = {0, 1, 2, 3, · · ·n − 1}I “Z mod n” puts each integerinto exactly one class

I loop around like a clock


Integers

Proof of Transitivity

Theorem: if a ≡ b (mod n) and b ≡ c (mod n)then a ≡ c (mod n)Proof: Assume that a ≡ b (mod n).Then by definition a − b = nq1 for some q1 ∈ Z.Similarly b − c = nq2 for some q2 ∈ Z.∴ (therefore)

a−c = (a−b)+(b−c) = nq1+nq2 = n(q1+q2)

and hence by definition n | (a − c) anda ≡ c (mod n). �


Integers

Algrebra of CongruenceTheorem: if a ≡ b and c ≡ d ,then a + c ≡ b + d and ac ≡ bd .Proof: By assumption, and the definition of congruence,a − b = q1n and c − d = q2n . Therefore,

(a + c) − (b + d) = (a − b) + (c − d)

= q1n + q2n = (q1 + q2)n

ac − bd = (b + q1n)c − b(c − q2n) = (q1c + q2b)n

Thus by definition, a + c ≡ b + d and ac ≡ bd . �


Integers

Zn Algebra

Hence we may develop algebra of + and × onZn . See the Affine Cipher Project.I addition / [multiplication tables]I inversesI solving equations


http://massey.limfinity.com/207/multtable.php

Integers

Problems

1. Compute 75 (mod 11), reducing as you go.2. What day of the week will it be 10000 days

from now?3. Solve 4(6x − 5) ≡ 5 (mod 7).


Integers

Affine Cipher

The affine cipher transforms each character inan alphabet by the formula:

y = f (x ) = (ax + b) (mod n)

I Experiment with the [web app].I Using the alphabet (space, A-Z), decrypt“VWBTU” if y = 5x + 2 (mod 27).


http://massey.limfinity.com/207/affinecipher.php

Integers

Invertibility

I If n = 27, a = 3, and b = 5then ’J’ (letter 10) gets tranformed to ’H’,but so does ’A’

I To be decipherable, a cipher function mustbe invertible (reversible). Each charactermust be mapped to exactly once.

I If n = 27, what values of a make the affinecipher an invertible function?

I What if n = 28 ?


Integers

Affine De-Cipher

I if y = f (x ) = ax + b thenx = f −1(y) = a−1(y − b)

I y = ax + b is invertible iff a−1 existsI find 7−1 (mod 27) by writing {7k (mod 27)}I 7−1 (mod 28) DNEI inverses come in pairs (some may be theirown inverse)

I hypothesize about when a−1 (mod n) exists


Integers

A Weird Function

Let δ : N+ → N be defined by

δ(n) = |{q ∈ N | q | n}|

I Describe the function in words.I Compute δ(21).


Integers

Prime Numbers

Let n > 1 be a natural number.I Definition: n is prime if its only naturalnumber divisors are 1 and n itself.

I n is prime iff δ(n) = .I If n is not prime, then it is composite.I If n is composite, then it has a prime factorwhich is ≤

√n .

I Sieve of Eratosthenes


Integers

Fundamental Theorem of Arithmetic

Fundamental Theorem of Arithmetic:each n > 1 can be written uniquely (up toordering) as the product of prime factors

n =∏

pαii

e.g. the prime factorization of 21560 is

21560 = 23 · 5 · 72 · 11


Integers

Prime Number Theorem

Let π(n) be the number of primes p ≤ n .

π(n) ≈ nln(n)

This estimate was proved to be asymptotic in1896.

limn→∞ π(n)

n/ ln(n)= 1


Integers

Greatest Common Divisor

I Define the greatest common divisor ofa , b ∈ Z to be the largest integer that is adivisor of both a and b.

I e.g. gcd(36, 84) = 12 (can use prime factors)I If a 6= 0, then gcd(a , 0) = |a |.I a and b are relatively prime (coprime) ifgcd(a , b) = 1


Integers

Least Common Multiple

I Define the least common multiple ofa , b ∈ Z to be the smallest natural numberthat is a multiple of both a and b.

I e.g. lcm(14, 35) = 70 (can use prime factors)I If ` = lcm(a , b) then a | ` and b | `

I Define lcm(a , 0) = 0 as a special case.


Integers

GCD and LCM

I gcd and lcm are commutative functionsI Use prime factors to find gcd(24, 30) andlcm(24, 30).

I Convince yourself thatgcd(a , b) · lcm(a , b) = |ab|.


Integers

Proving or Disproving

Consider a general if-then statement.I To be true, it must hold for any conditionthat satisfies the “if” part. No exceptions.

I You can’t prove the statement bycherry-picking examples that work.

I You can disprove the statement by findingone counter-example: a case where the “if”part is true, but the “then” part is false.


Integers

Disprove

Find a counter-example to the following:1. if n > 0 then 22n

+ 1 is prime225

+ 1 = 4294967297 = (641)(6700417)2. if c | (ab) then either c | a or c | b3. gcd(ma ,nb) = gcd(m ,n) gcd(a , b)


Integers

Prove

1. If x ≡ 3 (mod 8) then x 2 ≡ 9 (mod 16).2. The product of two consecutive integers is

even. (rewrite as if/then)3. If n is odd, then 8 | (n2 − 1).4. If c | a and c | b, then for any m ,n ∈ Z,

c | (ma + nb).5. If n ∈ N, then gcd(n ,n + 1) = 1.


Integers

Prove or disprove

1. If p is prime, then p + 7 is composite.2. If n > 0, then 3n + 2 is prime.3. If a ≡ b (mod n) then a2 ≡ b2 (mod n2)

4. If a ≡ b (mod n) then a2 ≡ b2 (mod n)


Integers

Mathematical Statements

I conjecture: believed to be true, butunproven

I lemma: a precursor to a more noteworthytheorem

I theorem: has been provedI corollary: follows as a special case of atheorem


Integers

Fun Fact

(Liouville) Pick any n ∈ N+ and write down itsdivisors, e.g. if n = 50 then {1, 2, 5, 10, 25, 50}.Then count how many divisors each divisor has.C = {1, 2, 2, 4, 3, 6}. Then (

∑C x )2 =

∑C x 3.

(1+2+2+4+3+6)2 = 1+8+8+64+27+216

Proof is beyond the scope of this class.


Integers

E.A. Lemma 1

Lemma: For any k ∈ Z, x divides both a and bif and only if (iff) x divides both a + kb and bProof: Suppose x | a and x | b.Then a = xq1 and b = xq2.Therefore a + kb = x (q1 + kq2), and sox | (a + kb).

Now suppose x | (a + kb) and x | b.Then a + kb = xq1 and b = xq2.Therefore a = xq1 − kb = x (q1 − kq2), and sox | a . �


Integers

E.A. Lemma 2

Lemma: For any k ∈ Z,

gcd(a , b) = gcd(b, a + kb)

Proof: By the previous lemma, the set ofcommon divisors of a and b is identical to theset of common divisors of a + kb and b.Therefore the GCD’s must be the same. �


Integers

Euclidean Algorithm

Euclidean Algorithm Theorem:

gcd(a , b) = gcd(b, a (mod b))

Proof: Use the division algorithm towrite a = bq + r , where r = a (mod b).Then r = a − bq . Taking k = −q , the previouslemma says that

gcd(a , b) = gcd(b, a − bq) = gcd(b, r)

�


Integers

E.A. ExampleThe Euclidean algorithm repeatedly uses theprevious theorem to find the GCD of twonumbers.

gcd(952, 210) = gcd(210, 112)= gcd(112, 98)= gcd(98, 14)= gcd(14, 0)= 14


Integers

E.A. Exercises

1. Use the EA to find gcd(12540, 2002), thenfind the lcm.

2. Show that 495 and 392 are relatively prime.3. True or false?

gcd(a , b, c) = gcd(gcd(a , b), c)4. Draw a line from (0, 0) to (a , b) in the

Cartesian plane. How many integer latticepoints does it cross?


Integers

GCD as a Linear Combination

Back-substitution allows us to write the gcd as alinear combination of the two arguments, that isgcd(a , b) = c1a + c2b.

952 = (210)(4) + 112

210 = (112)(1) + 98

112 = (98)(1) + 14

98 = (14)(7) + 0

112 = 952(1) + (210)(−4)

98 = (952)(−1) + (210)(5)

14 = (952)(2) + (210)(−9)


Integers

GCD as a Linear CombinationTheorem a and b are relatively prime iff 1 canbe written as a linear combination of a and b.Proof First suppose a and b are relativelyprime. Then gcd(a , b) = 1, and the EA withback-substitution allows us to writec1a + c2b = gcd(a , b) = 1.

Now suppose c1a + c2b = 1, and suppose x is adivisor of both a and b. Say a = xq1 andb = xq2. Then x (c1q1 + c2q2) = 1, and sinceeverything is an integer, it must be that |x | = 1.Therefore gcd(a , b) = 1. �


Integers

Inverses in Zn

Theorem If gcd(a ,n) = 1, then a has an inverse in Zn .Proof Since gcd(a ,n) = 1, we can write c1a + c2n = 1.Therefore c1a ≡ 1 (mod n), and c1 is the candidateinverse. (existence)To be the inverse, c1 must be the unique solution toax ≡ 1 in Zn .If ax ≡ 1 then ac1 − ax ≡ 0, and thereforec1a(c1 − x ) ≡ 0.But since c1a ≡ 1,we get c1 − x ≡ 0 or x ≡ c1. (uniqueness)∴ a−1 = c1�


Integers

Inverses in Zn

Theorem The converse is also true:if a has in inverse in Zn , then gcd(a ,n) = 1.Proof Suppose ab ≡ 1 (mod n), so that ab = 1+ kn .Let d be a divisor of both a and n , so that a = dq1 andn = dq2.Then dq1b = 1+ kdq2, and d(q1b − kq2) = 1. The onlyinteger possibility is d = ±1. �


Integers

Finding an inverse in Zn

1. Show that a = 28 and n = 135 arerelatively prime, and write 1 as a linearcombination of a and n .

2. Find 28−1 (mod 135).3. Find 4−1 in Z27 (easy by inspection)4. Find 8−1 in Z27 (use EA)5. Solve 10(x − 7) = 6+ 2x in Z27


Integers

Affine Cipher Decryption

Recall the affine cipher function and its inverse:

y = ax + b (mod n) x = a−1(y − b)

I We now know that a−1 exists iffgcd(a ,n) = 1.

I In general, use the EA to writec1a + c2n = 1, and then a−1 = c1.

I If n = 27, a = 2, and b = 5, theny = 2x + 5 (mod 27)x = 14(y − 5).


CRT

Pirate Treasure

A band of ten pirates captured a treasure chestfull of gold coins. When they attempted todivide the coins evenly, 3 were left over. A fightensued and one pirate was killed. Again theyattempted to divide the coins, but now therewere 8 left over. This time two pirates werekilled fighting over the extras. On the third try,the coins divided evenly. How many coins werein the chest?


CRT

Pirate Treasure

Solve the system of congruences:

x ≡ 3 (mod 10)x ≡ 8 (mod 9)x ≡ 0 (mod 7)


CRT

Chinese Remainder Theorem

Chinese Remainder Theorem: if m1 and m2 arerelatively prime, then

x ≡ r1 (mod m1)

x ≡ r2 (mod m2)

has a unique solution (mod m1m2).This theorem can be extended to 3 or morecongruences by induction.


CRT

CRT Algorithm

Idea: absorb each congruence sequentially:

x = 3+ 10n

3+ 10n ≡ 8 (mod 9)

1n ≡ 5 (mod 9)

n ≡ 1−15 (mod 9)

n = 5

x = 53 (mod 90)

53+ 90n ≡ 0 (mod 7)

4+ 6n ≡ 0 (mod 7)

6n ≡ 3 (mod 7)

n = 6−13 (mod 7)

n = 4

x = 413 (mod 630)


CRT

Football Scores

Find all scores less than 100 that satisfy:

x ≡ 1 (mod 2)

x ≡ 2 (mod 3)

x ≡ 3 (mod 7)


CRT

CRT ConditionsThis system has solutions

x ≡ 3 (mod 4)x ≡ 5 (mod 6)

but this one is inconsistent, and has no solution.

x ≡ 2 (mod 4)x ≡ 5 (mod 6)


CRT

Eggs

An old woman goes to market and a horse steps on herbasket and crushes the eggs. The rider offers to pay forthe damages and asks her how many eggs she hadbrought. She does not remember the exact number, butwhen she had taken them out two at a time, there wasone egg left. The same happened when she picked themout three, four, five, and six at a time, but when she tookthem seven at a time they came out even. What is thesmallest number of eggs she could have had?


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C-N Math 207 Discrete Math