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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.

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Presentation on theme: "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."— Presentation transcript:

1 The Fundamental Theorem of Arithmetic

2 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

3 Primes There is no other factoring!

4 Euclid of Alexandria 325BC to 265BC

5 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

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7 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

8 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?

9 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 th ed, p th ed)

10 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

11 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

12 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

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

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

15 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);

16 Primes N p rootN print print print print = 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);

17 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

18 Greatest common divisor gcd(a,b) gcd(120,500) prime factorisation of 120 is prime factorisation of 500 is … but there is a better algorithm (wots an algorithm?)

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20 Lowest/least common multiple lcm(a,b) = smallest x divisible by a and by b lcm(95256,432) prime factorisation of is prime factorisation of 432 is


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