1. The Integers & Division

2. Division
a divides b if
a is not zero
there is a m such that a.m = b
“a is a factor of b”
“b is a multiple of a”
a|b

3. Division
If a|b and a|c then a|(b+c)
“If a divides b and a divides c then a divides b plus c”
a|b  a.x = b
a|c  a.y = c
b+c = a.x + a.y
= a(x + y)
and that is divisible by a

4. Division
a|b  a.m = b
b.c = a.m.c
which is divisible by a

5. Division
a|b  a.x = b
b|c  b.y = c
c = a.x.y
and that is divisible by a

6. Division
Theorem 1 (page 202, 6th ed, page 154, 5th ed)

7. Every positive integer can be expressed as a unique
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

8. Primes
There is no other factoring!

9. 325BC to 265BC
Euclid of Alexandria

10. 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

12. 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

13. 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?

14. PRIMES
Put another way
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)

15. 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

16. 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

17. 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

18. Is 51 prime?
Consider prime divisors 2, 3, 5, 7 only 

19. Every positive integer can be expressed as a unique
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

20. 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 <= 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);
print(p);

21. Primes
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);
print(p);
N p rootN
print 7
print 11
print 13
7007 =

22. a divided by d = q remainder r
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 d = q remainder r
dividend
divisor
remainder
quotient
NOTE: remainder r is positive and divisor d is positive

23. Division
a = d.q + r and 0 <= r < d
a = -11 and d = 3 and 0 <= r < 3
-11 = 3q + r
q = -4 and r = 1
a = d.q + r and 0 <= r < d
a = -63 and d = 20 and 0 <= r <= 20
-63 = 20q + r
q = -4 and r = 17
a = d.q + r and 0 <= r < d
a = -25 and d = 15 and 0 <= r < 15
-25 = 15.q + r
q = -2 and r = 10

24. Division
a = d.q + r and 0 <= r < d
a = -11 and d = 3 and 0 <= r < 3
-11 = 3q + r
q = -4 and r = 1
Troubled by this?
Did you expect q = -3 and r = -2?
What if 3 of you went to a café and got a bill for £11?
Would you each put £3 down and then leg it?
Or £4 each and leave £1 tip?

25. if gcd(a,b) = 1 then a and b are relative prime
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

26. 20 … but there is a better algorithm (wots an algorithm?)
Greatest common divisor gcd(a,b)
gcd(120,500)
prime factorisation of 120 is
prime factorisation of 500 is 20
… but there is a better algorithm (wots an algorithm?)

28. Lowest/least common multiple lcm(a,b) = smallest x divisible by a and by b
prime factorisation of is
prime factorisation of 432 is
190512

29. mod arithmetic
a mod m is the remainder of a divided by m
a mod m is the integer r such that
a = qm + r and 0 <= r < m
again, r is positive
Examples
17 mod 3 = 2
17 mod 12 = 5 (5 o’clock)
-17 mod 3 = 1

30. mod arithmetic
congruences
a is congruent to b modulo m if m divides a - b

31. mod arithmetic
a is congruent to b modulo m if m divides a - b

32. mod arithmetic
a is congruent to b modulo m if m divides a - b

33. mod arithmetic
a is congruent to b modulo m if m divides a - b

34. mod arithmetic
a is congruent to b modulo m if m divides a - b

35. Mod arithmetic
examples
-133 mod 9 = 2 (but in Claire?)
list 5 numbers that are congruent to 4 modulo 12
hash function h(k) = k mod 101
h( )
h( )
h( )
h( )

36. What have we done? (summary)
divisibility a | b
FTA
proof of FTA
test for primality
computation of prime factorisation
gcd and lcm
Division algorithm
aint no algorithm
Mod arithmetic
congruences

37. Division and multiplication That’s all for now folks