Uploaded by Fimah Sichone

1-CSC2901-NUMBER-THEORY

advertisement
Discrete Mathematics
Number Theory
Reading
• Barnier & Chan
– Chapter 6
Outline
•
•
•
•
•
•
Laws of elementary algebra
Factorisation in the Integers
Greatest common divisor
Modular Arithmetic and Binary Numbers
Applications: Cryptography
Note; s.t stands for ‘such that’
Introduction
• Digital computers store information and data
as computer words, which are composed of
strings of 0’s and 1’s
• As a computer scientist knowledge of the
arithmetic of binary numbers is essential
• We start by looking at decimal numbers since
they have a lot in common with binary
numbers
Elementary Algebra
• Number theory is a vast and fascinating field
of mathematics, consisting
– properties of whole numbers.
– Primes and prime factorization
– functions such as the divisor function
– modular arithmetic.
Elementary Algebra
• Definition
– A set 𝑨 has closure (or is closed) under an
operation ( ) if performance of that operation on
members of the set 𝑨 always produces a member
of the set 𝑨. I.e. if π‘Ž,𝑏 οƒŽ 𝐴 , then π‘Ž  𝑏 οƒŽ 𝐴
• The set of integers is closed under addition,
subtraction and multiplication but not under
division.
Laws of Elementary Algebra
Example
• Show that the following equations hold.
– 2 x (3 + 4) = 2x3 + 2x4
– 4 x(6 x 2) = (4 x 6) x 2
– Show that the left hand side of the equation equal
to the right hand side, following the laws of
associativity and precedence.
Example
lhs
= 2 x (3 + 4)
=2x7
= 14
=6+8
=2x3 + 2x4
= rhs
Laws of Elementary algebra
• Let x, y, c be members of a set A and
operation o defined on A. We define the
following
– o is Reflexive if a o a holds
– o is symmetric if when a o b then b o a
– o is transitive if when a o b and b o c then a o c
• E.g. Equality on integers is reflexive,
symmetric, and transitive.
Factorisation
• Recap of rule of Algebra over 
– For x, y οƒŽ 
• x.y = 0 iff x = 0 or y = 0
• Since there is no integer multiplicative inverse
for each integer, it is not always possible to
find an integer x s.t xd = a, implying a is not
always divisible by d
– E.g. 3x = 2, 5x = 11, etc.
Integer divisibility
• Definition
– Let a, d οƒŽ , with d ο‚Ή 0, the integer d divides a written
d|a if there exists an integer m s.t a = md
– The integer a is a multiple of d and d is a divisor of a
when d|a
– A divisor is a factor.
• Theorem
– Let a, d ο‚Ή 0 οƒŽ . Then the following is true:
• d|0
• 1|a
• d|d
Integer Divisibility
• For any integer d, d|d
• Proof
– d = d.1,
– since 1 is an integer, d|d
Factorisation in the Integers
• Theorem
– Let a,b,d and e be integers, with dο‚Ή0 and e ο‚Ή0.
The following are true
i.
ii.
iii.
iv.
If d|a and d|b, then d|(a+b)
If d|a, then d|ab
If d|e and e|b then d|b
If d|e and e|d then d = e or d = -e
– Proof for ii)
• d|a, if a = md, m οƒŽ  οƒž ab = (md)b οƒž (mb)d οƒž d|ab
since mb is an integer
Prime Numbers
• Definition
– A positive integer 𝑝 is prime if 𝑝 has exactly two
positive factors
– I.e. for all integers r and s, if n = r.s then either r =
1, or s = 1
• The numbers 2, 3, 5 , 7, 11, 13, .. are but just a
few of the members in the sequence of
primes
• Primes are very important numbers…
Prime Numbers
• Definition
– An integer n > 1 is composite if n = r.s for some
integers r and s with r ο‚Ή 1 and s ο‚Ή 1.
• Example
– 10 = 5.2
– 15 = 5.3
– 9 = 3.3 are all composite numbers
Prime Numbers
• Theorem
– Every integer 𝑛 > 2 has a prime factor
• Theorem
– Every number is either a prime or a product of primes.
• Definition
– Two integers a and b are said to be coprime or
relatively prime to each other if the two numbers
have no common factors other than 1
Division Algorithm for 
• Algorithm
– Let a and d be integers, with d ο‚Ή 0. There exist
unique integers q and r such that
– a = qd + r and 0 ο‚£ r ο€Ό |d|
• E.g.
– For a = 20 and d = 7, q = 2 and r = 6
– I.e. 20 = 2.7 + 6
• When r = 0, d|a otherwise we say that d does
not divide a
Division Algorithm for Z
• Ex:
– Let 𝑑 and π‘˜ be positive integers. Suppose π‘ž is the
quotient and r is the remainder when π‘Ž is divided
by 𝑑. Prove that π‘ž is the quotient and π‘˜π‘Ÿ is the
remainder when π‘˜π‘Ž is divided by π‘˜π‘‘.
The Greatest Common Divisor
• Definition
– Given two integers a and b, the greatest common
divisor of a and b denoted by 𝑔𝑐𝑑(π‘Ž,𝑏) is the
greatest number which divides π‘Ž and 𝑏.
– In other words, the gcd is the greatest factor of
both π‘Ž and 𝑏.
• For example, let π‘Ž = 12 and 𝑏 = 20 then gcd(20,12) = 4
• F(20) = {1,2,4,5,10,20}
• F(12) = {1,2,3,4,12} and 4 is the greatest in the
intersection
The Greatest Common Divisor
• If g is the greatest common divisor of a and b
then
– g|a and g|b
– If any other number dο‚Ήg is s.t d|a and d|b then
d|g as well
• If a > b then
– b = gcd(a,b) if b|a and
– Gcd(a,b) = gcd(b, a-b) if b does not divide a
Greatest Common Divisor
• Example
– Find gcd of
• a = 15, b = 7
• a = 8, b= 64
• a = 6, b = 9
• If a < b then the gcd(a, b) = gcd(b,a)
Greatest Common Divisor
• Theorem: Euclidean method
– Let a and d be integers with a = qd + r, then
• gcd(a,d) = gcd(d,r)
– We will not provide proof for this theorem
• This theorem provides a mechanism for
iteratively finding the gcd of a and d using
long division.
• The 𝑔𝑐𝑑 is the remainder just before the zero
remainder
Greatest Common Divisor
• Example
• gcd(9,3) = 3, since 3|9
• gcd(15, 9), we apply long division as follows
– 15 = 9.1 + 6
– 9 = 6.1 + 3
– 6 = 3.2 + 0
• The r just before 0 is the gcd i.e. 3
Greatest Common Divisor
• Theorem
– If g is the gcd of a and b, then there exists integers
m and n such that π‘Žπ‘š + 𝑏𝑛 = 𝑔
– We provide no truth
• The values m and n are obtained by reversing
the long division method of deriving the gcd
of a and b
Greatest Common Divisor
• Example:
– Find m and n for the given a and b s.t am + bn = g
– a = 15, b = 9
• Solution
– We just saw that
– 15 = 9.1 + 6…(i)
– 9 = 6.1 + 3…(ii)
– 6 = 3.2 + 0
Greatest Common Divisor
 15 = 9.1 + 6…(i)
 9 = 6.1 + 3…(ii)
 6 = 3.2 + 0
οƒž3 = 9 – 6.1, from (ii)
οƒž3 = 9 – (15 – 9.1).1 from (i)
οƒž3 = 9 -15.1 + 9.1
οƒž3 = 9.2 – 15.1
οƒž3 = -15.1 +9.2,
οƒžm = -1 and n = 2
Greatest Common Divisor
• Definition
– Two integers a and p, with p being prime number.
Then p is relatively prime to a if the gcd(a, b) = 1
• Example
– The following pairs of a and b are relatively prime
– a = 10, b = 7
– a = 10, b = 7
– a = 9, b = 8
Modular Arithmetic
• Definition
– Let a, b and n be integers with n > 0. Then a is said to
be congruent to b modulo n denoted by
aο‚Ί
b(mod n) if n leaves the same remainder when divided
by a and by b.
– I.e. a%n = b%n
– Equivalently the difference of a and b is divisible by n
or a – b = nk, for some integer k.
• E.g.
– 8 ο‚Ί 5 (mod 3)
Modular Arithmetic
• Mod operator
– The domain of a(mod n) is the set of integers {0, 1,
2, .., n-1} regardless of the value of a.
– 8(mod 3) = 2
– 8(mod 9) = 8
– 21(mod 5) = 0 etc
– 82(mod 9) = ??
– 1023(mod 2) = ??
Modular Arithmetic
• Theorem
– Let a, b, c, d and n be integers such that n >0, and
that a ο‚Ί b(mod n), c ο‚Ί d(mod n), then
i.
ii.
iii.
iv.
v.
a+k ο‚Ί b+k(mod n)
ak ο‚Ί bk(mod n)
ac ο‚Ί bd(mod n)
a2 ο‚Ί b2(mod n)
ap ο‚Ί bp(mod n) for any integer p = 1, 2, 3,….
• We show proof for (i) and (iv)
Modular Arithmetic
• Let a ο‚Ί b(mod n), show that ak ο‚Ί bk(mod n) for any
integer k
• Proof
Premise:
a ο‚Ί b(mod n),….(i)
b – a = nr, for an integer r…(ii)
bk – ak
=
=
=
=
(b – a)k
(nr)k, from (ii)
n(rk),
ns, where s = rk is an integer
οƒž akο‚Ίbk(mod n)
Modular Arithmetic
• Definition
– If a.b ο‚Ί 1(mod n) then b is said to be an inverse of
a modulo n.
– E.g. 3 is inverse of 2 mod 5 since
3.2 = 6ο‚Ί1(mod 5)
• Example
– Solve for x in
• x ο‚Ί 3(mod 7)
• 7x ο‚Ί 3(mod 5)
• 5ο‚Ίx(mod 7)
Modular Arithmetic
• Theorem
– π‘Žπ‘ π‘šπ‘œπ‘‘ 𝑛 = [(π‘Ž(π‘šπ‘œπ‘‘ 𝑛)(𝑏 (π‘šπ‘œπ‘‘ 𝑛)](π‘šπ‘œπ‘‘ 𝑛)
– This theorem is used to simplify expressions, involving
really large numbers, without the need of knowing
their exact values
• E.g.
55.26(mod 4)
=(55(mod 4).26(mod 4))(mod 4)
=(1.3)(mod 4)
= 3(mod 4)
=4
Modular Arithmetic
• Example
– Find 355(mod 8)
– Solution
• Note 32 ο‚Ί 1(mod 8)
• 355 = [32]27 .3
• 355(mod 8)
ο‚Ί 127.3(mod 8)
ο‚Ί 3(mod 8)
=3
Modular Arithmetic
• Example:
• What is the last digit of 355?
– Solution
• Find 355(mod 10)
Modular Arithmetic: Applications
• Say Ann wishes to send Bob a message M.
• Danger is, a third party might intercept the
message and do whatever they wish with it.
• Solution
– encrypt the message M to C.
– Share the encryption key with receiver.
– Then send encrypted message C to receiver
– Receiver decrypts by reversing the encryption
procedure.
Cryptology
• Cryptology = cryptography + cryptanalysis
• Cryptology is the science of ciphering data or
messages (cryptography) and deciphering the
encrypted messages (cryptanalysis).
Caesar's Cipher
• Caesar’s method
– Shift the positions of the alphabetical letters by a
known step (key) E.g. a right shift of 3, the letter D
would be replaced by G, E would become H, and
so on. In this convention it implies that a new
alphabet is created as follows:
Caesar's Cipher
• Caesar’s method
– Use the ciphertext to generate message e.g.
• Without the knowledge of the key, (right shift of 3), it is
difficult for anyone to decipher this message.
– Alice sends this message and Bob will decipher by
reversing.
– Modular arithmetic is used here. Do you see how?
Caesar's Cipher
• Caesar’s method
– Let ο‘οƒŽM, then to get  its ciphered equivalent apply
•  = (+4)(mod 26)
– At the receiver, to get the plaintext, apply
•  = (-4)(mod 26)
• Note:
– -3(mod 26) = 26-3(mod 26) = 23(mod 26) = 23.
– To avoid negatives, Bob can use
•  = (26 +-4)(mod 26)
Caesar's Cipher
• Caesar’s method
– Problem
• This method is easy to crack
• In computer science, a more complex method
of encryption employing modular arithmetic is
used.
• We look at a very simplified version of the RSA
encryption algorithm
RSA Encryption
• Rivest-Shamir-Adleman (RSA) is a public-key
crytosystem, used in secure data transmission,
attributed to Ron Rivest, Adi Shamir, and
Leonard Adleman, who first publicly described
the algorithm in 1977.
• In transmitting data over networks, it is a good
practice to have the data transmission channel
secured.
RSA Encryption
• Operation
– To receive a message from Alice, Bob will have to
• create his public key and share it with Alice.
• Then he will also generate his private key to use to
decrypt messages from Alice
– These operations are dependent on huge numbers
for them to be almost untraceable by a guesser.
RSA Encryption
• Operation
– Bobs public key generation.
•
•
•
•
•
Bob selects 2 very large prime numbers p and q
Generates n = p*q
Generate the totient (n) = (p-1)(q-1)
Genarate e s.t 1 < e < (n) , e is coprime with (n)
(e, n) is the public key which he shares with Alice.
– Alice’s messege
• C(M) = Me(mod n)
RSA Encryption
• Example
– Bobs public key generation.
•
•
•
•
•
Bob selects 2 very large prime numbers p = 11, and q = 7
Generates n = p*q = 11 * 7 = 77
Generate the totient (n) = (p-1)(q-1) = 10*6 = 60
Genarate e s.t 1 < e < 60, e is coprime with (n) e = 7
(7, 77) is the public key which he shares with Alice.
– Alice’s message
• To send D = 4
• C(4) = 47(mod 77) = 60 and sends to Bob
RSA Encryption
• Example
– Decrypting 16.
• Bob calculates d s.t deο‚Ί1(mod (n))
• (d, n) is Bobs private key.
– To decrypt a messge from Alice
• M(C) = Cd(mod n)
RSA Encryption
• Example
– Bobs private key generation.
• Bob calculates d s.t 7dο‚Ί1(mod 60)
• d = 43
• (43, 77) is Bob’s private key.
– To decrypt a message from Alice
• M(C) = Cd(mod n) = 6043(mod 77) = 4
Exercise
• Alice distributed her public key based on the
following values: p = 3, q = 5. She receives
message “F” from Bob. Decode this message.
Binary Numbers
• Binary numbers are numbers composed of
digits in base 2 – i.e. in the set {0,1}
• Like decimal numbers (base 10), binary
numbers are positional numbers.
– Each digit in the position of the number has a
value of 2 raised to its position, with the leftmost
position being a 0
– …2, 1, 0
Binary Numbers
• For simplicity we stick to 8 digit binary
numbers.
• In this case the positional or place values are
27 26 25 24 23 22 21 20
128 64 32 16 8
4
2
1
Operations on Binary numbers are just like in
base 10 but use base 2
Binary Numbers
• Find 10112 + 112
Conversions between Binary to
Decimal
• Decimal to binary
– Repeat long division with d = 2
– Reverse the r terms to get the binary representation
• E.g. to convert 11
–
–
–
–
–
11 = 2.5 + 1
5 = 2.2 + 1
2 = 2.1 + 0
1 = 2.0 + 1
We get 10112
Conversions between Binary to
Decimal
• Binary to decimal
– Add the products of the place values and their
respectively placed digits.
– E.g. 1012
• 4*1 + 1 = 5
– Points to note
• If you have n bits, you have a range from 0 to 2n -1
• E.g. 8 bits has 0 to 255
Binary Numbers
• Examples
– Convert to decimal
• 11111112
• 100000002
• 111011102
– Convert to binary
• 191
• 253
Binary Numbers
• Data and instructions in computers are stored
as binary numbers
– BCD, EBCDIC
• Machine code are instructions that are
executed by machines.
Summary
• Number theory discusses important topics
including
– fundamental laws of algebra,
– division algorithm
– greatest common divisor
– modular Arithmetic
• We also looked at binary numbers.
???QUESTIONS???
Download