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

k

i

eiipn

1

My name is Euclid

Primes

5.3.25.3.2.260 2

There is no other factoring!

22 5.25.5.2.2100

Euclid of Alexandria

325BC to 265BC

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

a.bn n)composite(

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

n divisor prime a has n composite is n If

a.bn n)composite(

nb na

n divisor prime a hasn then composite isn if

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

) nb n(a

Put another way

(p 211 6th ed, p 155 5th ed)

PRIMES

n divisor prime a hasn then composite isn if

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 divisorless than or equal to n

Therefore we can test if n is divisible by primes in the range2 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

6 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

8 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?

751

Consider prime divisors 2, 3, 5, 7 only

17351

Primes Compute the prime factorisation of n

The Fundamental Theorem of Arithmetic

Every positive integer can be expressed as a unique product of primes

k

i

eiipn

1

Revisited

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 primerootN := floor(sqrt(N)) // where we stopwhile p <= rootNdo 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 83print 7 1001 7 31print 7 143 7 11 143 11 11print 11 13 11 3print 13

7007 = 7.7.11.13

p := 2; // the 1st primerootN := floor(sqrt(N)) // where we stopwhile p <= rootNdo 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

),min(),min(2

),min(1 ...),gcd( 2211 nn ba

nbaba pppba

Greatest common divisor gcd(a,b)

),min(),min(2

),min(1 ...),gcd( 2211 nn ba

nbaba pppba

• gcd(120,500)• prime factorisation of 120 is 2.2.2.3.5• prime factorisation of 500 is 2.2.5.5.5

)3,1min()0,1min()2,3min( 532)500,120gcd(

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

),max(),max(2

),max(1 ...),( 2211 nn ba

nbaba pppbalcm

• 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

)0,2max()0,0max()3,5max()4,3max( 7532)432,95256( lcm

190512