# The Fundamental Theorem of Arithmetic. Primes p > 1 is prime if the only positive factors are 1 and p if p is not prime it is composite The Fundamental.

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The Fundamental Theorem of Arithmetic

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!

Euclid of Alexandria 325BC to 265BC

Primes Euclids 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 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

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 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)

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

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

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);

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);

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

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 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

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