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:" 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

3 CMSC 250 Some definitions l Z is the integers l Q is the rational numbers (quotients of integers) r  Q  (  a,b  Z) [(r = a / b)  (b  0)] l Irrational numbers are those which are not rational l R is the real numbers l A superscript of + indicates the positive portion only of one of these sets of numbers l A superscript of – indicates the negative portion only of one of these sets of numbers l Other superscripts can be used, such as Z even, Z odd, Q >5 l We can define the closure of these sets for an operation Z is closed under what operations?

4 CMSC 250 Integer definitions l An even integer n  Z even  (  k  Z) [n = 2k] l An odd integer n  Z odd  (  k  Z) [n = 2k + 1] l A prime integer (Z >1 ) n  Z prime  (  r,s  Z + ) [ (n = r  s)  (r = 1)  (s = 1)] l A composite integer (Z >1 ) n  Z composite  (  r,s  Z + ) [(n = r  s)  (r  1)  (s  1)]

5 CMSC 250 Constructive proof of existence l How can we prove the following?

6 CMSC 250 Discrete Structures CMSC 250 Lecture 13 February 25, 2008

7 CMSC 250 Methods of proving universally quantified statements l 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)]] l 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  Z even  r 2  Z even ]

8 CMSC 250 Examples of generalizing from the generic particular l The product of any two odd integers is also odd. –(  m,n  Z) [(m  Z odd  n  Z odd )  m  n  Z odd ] l The product of any two rationals is also rational. –(  m,n  Q) [m  n  Q]

9 CMSC 250 Discrete Structures CMSC 250 Lecture 14 February 27, 2008

10 CMSC 250 Disproof by counterexample l (  r  Z) [r 2  Z +  r  Z + ] –counterexample: r 2 = 9  r = –3 r 2  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 l 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

11 CMSC 250 Divisibility l 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 l Results involving divisibility: (  a, b  Z >1 )[a | b → a | (b + 1)] (  x  Z >1 )(  p  Z prime )[p | x] –note another proof method would be used here, proof by division into cases

12 CMSC 250 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]

13 CMSC 250 Discrete Structures CMSC 250 Lecture 15 February 29, 2008

14 CMSC 250 Proof by contradiction l The number of primes is infinite l Assume this is false, that there is some largest prime number, so there are only n different prime numbers l Consider the number

15 CMSC 250 Prime factored form l The Unique Factorization Theorem (Theorem 3.3.3) Given any integer n > 1, (  k  Z)(  p 1,p 2,…p k  Z prime )(  e 1,e 2,…e k  Z + )[n = p 1 e1 × p 2 e2 × p 3 e3 × …× p k ek ] where the p’s are distinct and any other expression of n is identical to this except maybe in the order of the factors. l Standard factored form n = p 1 e1 × p 2 e2 × p 3 e3 × … × p k ek p i < p i+1 l (  m  Z)[8×7×6×5×4×3×2×m = 17×16×15×14×13×12×11×10] –does 17 | m ??

16 CMSC 250 More integer definitions l div and mod operators n div d- integer quotient for n mod d- integer remainder for (n div d = q) ^ (n mod d = r)  n = d × q + r where n  Z  0, d  Z +, r  Z, q  Z, 0  r < d l Relating “mod” to “divides”: d | n  0 = n mod d  0  d n l 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

17 CMSC 250 Discrete Structures CMSC 250 Lecture 16 March 3, 2008

18 CMSC 250 Modular arithmetic (Theorem ) l 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.a m  c m (mod n) for all integers m l Proof using this definition: (  m  Z + )(  a,b  Z)[a  m b  (  k  Z) [a = b + km]]

19 CMSC 250 Discrete Structures CMSC 250 Lecture 18 March 7, 2008

20 CMSC 250 Floor and ceiling l 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 l Proofs using floor and ceiling: (  x,y  R) [  x+y  =  x  +  y  ] (  x  R)(  y  Z)[  x+y  =  x  + y ]

21 CMSC 250 More proof by division into cases The floor of (n/2) is either a) n/2 when n is even or b) (n–1)/2 when n is odd

22 CMSC 250 Discrete Structures CMSC 250 Lecture 19 March 10, 2008

23 CMSC 250 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 +

24 CMSC 250 Proof by division into cases again (  n  Z) [3 | n  n 2  3 1] or, alternatively, (  n  Z) [3 | n  n 2  1 (mod 3)]

25 CMSC 250 Steps toward proving the unique factorization theorem l Every integer greater than or equal to 2 has at least one prime that divides it l For all integers greater than 1, if a | b, then a | (b+1) l There are an infinite number of primes

26 CMSC 250 Discrete Structures CMSC 250 Lecture 20 March 12, 2008

27 CMSC 250 Using the unique factorization theorem l Prove that (  a  Z + )(  q  Z prime ) [q | a 2  q | a] l Prove that

28 CMSC 250 Discrete Structures CMSC 250 Lecture 21 March 14, 2008

29 CMSC 250 Summary of proof methods l Constructive proof of existence l Proof by exhaustion l Proof by generalizing from the generic particular l Proof by contraposition l Proof by contradiction l Proof by division into cases

30 CMSC 250 Errors in proofs l Arguing from example for universal proof l Misuse of variables l Jumping to the conclusion (missing steps) l Begging the question l Using "if" about something that is known