1. The Integers & Division

2. a divides b if a is not zero there is a m such that a.m = b “a is a factor of b” “b is a multiple of a” a|b Division

3. If a|b and a|c then a|(b+c) “ If a divides b and a divides c then a divides b plus c ” a|b  a.x = b a|c  a.y = c b+c = a.x + a.y = a(x + y) and that is divisible by a Division

4. a|b  a.m = b b.c = a.m.c which is divisible by a Division

5. a|b  a.x = b b|c  b.y = c c = a.x.y and that is divisible by a Division

6. Theorem 1 (page 202, 6 th ed, page 154, 5 th ed)

7. Primes p > 1 is prime if the only positive factors are 1 and p if p is not prime it is composite The Fundamental Theorem of Arithmetic Every positive integer can be expressed as a unique product of primes My name is Euclid

8. Primes There is no other factoring!

9. Euclid of Alexandria 325BC to 265BC

10. Primes Euclid’s words “if a number be the least that is measured by prime numbers, it will not be measured by any other prime except those originally measuring it “ Where “measuring” is “dividing” The Elements

12. Proof of Fundamental Theorem of Arithmetic Well Ordering Principle (WOP) every non-empty set of positive integers has a least element RTP: Every integer n > 1 can be written as a product of primes If n is prime we are done n is composite and has a positive divisor 1 < p < n let p 1 be the smallest of these divisors p 1 must be prime otherwise there is an integer k, 1 < k < p 1, and k divides p 1 consequently n = n 1 times p 1 (i.e. n 1 = n  p 1 ) where p 1 is prime and n 1 < n repeat the argument with n 1 If n 1 is prime we are done otherwise n 1 = n 2 times p 2 where p 2 is prime and n 2 < n 1 and p 2  p 1 … this process terminates due to the WOP

13. PRIMES The dumb way to test if n is prime if n is divisible by 2 return(“composite”) if n is divisible by 3 return(“composite”) if n is divisible by 4 return(“composite”) … if n is divisible by n-1 return(“composite”) return(“prime”) Question: is n (n > 2) ever divisible by n-1?

14. PRIMES Therefore, the divisor a or b is either prime or due to the fundamental theorem of arithmetic, can be expressed as a product of primes Put another way (p 211 6 th ed, p 155 5 th ed)

15. PRIMES We now have a test for primality If a number is not composite it is prime If a number is prime then it does NOT have a prime divisor less than or equal to  n Therefore we can test if n is divisible by primes in the range 2 to  n If none are found n must be prime

16. Primes Prove that 41 is prime To be prime, 41 must not be composite If composite 41 has a divisor less than or equal to square root of 41 The only primes not exceeding 6 are 2, 3, and 5 None of these divides 41 Therefore 41 is not composite, it is prime Remember: floor(x)  x  the largest integer smaller than x

17. Primes for the class! Prove that 67 is prime To be prime, 67 must not be composite If composite 67 has a divisor less than or equal to square root of 67 The only primes not exceeding 8 are 2, 3, 5, and 7 None of these divides 67 Therefore 67 is not composite, it is prime

18. Is 51 prime? Consider prime divisors 2, 3, 5, 7 only 

19. PrimesCompute the prime factorisation of n The Fundamental Theorem of Arithmetic Every positive integer can be expressed as a unique product of primes Revisited

20. PrimesCompute the prime factorisation of n assume nextPrime(i) delivers next prime number greater than i nextPrime(7) = 11 and nextPrime(nextPrime(7)) = 13 floor(sqrt(n)) delivers largest integer  square root of n floor(sqrt(97)) = 9 p := 2; // the 1st prime rootN := floor(sqrt(N)) // where we stop while p <= rootN do begin if p|n then begin print(p); // p is a prime divisor n := n/p; rootN := floor(sqrt(n)); end else p := nextPrime(p); end print(p);

21. Primes N p rootN 7007 2 83 7007 3 83 7007 5 83 7007 7 83 print 7 1001 7 31 print 7 143 7 11 143 11 11 print 11 13 11 3 print 13 7007 = 7.7.11.13 p := 2; // the 1st prime rootN := floor(sqrt(N)) // where we stop while p <= rootN do begin if p|n then begin print(p); // p is a prime divisor n := n/p; rootN := floor(sqrt(n)); end else p := nextPrime(p); end print(p);

22. The Division Algorithm (aint no algorithm) a is an integer and d is a positive integer there exists unique integers q and r, 0  r  d a = d.q. + r a divided by d = q remainder r dividend divisor quotient remainder NOTE: remainder r is positive and divisor d is positive

23. Division a = d.q + r and 0 <= r < d a = -11 and d = 3 and 0 <= r < 3 -11 = 3q + r q = -4 and r = 1 a = d.q + r and 0 <= r < d a = -63 and d = 20 and 0 <= r <= 20 -63 = 20q + r q = -4 and r = 17 a = d.q + r and 0 <= r < d a = -25 and d = 15 and 0 <= r < 15 -25 = 15.q + r q = -2 and r = 10

24. Division a = d.q + r and 0 <= r < d a = -11 and d = 3 and 0 <= r < 3 -11 = 3q + r q = -4 and r = 1 Troubled by this? Did you expect q = -3 and r = -2? What if 3 of you went to a café and got a bill for £11? Would you each put £3 down and then leg it? Or £4 each and leave £1 tip?

25. Greatest common divisor gcd(a,b) and Least common multiple gcd(a,b) is largest d such that d|a and d|b if gcd(a,b) = 1 then a and b are relative prime lcm(a,b) is the smallest/least x such that a|x and b|x 3 Naïve algorithms for gcd(a,b) start with x at 1 up to min(a,b) testing if x | a and x |b remember the last (largest) successful value start with x at min(a,b) and count down to 1 testing if x|a and x|b stop when the first value of x is found compute the prime factorisation of a and of b and then see below

26. Greatest common divisor gcd(a,b) gcd(120,500) prime factorisation of 120 is 2.2.2.3.5 prime factorisation of 500 is 2.2.5.5.5 20 … but there is a better algorithm (wots an algorithm?)

28. Lowest/least common multiple lcm(a,b) = smallest x divisible by a and by b lcm(95256,432) prime factorisation of 95256 is 2.2.2.3.3.3.3.3.7.7 prime factorisation of 432 is 2.2.2.2.3.3.3 190512

29. mod arithmetic a mod m is the remainder of a divided by m a mod m is the integer r such that a = qm + r and 0 <= r < m again, r is positive Examples 17 mod 3 = 2 17 mod 12 = 5 (5 o’clock) -17 mod 3 = 1

30. mod arithmetic a is congruent to b modulo m if m divides a - b congruences

31. mod arithmetica is congruent to b modulo m if m divides a - b

32. mod arithmetica is congruent to b modulo m if m divides a - b

33. mod arithmetica is congruent to b modulo m if m divides a - b

34. mod arithmetica is congruent to b modulo m if m divides a - b

35. Mod arithmetic -133 mod 9 = 2 (but in Claire?) list 5 numbers that are congruent to 4 modulo 12 hash function h(k) = k mod 101 h(104578690) h(432222187) h(372201919) h(501338753) examples

36. What have we done? (summary) divisibility a | b FTA proof of FTA test for primality computation of prime factorisation gcd and lcm Division algorithm aint no algorithm Mod arithmetic congruences

37. Division and multiplication That’s all for now folks