1 CMSC 250 Chapter 3, Number Theory. 2 CMSC 250 Introductory number theory l A good proof should have: a statement of what is to be proven Proof:

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Chapter 3, Number Theory

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Introductory number theory

A good proof should have:– a statement of what is to be proven– "Proof:" to indicate where the proof starts– a clear indication of flow– a clear indication of the reason for each step– careful notation, completeness and order– a clear indication of the conclusion

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Some definitions Z is the integers Q is the rational numbers (quotients of integers)

r Q (a,b Z) [(r = a / b) (b 0)] Irrational numbers are those which are not rational R is the real numbers A superscript of + indicates the positive portion only of

one of these sets of numbers A superscript of – indicates the negative portion only of

one of these sets of numbers Other superscripts can be used, such as Zeven, Zodd , Q>5

We can define the closure of these sets for an operationZ is closed under what operations?

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Integer definitions An even integer

n Zeven (k Z) [n = 2k] An odd integer

n Zodd (k Z) [n = 2k + 1] A prime integer (Z>1)

n Zprime (r,s Z+) [ (n = r s) (r = 1) (s = 1)] A composite integer (Z>1)

n Zcomposite (r,s Z+) [(n = r s) (r 1) (s 1)]

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Constructive proof of existence

How can we prove the following?

]srnqp n

s q r q spr p s r q [p )Zsr,q,p,Z,n(

3333

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Discrete StructuresCMSC 250Lecture 13

February 25, 2008

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Methods of proving universally quantified statements

Method of exhaustion– prove the statement is true for each and every member of the

domain– (r Z+)[23 < r < 29 ( p,q Z+) [(r = p q) (p q)]]

Generalizing from the generic particular– suppose that x is a particular but arbitrarily-chosen element of

the domain– show that x satisfies the property– then the result holds for every element of the domain– example: (r Z) [r Zeven r2 Zeven ]

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Examples of generalizing from the generic particular

The product of any two odd integers is also odd.– (m,n Z) [(m Zodd n Zodd) m n Zodd ]

The product of any two rationals is also rational.– (m,n Q) [m n Q]

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Discrete StructuresCMSC 250Lecture 14

February 27, 2008

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Disproof by counterexample

( r Z) [r2 Z+ r Z+]– counterexample: r2 = 9 r = –3

• r2 Z+ since 9 Z+ so the antecedent is true• but r Z+ since –3 Z+ so the consequent is false• this means the implication is false for r = –3, so this is a valid

counterexample When you give a counterexample you must justify that it

is a valid counterexample, by showing the algebra (or other interpretation needed) to support your claim

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Divisibility

Definition: d | n (k Z)[n = d × k]– n is divisible by d– n is a multiple of d– d is a divisor of n– d divides n

Results involving divisibility:(a, b Z>1)[a | b→ a | (b + 1)](x Z>1)(p Zprime)[p | x]– note another proof method would be used here, proof by

division into cases

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Proof by contrapositive

For all positive integers n, if n does not divide a number of which d is a factor, then n can not divide d.

(n,d,c Z+) [n | dc n | d](n,d,c Z+) [n | d n | dc]

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Discrete StructuresCMSC 250Lecture 15

February 29, 2008

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Proof by contradiction

The number of primes is infinite Assume this is false, that there is some largest prime

number, so there are only n different prime numbers Consider the number

1)pp...p(pk n1n21

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Prime factored form The Unique Factorization Theorem (Theorem 3.3.3)

Given any integer n > 1,(kZ)(p1,p2,…pkZprime)(e1,e2,…ekZ+)[n = p1

e1 × p2e2 × p3

e3 × …× pkek]

where the p’s are distinct and any other expression of n is identical to this except maybe in the order of the factors.

Standard factored formn = p1

e1 × p2e2 × p3

e3 × … × pkek

pi < pi+1

(mZ)[8×7×6×5×4×3×2×m = 17×16×15×14×13×12×11×10]– does 17 | m ??

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More integer definitions div and mod operators

n div d- integer quotient forn mod d- integer remainder for(n div d = q) ^ (n mod d = r) n = d × q + r

where n Z0, d Z+, r Z, q Z, 0 r < d Relating “mod” to “divides”:

d | n 0 = n mod d

0 d n

Definition of equivalence in a mod:x d y d | (x–y) (note: their remainders are equal)sometimes written as x y mod d, meaning (x y) mod d

dn

dn

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Discrete StructuresCMSC 250Lecture 16

March 3, 2008

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Modular arithmetic (Theorem 10.4.3)

Let a, b, c, d, n Z, and n > 1. Suppose a c (mod n) and b d (mod n). Then:

1. (a + b) (c + d) (mod n)2. (a – b) (c – d) (mod n)3. ab cd (mod n)4. am cm (mod n) for all integers m

Proof using this definition:(m Z+)(a,b Z)[a m

b (k Z) [a = b + km]]

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Discrete StructuresCMSC 250Lecture 18

March 7, 2008

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Floor and ceiling

Definitions:– n is the floor of x where x R n Z

x = n n x < n+1– n is the ceiling of x where x R n Z

x = n n–1 < x n Proofs using floor and ceiling:

(x,y R) [ x+y = x + y ](x R)(y Z)[ x+y = x + y ]

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More proof by division into cases

The floor of (n/2) is eithera) n/2 when n is even

or b) (n–1)/2 when n is odd

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Discrete StructuresCMSC 250Lecture 19

March 10, 2008

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The quotient remainder theorem

(n Z)(d Z+)(q,r Z)[(n = dq + r) (0 r < d)]

Proving definition of equivalence in a mod using the quotient remainder theorem

This means prove that if [m d

n], then [d | (n-m)]where m,n Z and d Z+

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Proof by division into cases again

(n Z) [3 | n n2 3 1]or, alternatively, (n Z) [3 | n n2 1 (mod 3)]

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Steps toward proving the unique factorization theorem

Every integer greater than or equal to 2 has at least one prime that divides it

For all integers greater than 1, if a | b, then a | (b+1)

There are an infinite number of primes

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Discrete StructuresCMSC 250Lecture 20

March 12, 2008

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Using the unique factorization theorem

Prove that (a Z+)(q Zprime) [q | a2 q | a]

Prove that Q3

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Discrete StructuresCMSC 250Lecture 21

March 14, 2008

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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

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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