The Integers & Division

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

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

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

Every positive integer can be expressed as a unique 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

Primes There is no other factoring!

325BC to 265BC Euclid of Alexandria

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

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 p1 be the smallest of these divisors p1 must be prime otherwise there is an integer k, 1 < k < p1, and k divides p1 consequently n = n1 times p1 (i.e. n1 = n  p1) where p1 is prime and n1 < n repeat the argument with n1 If n1 is prime we are done otherwise n1 = n2 times p2 where p2 is prime and n2 < n1 and p2  p1 … this process terminates due to the WOP

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?

PRIMES Put another way 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 (p 211 6th ed, p 155 5th ed)

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

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

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

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

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

Primes Compute 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); print(p);

Primes 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); print(p); N p rootN 7007 2 83 7007 3 83 7007 5 83 7007 7 83 print 7 1001 7 31 143 7 11 143 11 11 print 11 13 11 3 print 13 7007 = 7.7.11.13

a divided by d = q remainder r 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 remainder quotient NOTE: remainder r is positive and divisor d is positive

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

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?

if gcd(a,b) = 1 then a and b are relative prime 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

20 … but there is a better algorithm (wots an algorithm?) 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?)

Lowest/least common multiple lcm(a,b) = smallest x divisible by a and by b 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

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

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

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

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

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

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

Mod arithmetic examples -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)

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

Division and multiplication That’s all for now folks