AF2

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AF2
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Turn off your phones
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Primes, gcd, some examples, reading
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Primes

• pgt1 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
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• How many prime numbers are there?
• Is there a largest prime number?

• assume we know all the prime numbers
• let p be the largest prime
• multiply all the primes together
• add 1
• call this n
• n is not divisible by 2, we get remainder 1
• n is not divisible by 3, we get remainder 1
• n is not divisible by 5, we get remainder 1
• n is not divisible by p, we get remainder 1
• But, by the FTA we know n has a prime
  factorisation
• Therefore n must have a prime divisor gt p
• Therefore there is no greatest prime

Due to Euclid
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• Are there an infinite number of primes?

• Reformulate this as
• For any n, is there a prime greater than n?
• compute a new number x n! 1
• i.e. x 1.2.3.4.5.6n-1.n 1
• x is not divisible by any number in the range 2
  to n
• we always get remainder 1
• the FTA states x is a product of primes
• x has a prime divisor
• xs smallest prime divisor is greater than n
• Consequently for any n there is a prime greater
  than n

Rosen, page 180, example 24
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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
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How big a prime do we want?
Page 148, for RSA encryption The most efficient
factorisation methods (1999) require billions of
years to factor 400 digit integers. Consequently
when 200 digit primes messages encrypted
cannot be found in a reasonable time. When
new factorisation techniques are found it will
be necessary to use larger primes to ensure
secrecy/security of messages.
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Division

• a divides b if
• a is not zero
• there is a c such that a.c b
• a is a factor of b
• b is a multiple of a
• ab

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Division

• If ab and ac then a(bc)
• If a divides b and a divides c then a divides b
  plus c

• ab ? a.x b
• ac ? a.y c
• bc a.x a.y
• a(x y)
• and that is divisible by a

Note also if a b and ac then a(b-c)
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Division

• ab ? a.x b
• b.c a.x.c
• which is divisible by a

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Division

• ab ? a.x b
• bc ? b.y c
• c a.x.y
• and that is divisible by a

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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 b q remainder r
NOTE r is positive and d is positive
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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
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gcd, a geometric view

• We have a floor to tile
• The floor is 57 units wide and 152 units long
• What is the largest square tile we can use
• to tile the floor
• this will be the least number of square tiles!

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gcd, a geometric view

• Find gcd(57,152)
• 19 is the answer
• we need 24 tiles

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• Theorem
• when a b.q r
• greatest common divisor of a and b
• is also
• greatest common divisor of b and r
• i.e. gcd(a,b) gcd(b,r)

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A proof follows The proof is based on what we
already know about division
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RTP gcd(a,b) gcd(b,r) where a b.q r
Sketch of the proof Show that common divisors of
a and b are also common divisors of b and r
(1) show that if some da and db then dr (2)
show that if some db and dr then da (3)
conclude that a common divisor of a and b is
also a common divisor of b and r (4)
consequently gcd(a,b) gcd(b,r)
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RTP gcd(a,b) gcd(b,r) where a b.q r
(1) show that if some da and db then dr

• a b.q r
• ?a - b.q r
• We have already proved that
• if db then db.q
• if da and db.q then d(a - b.q)
• since da and db.q it follows that d(a - b.q)
• since (a - b.q) r it follows that dr
• Consequently a common divisor of a and b also
• divides r

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RTP gcd(a,b) gcd(b,r) where a b.q r
(2) show that if some db and dr then da

• a b.q r
• We have already proved that
• if db then db.q
• if db.q and dr then d(b.q r)
• since db.q and dr it follows that d(b.q r)
• since (b.q r) a it follows that da
• Consequently a common divisor of b and r also
• divides a

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RTP gcd(a,b) gcd(b,r) where a b.q r
(3) conclude that a common divisor of a and b
is also a common divisor of b and r

• From (1) and (2) we have proved that
• a common divisor of a and b is also a divisor of
  r
• a common divisor of b and r is also a divisor of
  a
• consequently a common divisor of a and b is also
• a common divisor of b and r

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RTP gcd(a,b) gcd(b,r) where a b.q r
(4) consequently gcd(a,b) gcd(b,r)
From (3) we can conclude that the greatest common
divisor of a and b is also a greatest common
divisor of b and r
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• This suggests an algorithm
• given a and b
• if a 0 then b is the gcd
• if b 0 then a is the gcd
• otherwise
• a ?b q remainder r
• we have established that we can substitute
• gcd(a,b) with gcd(b,r)
• now compute gcd(b,r)
• note, r is decreasing!

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gcd

• We have shown that when a b.q r
• any divisor of a and b is also a divisor of b
  and r
• consequently gcd(a,b) gcd(b,r)
• We state that gcd(0,b) b and gcd(a,0) a
• therefore, there is nothing to do if a or b is
  zero
• Due to our theorem
• we can reduce gcd(a,b) to a simpler problem
  gcd(b,r)
• at each iteration we replace
• b with r where 0 ? r ? b
• at each iteration b decreases
• eventually this must terminate with b 0
• and as we said, we are then done
• and a is the answer!

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gcd(a,b)

• gcd(a,b)
• if b 0
• then a
• else gcd(b,a mod b)

• (0) gcd(a,b)
• (1) if b 0 goto 6
• (2) let r a mod b
• (3) a b
• (4) b r
• (5) goto (1)
• (6) return(a)

• gcd(a,b)
• while b ? 0
• do begin
• r a mod b
• a b
• b r
• end
• a

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(No Transcript)
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gcd(ainteger,binteger) integer -gt if (b
0) a else gcd(b,a mod b)
relativePrime(ainteger,binteger) boolean -gt
gcd(a,b) 1

• Try
• gcd(414,662) and then gcd(662,414)
• gcd(120,500) and gcd(500,120)
• list all numbers relative prime to 22

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Numbers

• A number to the base b
• it has k1 terms
• the as are all in the range 0 to b-1

• what is
• 10101101 base 2 in base 10
• peter base 26 in base 10

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Numbers

• Can you do arithmetic in different bases?
• I expect so (see pages 169-177)
• Can you do addition, multiplication,
  subtraction, division
• in different bases? (I expect so)
• Do you know the algorithm for this?
• Again, I expect so.

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Reading

• Please read the following
• Cryptology, pages 165-166
• Representation of Integers, pages 169-177
• and the Footnote on page 177