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Module #9 – Number Theory
3. Algorithms, The Integers
and Matrices
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Module #9 – Number Theory
Section 3.4: The Integers and
Division
Division:
Let a, b  Z with a  0.
• a |b  “a divides b”.
We say that “a is a factor of b”, “a is a divisor of
b”, and “b is a multiple of a”.
• a does not divide b is denoted by a | b.
• We can express a | b using the quantifier
 c (b = ac), Domain = Z.
3  12  True, but 3  7  False.
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Module #9 – Number Theory
The Divides Relations
• For every a, b, c  Z, we have
1. If a | b and a | c , then a | (b + c).
2. If a | b , then a | bc .
3. If a | b and b | c , then a | c .
Examples:
 3  12 and 3  9  3  (12 + 9)  3  21 (21 ÷ 3 = 7)
 2  6  2  (6 × 3)  2  18 (18 ÷ 2 = 9)
 4  8 and 8  64  4  64 (64 ÷ 4 = 16)
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Module #9 – Number Theory
The Division Algorithm
• let a be an integer and d a positive integer, then
there exist unique integers q and r such that:
a = d ×q + r ,0r<d.
• d is called divisor and a is called dividend.
• q is the quotient and r is the remainder (must be
positive integer).
q = a div d , r = a mod d .
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Module #9 – Number Theory
Examples
• What are the quotient and remainder when 101 is
divided by 11?
101 = 11 × 9 + 2,
q = 101 div 11 = 9, r = 101 – 11 × 9 = 2 = 101 mod 9
• What are the quotient and the remainder when −11
is divided by 3?
−11 = 3 × (−4) + 1 , q = −4 , r = 1
  11
q  11 div 3  
  3.6  4

 3 
r  11  (4)  3  1  11 mod 3
• Note:
− 11  3 × (− 3) − 2 because r can't be negative.
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Module #9 – Number Theory
Examples
• Find a and b if :
2a + b = 46 mod 7 and a + 2b = 47 div 9.
46 = 6 × 7 + 4 and 47 = 5 × 9 + 2
46 mod 7 = 4 and 47 div 9 = 5
2a + b = 4 (1)
a + 2b = 5 (2)
(1) – 2×(2) → – 3b = – 6
b = 2 and a = 1.
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Module #9 – Number Theory
Modular Congruence
Theorem: Let a, b  Z, m  Z+. We say that a is
congruent to b modulo m written
a  b (mod m),
if and only if
1) a mod m = b mod m , or
2) m | (a  b ) i.e. (a  b) mod m = 0.
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Module #9 – Number Theory
Examples
Is 17 congruent to 5 modulo 6?
17 mod 6 = 5 and 5 mod 6 = 5,
also 6 | (17 − 5)  6 | 12 where 12 ÷ 6 = 2,
then 17 ≡ 5 (mod 6)
Is 24 congruent to 14 modulo 6?
Since, 24 mod 6 = 0 and 14 mod 6 = 2,
also 6 | (24 − 14)  6 | 10,
then, 24 ≡ 14 (mod 6).
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Module #9 – Number Theory
Useful Congruence Theorems
Let a, b, c, d  Z, m  Z+.
If a  b (mod m) and c  d (mod m), then
1) a + c  b + d (mod m), and
2) ac  bd (mod m)
e.g. Let 7 ≡ 2 (mod 5) and 11 ≡ 1 (mod 5) then:
(7 + 11) ≡ (2 + 1) (mod 5)  18 ≡ 3 (mod 5) and
(7 × 11) ≡ (2 × 1) (mod 5)  77 ≡ 2 (mod 5).
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Module #9 – Number Theory
Cryptology
• Caesar’s encryption method can be represented by the
function f (p) = (p + 3) mod 26, 0 ≤ p ≤ 25. The letter “A”
is replaced by 0, “B” by 1, …, and “Z” by 25.
e.g. What is the secret message produce from the message “GOOD
MORNING ZERO”?
First replace the letters in the message with numbers
p : 6 14 14 3 12 14 17 13 8 13 6 25 4 17 14
Second replace p by f (p)
f (p) : 9 17 17 6 15 17 20 16 11 16 9 2 7 20 17
Translate this back to letters produces the encrypted message
“JRRG PRUQLQJ CHUR ”
• To recover the original message from a secrete message.
The inverse function f -1(p) = (p – 3) mod 26 can be used.
This process is called decryption.
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Module #9 – Number Theory
Hashing Functions
Suppose that a computer has only the 20 memory locations 0,
1, 2, …, 19. Use the hashing function h where
h(x) = (x + 5) mod 20
to determine the memory locations in which 57, 32, and 98
are stored.
•
•
•
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h(57) = (57 +5) mod 20 = 62 mod 20 = 2
h(32) = 37 mod 20 = 17
h(98) = 103 mod 20 = 3
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Module #9 – Number Theory
Section 3.5: Primes and Greatest
Common Divisors
• A positive integer p > 1 is prime if the only
positive factors of p are 1 and p.
Some primes: 2, 3, 5, 7, 11, 13, 17, ...
• Non-prime integer greater than 1 are called
composite, because they can be composed by
multiplying two integers greater than 1.
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Module #9 – Number Theory
Fundamental Theorem of Arithmetic
• Every positive integer greater than 1 has a unique
representation as a prime or as the product of two
or more primes where the prime factors are written
in order of non-decreasing size. (tree or division)
100 = 2·2·5·5 = 2252
13 = 13
1024 = 2·2·2·2·2·2·2·2·2·2 = 210
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Module #9 – Number Theory
Theorem
• If n is a composite integer, then n has prime
divisor ≤ n .
e.g. 49  prime numbers less than 49 are 2, 3, 5, 7
16  prime numbers less than 16 are 2, 3.
• An integer n is prime if it is not divisible by any
prime ≤ n .
e.g. 13 where 13 = 3.6 so the prime numbers are 2, 3
but non of them divides 13 so 13 is prime.
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Module #9 – Number Theory
Prime Factorization Technique
• To find the prime factor of an integer n :
1. Find n .
2. List all primes ≤ n ,
2, 3, 5, 7, … , up to n .
3. Find all prime factors that divides n.
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Module #9 – Number Theory
Example 1
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Module #9 – Number Theory
Example 2
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Module #9 – Number Theory
Example 3
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Module #9 – Number Theory
Greatest Common Divisors
• The greatest common divisor gcd(a, b) of integers
a, b (not both 0) is the largest (most positive) integer
d that is a divisor both of a and of b.
• To find gcd: 1. Find all positive common divisors
of both a and b, then take the largest divisor:
e.g Find gcd (24, 36)?
Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Common divisors: 1, 2, 3, 4, 6, 12
Maximum = 12, so gcd (24, 36) = 12
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Module #9 – Number Theory
Ways to Find GCD
2. Use prime factorization:
Take the minimum power of common factors
Example: Find gcd(24, 180)
24 = 2×2×2×3 = 23×3
180 = 2×2×3×3 = 22×32×5
gcd(24, 36) = 2min(3,2)×3min(1,2)×5min(0,1) = 22×31 = 12
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Module #9 – Number Theory
Examples
• Find gcd (360, 3500)?
360 = 23 × 32 × 5
3500 = 22 × 53 × 7
 gcd (360, 3500) = 22 × 5 = 20
• Find gcd(17, 22)?
No common divisors so gcd(17, 22) = 1 so, the
numbers 17 and 22 are called relatively prime.
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Module #9 – Number Theory
Least Common Multiple
• lcm(a, b) of positive integers, a and b, is the
smallest positive integer that is a multiple both of
a and of b.
• Find lcm(6, 10)?
Take the maximum power of all factors
6 = 2 × 3 , 10 = 2 × 5
lcm(6, 10) = 2max(1,1)×3max(1,0)×5max(0,1) = 2×3×5 = 30
• Find lcm (24, 180)?
24 = 23 × 31 , 180 = 22 × 32 × 5
 lcm (24, 36) = 23 × 32 × 5 = 360.
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Module #9 – Number Theory
Section 3.6: Integers and Algorithms
Introduction:
The term algorithm originally referred to
procedures for performing arithmetic operations
using the decimal representations of integers.
These algorithms, adapted for use with binary
representations, are the basis for computer
arithmetic.
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Module #9 – Number Theory
Representations of Integers
In everyday life we use decimal notation to express
integers.
For example, 965 is used to denote
9·102 + 6·10 + 5 .
However, it is often convenient to use bases other
than 10.
Module #9 – Number Theory
Representations of Integers
Computers usually use binary notation (with 2 as
the base) when carrying out arithmetic, and octal
(base 8) or hexadecimal (base 16) notation when
expressing characters, such as letters or digits. In
fact, we can use any positive integer greater than 1
as the base when expressing integers.
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Module #9 – Number Theory
Binary Expansions (Binary to Decimal)
What is the decimal expansion of the integer that has
(101011111)2 as its binary expansion?
Solution: We have
(l 01011111)2 = 1·28 + 0·27 + 1·26 + 0·25 + 1·24
+ 1·23 + 1· 22 + 1·21 + 1·20
= 256 + 64 + 16 + 8 + 4 + 2 + 1
= 351.
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Module #9 – Number Theory
Hexadecimal Expansions
Decimal system
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
0
1
2
3
4
5
6
7
8
9
A
Hexadecimal system
B
C
D
E
F
Module #9 – Number Theory
Any to Decimal
• What is the decimal expansion of the hexadecimal
expansion of (2AE0B)16 ?
Solution: We have
(2AE0B)16 = 2·164 + 10·163 + 14·162 + 0·16 + 11
= (175627)10 .
• (3016)7 = 3·73 + 0·72 + 1·7 + 6 = (356)10 .
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Module #9 – Number Theory
Base Conversion (Decimal to Any)
Find the binary expansion of (241)10 .
Solution: First divide 241 by 2 to obtain
241 = 2 . 120 + 1 .
Successively dividing quotients by 2 gives
120 = 2 · 60 + 0
60 = 2 · 30 + 0
30 = 2 · 15 + 0
15 = 2 · 7 + 1
7=2·3+1
3=2·1+1
1=2·0+1
Therefore, (241)10 = (11110001)2
241  2
1 120
0 60
0 30
0 15
1
7
1
3
1 1
1
0
 
mod div
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Module #9 – Number Theory
Decimal to Any
Find the base 8 expansion of (12345)10 .
Solution:
First, divide l2345 by 8 to obtain
12345 = 8 · 1543 + 1.
Successively dividing quotients by 8 gives:
12345
1543 = 8 · 192 + 7,
1 1543
192 = 8 · 24 + 0,
7 192
24 = 8 · 3 + 0,
0 24
3 = 8 · 0 + 3.
0 3
Therefore, (12345)10 = (30071)8.
3 0
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Module #9 – Number Theory
Example
Find the hexadecimal expansion of (177130)10.
Solution:
177130 = 16 · 11070 + 10
177130
11070 = 16 · 691 + 14 ,
10 11070
14 691
691 = 16 · 43 + 3 ,
3 43
43 = 16 · 2 + 11 ,
11 2
2 = 16 · 0 + 2.
2 0
Therefore, (177130)10 = (2B3EA)16
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Module #9 – Number Theory
Binary  Octal  Hexadecimal
Hexadecima l Binary
Octal Binary
0
000
0
0000
1
0001
2
0010
3
0011
4
0100
1
001
2
010
5
0101
3
011
6
0110
4
100
7
0111
5
101
8
1000
9
1001
6
110
A
1010
7
111
B
1011
C
1100
D
1101
E
1110
F
1111
 (307)8  (011 000 111) 2
(5B7)16  (0101 1011 0111) 2
 (10100111101) 2
 (0101, 0011,1101) 2
 (53D)16
 (607)8  (110,000,111) 2
 (0001,1000,0111) 2  (1A7)16
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Module #9 – Number Theory
Section 3.8: Matrices
• An m×n matrix is a rectangular array of mn
objects (usually numbers) arranged in m horizontal
rows and n vertical columns.
• An nn matrix is called a square matrix, whose
order is n.
2 3 
5  1


7 0 
32
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2 1
3 1

 22
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Module #9 – Number Theory
Matrix Equality
• Two matrices A and B are equal if and only if they
have the same number of rows, the same number
of columns, and all corresponding elements are
equal.
 3 2  3 2 0
  1 6     1 6 0

 

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Module #9 – Number Theory
Row and Column Order
• The rows in a matrix are usually indexed 1 to m
from top to bottom. The columns are usually
indexed 1 to n from left to right. Elements are
indexed by row, then column.
 a11
a
21

A  [aij ] 
 

am1
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a12
a22

am 2
 a1n 
 a2 n 
  

 amn 
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Module #9 – Number Theory
Matrix Sums
• The sum A + B of two matrices A, B (which must
have the same number of rows, and the same
number of columns) is the matrix given by adding
corresponding elements.
• A + B = [aij + bij]
6  7  5 9
1
2
0  8  4  1  4  9

 
 

1
2 3
6  4
8 
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Module #9 – Number Theory
Matrix Products
For an mk matrix A and a kn matrix B, the
product AB is the mn matrix:
k

AB  C  [cij ]   aipbpj 
 p 1

i.e. The element (i, j ) of AB is given by the vector
dot product of the ith row of A and the jth column
of B (considered as vectors).
Note: Matrix multiplication is not commutative!
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Module #9 – Number Theory
Matrix Product Example
0  1 1
0 1  1 
2 0 3  2 0  2

 1 0 3

2×3
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3×4
0
0  5  1
1

0  

3  2 11 3

1
2×4
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Module #9 – Number Theory
Identity Matrices
• The identity matrix of order n, In , is the order-n
matrix with 1’s along the upper-left to lower-right
diagonal and 0’s everywhere else.
• A In = A
1 0  0


1 if i  j  0 1  0
I n  

0 if i  j       


0 0  1 
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Module #9 – Number Theory
Powers of Matrices
• If A is an nn square matrix and p  0, then:
Ap  AAA ··· A (A0  In)
p times
3
 2 1  2 1  2 1  2 1
  1 0    1 0   1 0   1 0

 



2  4
3
 2 1  3







1
0

2

1

3

2


 

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Module #9 – Number Theory
Matrix Transposition
• If A is an mn matrix, then the transpose of A is
the nm matrix AT given by interchanging the
rows and the columns of A.
0
2
3
2 1


T
A
 A  1  1 

0  1  2
3  2
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Module #9 – Number Theory
Symmetric Matrices
• A square matrix A is symmetric if and only if AT
= A.
• Which is symmetric?
A
1 1
1 1


1 1
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B
1
3
 2
 1

0 1


 3  1 2 
C
1
3 0
0 2  1 


1 1  2
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Module #9 – Number Theory
Zero-One Matrices
All elements of a zero-one matrix are 0 or 1,
representing False & True respectively.
• The join of A and B (both mn zero-one matrices)
is
A  B : [aij  bij].
• The meet of A and B is:
A  B  [aij  bij].
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Module #9 – Number Theory
Example
1 0 1
A= 

0 1 0 
0 1 0 
B = 

1 1 0 
1 1 1 
• The join between A and B is A  B = 

1
1
0


0 0 0
• The meet between A and B is A  B = 0 1 0 


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Module #9 – Number Theory
Boolean Products
• Let A be an m k zero-one matrix and B be a
k n zero-one matrix,
• The Boolean product A⊙B of A and B is like
normal matrix product, but using  instead +
and using  instead
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.
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Module #9 – Number Theory
Boolean Powers
• For a square zero-one matrix A, and any k  0, the
kth Boolean power of A is simply the Boolean
product of k copies of A.
A[k]  A⊙A⊙…⊙A
k times
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Module #9 – Number Theory
Example
Let A =
1
0

1
0
1
0
and B =
(1  1)  (0  0)
A⊙B = (0  1)  (1  0)

(1  1)  (0  0)
1 1 0
0 1 1


(1  1)  (0  1) (1  0)  (0  1)
(0  1)  (1  1) (0  0)  (1  1)
(1  1)  (0  1) (1  0)  (0  1)




1 1 0
 0 1 1
1 1 0
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