The Fundamental Theorem of Arithmetic

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The Fundamental Theorem of Arithmetic
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Primes

• p gt 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
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Primes
There is no other factoring!
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325BC to 265BC
Euclid of Alexandria
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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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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 gt 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 lt p
  lt n
• let p1 be the smallest of these divisors
• p1 must be prime otherwise
• there is an integer k, 1 lt k lt p1, and k divides
  p1
• consequently n n1 times p1 (i.e. n1 n ?
  p1)
• where p1 is prime and n1 lt n
• repeat the argument with n1
• If n1 is prime we are done
• otherwise n1 n2 times p2
• where p2 is prime and n2 lt n1 and p2 ? p1
• this process terminates due to the WOP

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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 gt 2) ever divisible by n-1?
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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
(p 211 6th ed, p 155 5th ed)
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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
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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
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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
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Is 51 prime?
Consider prime divisors 2, 3, 5, 7 only
?
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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
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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 lt rootN do begin
if pn then begin print(p) //
p is a prime divisor n n/p
rootN floor(sqrt(n)) end
else p nextPrime(p) end print(p)
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Primes
p 2 // the 1st prime rootN floor(sqrt(N))
// where we stop while p lt rootN do begin
if pn then begin print(p) //
p is a prime divisor n n/p
rootN floor(sqrt(n)) end
else p nextPrime(p) end print(p)
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
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Greatest common divisor gcd(a,b) and Least common
multiple

• gcd(a,b) is largest d such that da and db
• if gcd(a,b) 1 then a and b are relative prime
• lcm(a,b) is the smallest/least x such that ax
  and bx

• 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 xa and xb
• stop when the first value of x is found
• compute the prime factorisation of a and of b
• and then see below

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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

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

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