cmsc 250 discrete structures number theory. 20 june 2007number theory2 exactly one car in the plant...

CMSC 250Discrete Structures

Number Theory

20 June 2007 Number Theory 2

Exactly one car in the plant has color

H(a) := “a has color” xCars

– H(x) aCars

– a x ~ H(a)

H(a,b) := “a has color b” xCars

yColors H(x,y)

aCars, bColors a x ~ H(a,b)


At most one car in the plant has color

H(a,b) := “a has color b” x,aCars

y,bColors [H(x,y) H(a,b)] x = a


At least two cars in the plant have color

H(a,b) := “a has color b” x,aCars

y,bColors H(x,y) H(a,b) x a


Existential Gen/Inst Existential Generalization

– P(value)– value DxD such that P(x)

Existential Instantiation xD such that P(x)P(a), aD such that P(a) is true


Proofs Must Have! Clear statement of what you are proving Clear indication you are starting the proof Clear indication of flow Clear indication of reason for each step Careful notation, completeness and order Clear indication of the conclusion and why

it is valid.

Suggest pencil and good erasure when needed


Mathematical Proofs For any real number x, x – 1 = x – 1

– Can you prove that?

For any real number x, x + y = x + y– Can you prove that?


Domains Z – integers Q – rational numbers (quotients of

integers)– rQ a,bZ, (r = a/b) (b 0)– Irrational = not rational

R – real numbers Superscripts: Z+, Z-, Zeven, Zodd, Q>5


Closure of Sets (Integers) Addition

– If aZ and bZ, then (a + b) Z Subtraction

– If aZ and bZ, then (a – b) Z Multiplication

– If aZ and bZ, then (a * b) Z– If aZ and bZ, then abZ


Integer Definitions Even integer

– nZeven kZ, n = 2k Odd integer

– nZodd kZ, n = 2k+1 Prime integer (Z>1)

– nZprime r,sZ+, (n=r*s) (r=1) (s=1) Composite integer (Z>1)

– nZcomposite r,sZ+, (n=r*s) (r1) (s1)


Examples Prove 4 is even Prove 5100 is even Is 0 even? Is -301 even? If aZ and bZ, then is 6a2b even? Is every integer either even or odd? Prove? Is 1 prime? What are the first 6 primes? Is it true that every integer greater than 1

is either composite or prime? Prove?


Constructive Proofs of Existence

Proving an xD, such that Q(x):– Finding an x in D that makes Q(x) true– Giving a set of directions (such as a

formula or algorithm) that will give an x in D that makes Q(x) true


kZ such that 22r + 18s = 2k

Where rZ and sZ k,r,sZ, such that 22r + 18s = 2k Prove?


Constructive Proof of Existence

If we want to prove: nZeven, p,q,r,sZprime

n=p+q n=r+s pr ps qr qs

– Let n=10 n Zeven by definition of even

– Let p = 5 and the q = 5 p,q Zprime by definition of prime 10 = 5+5

– Let r = 3 and s = 7 r,s Zprime by definition of prime 10 = 3+7

– And all of the inequalities hold


Proving Universal Statements

Proof by exhaustion– Can only be used on finite domains

rZ+ where 23<r<29 p,qZ+ (r=pq) (p q) But not even all of those

Proof by Generalizing from the Generic Particular– Let x represent a particular but arbitrarily chosen

element in the domain– Show that x satisfies the predicate– This does not mean you choose an element at

random


The sum of any two integers is even

Formally m,nZeven, (m + n)Zeven m,nZ, (mZeven nZeven) (m + n)Zeven

Proof:– Start

Let m be a generic particular even number Let n be a generic particular even number

– Show that (m + n)Zeven (on the board)


Proofs Must Have! Clear statement of what you are proving Clear indication you are starting the proof Clear indication of flow Clear indication of reason for each step Careful notation, completeness and order Clear indication of the conclusion and why

it is valid.

Suggest pencil and good erasure when needed


An even number times an integer yields an even number


The product of any two odd integers is odd


Prove Universal False by Counterexample

a,bR, a2=b2 a=b

Let a = 2; b = -2 a2=b2

22 = (-2)2

4 = 4 – is TRUE a=b 2 = -2 – is FALSE TRUE FALSE FALSE


Rational Numbers Q – rational numbers (quotients of integers)

– rQ a,bZ, (r = a/b) (b 0)– Irrational = not rational

Which of the following are rational?– 10/3– 0.281– 7– 0– 2/0– 0.1212– 5.1212


Prove 7 is a rational number rQ a,bZ, (r = a/b) (b 0)

Let a = 7 Let b = 1 7 = 7/1 (by algebra) 7 Z 1 Z 1 0


Prove nZ, n is rational (i.e. nQ)


Sum of any two rational numbers is rational

r,sQ, r + s Q Let r = a/b Let s = c/d r + s = a/b + c/d = (ad + cb)/bd

Theorem– Statement that is known to be true because it

has been proved. Corollary

– Statement whose truth can be immediately deduced from a proved theorem

– E.g.: The double of a ration number is rational


Division Definitions If n and d are integers, then

– N is divisible by d if, and only if, n=dk for some integer k

– d|n kZ, n=dk (read – “d divides n”) Alternatively, we say that

– n is a multiple of d, or– n is divisible by d, or– d is a factor of n, or– d is a divisor of n, or– d divides n.

32 a multiple of -16? 21 divisible by

3?7 a factor of -7?

5 divide 40?7|42?


Transitivity of Divisibility a,b,cR (a|b b|c) a|c Need to show: a|c c = a*k, where a 0 Choose generic particulars a,b,c s.t. a|b b|c

– a|b b = ar, where a,rZ and a 0– b|c c = bs, where b,sZ and b 0

Substitution– c = bs– c = (ar)s– c = a(rs)– k = (rs)– c = ak


Proof Using Contrapositive For all positive integers, if n does not

divide a number to which d is a factor, then n cannot divide d.

n,d,cZ+, ndc nd n,d,cZ+, n|d n|dc (Contrapositive)

Prove …


a,bZ, (a|b b|a) a=b Choose general particular a,bZ s.t. a|b b|a a|b b = a*r, where a,rZ and a 0 b|a a = b*s, where b,sZ and b 0 Algebra

– a = bs– a = (ar)s– a = a(rs)– 1 = rs (since a 0)

Is there a unique solution?– No; r=s=1, r=s=-1

Substitution– a=b or a=-b


Proof by Contradiction Suppose the statement to be proven

is FALSE Show that this leads to a logical

contradiction Conclude the original statement is

TRUE

We can do this since every statement is TRUE or FALSE, but not BOTH.


There is no largest integer. Suppose there is. Let P represent that integer. This means that nZ P ≥ n Show this leads to a contradiction:

– Let m = P + 1– mZ by closure of addition– m > p, by algebra; CONTRADICTION– So P, is not the largest integer.


Sum of any rational number and any irrational number is

irrational Suppose there exists a rational number r and an irrational number s such that r + s is rational

This means:– r = a/b for a,bZ b0– r + s = c/dZ for d0– r + s = c/dZ fo– a/b + s = c/d– s = c/d – a/b– s = (cb – ad)/bd, num integer, denom int0– Contradiction!


Proof by Contradiction Every integer is rational Suppose every integer is irrational From supposition, 1 is irrational, but 1 =

1/1 which is rational Since our supposition led to a contradiction,

then our original statement must be true

ERROR nZ, nQ nZ, nQ– There is some integer that is irrational


Unique Factorization

Example:– 72– 2 2 2 3 3– 2 3 3 2 2– 3 2 2 3 2– 32 23

– 23 32 (Standard Factored Form)

kek

eee ppppn 321321


Q2


Q 231


More Integer Definitions Div and mod operators

– n div d – integer quotient for n d– n div d – integer remainder for n d– (n div d = q) (n mod d = r) n = qd +

r where nZ0, dZ+, rZ, qZ, 0r<d– (Quotient Remainder Theorem)

Relating “mod” to “divides”– d|n 0 = n mod d

– d|n 0 d n


Modular Notation For p,q,rZ Equivalent notations:

– p x q

– (p mod x) = (q mod x)– x|(p – q)


Proofs Using this Notation mZ+, a,bZ

a m b kZ a = b + km

mZ+, a,b,c,dZ (a m b) (c m d) (a + c) m (b + d)


Proof by Division into Cases nZ, 3n n2 3 1


The square of any integer has form 4k or 4k + 1 for some

integer k


Floor & Ceiling Definitions n is the floor of x where xR nZ

x = n n x < n + 1

n is the ceiling of x where xR nZ x = n n – 1 < x n


Floor Proofs x,yR x + y = x + y

xR yZ x + y = x + y


Another Floor Proof The floor of n/2 is either:

– n/2 when n is even– (n-1)/2 when n is odd

Prove by division into cases


nZ, pZprime, p|n p(n + 1)

Suppose n,p s.t. p|n p|(n + 1) Let n = pi, n + 1 = pj pj = pi + 1 pj – pi = 1 p(j – i) = 1 p|1 p can be 1 or -1, neither of which is

prime


Is the set of primes infinite? Assume finite set of primes of size N

– {p1, p2, ,pN}

Construct x = p1 p2 … pN

Prime factorization says: pi, such that pi|(x+1)

That pi must be in {p1, p2, ,pN} so pi|x

pi|x pi|(x+1) Which contradicts the previous theorem


Summary of Proof Methods Constructive Proof of Existence

Proof by Exhaustion Proof by Generalizing from the

Generic Particular Proof by Contraposition Proof by Contradiction Proof by Division into Cases


Errors in Proofs Arguing from example for universal

proof. Misuse of Variables Jumping to the Conclusion (missing

steps) Begging the Question Using "if" about something that is

known


Applications Programming

– If … then …– Loops …– Algorithms (e.g. gcd) …

More examples– Calculator

Sqrt(2) = 1.414213562 40.72727272727

= ?– Cryptography …


Using the Unique Prime Factorization Theorem

Prove: aZ+ 3 | a2 3 | a

Prove: aZ+qZprime q|a2 q|a


nZ, n2ZoddnZodd


nZ, a,bZodd, n=a+b

cmsc 250 discrete structures number theory. 20 june 2007number theory2 exactly one car in the plant...

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