Lec-9gdfweerwertwertvgwertwertevrgwgtwertertgdfg.pdf

Discrete Mathematics
Lecture- 09
Number Theory

Prime Numbers: A prime number is one, that has exactly two
factors, which means, it can be divided by only “1” and itself.
But “1” is not a prime number.
Example: 3 is a prime number because 3 can be divided by
only two numbers i.e. 1 and 3 itself.
Composite Numbers: A composite number has more than two
factors, which means apart from getting divided by the
number 1 and itself, it can also be divided by at least one
integer or number. We don’t consider ‘1’ as a composite
number.
Example: 12 is a composite number because it can be divided
by 1, 2, 3, 4, 6 and 12. So, the number ‘12’ has 6 factors.

Types of Composite Numbers: There are two types of composite
numbers:
1. Even composite numbers
2. Odd composite numbers
Odd Composite numbers: The odd positive integers or the odd
numbers that are not prime numbers are called odd composite
numbers. For example, 9, 21, 33, 45, etc., are odd composite
numbers.
Even Composite numbers: The even numbers that are not prime
numbers are called even composite numbers. For example, 4,
10, 16, 28, 56, etc., are even composite numbers.

Prime Factorization: Prime factorization is the process of writing a number as the
product of prime numbers. Prime numbers are the numbers that have only two factors,
1 and the number itself.
✓ Prime factorization of 12 is 2 × 2 × 3 = 22 × 3
✓ Prime factorization of 24 is 2 × 2 × 2 × 3 = 23 × 3
✓ Prime factorization of 36 is 2 × 2 × 3 × 3 = 2² × 3²

Greatest Common Divisor(GCD) Or HCF: Given two non-negative integers a
and b, we have to find their GCD, i.e. the largest number that is a divisor of
both a and b. It's commonly denoted by GCD(a, b).
Procedure: Example: Find the GCD of 33 & 12
Step1: Start with input A & B where (A>B)
Step2: Quotient(Q) = A/B
Remainder = A % B
Step3: A = B and B = R
Step4: Repeat 2 & 3 until A > 0
Step5: CGD = A
So, the GCD of 33 and 12 = 3
Q A B R
2 33 12 9
1 12 9 3
3 9 3 0
3 0

Least Common Multiple(LCM): LCM or the Least Common
Multiple of two given numbers A and B is the Least number that
can be divided by both A and B, leaving the remainder 0 in
each case.
LCM(A, B) = (a * b) / GCD(A, B)
Example: Find the LCM of 33 & 12.
Solution:
The GCD of 33 & 12 is 3(from previous slide)
So, the LCM(33,12) = (33 * 12) / 3 = 132

Factorial: A factorial is a mathematical operation that you write like this:
n!.
It represents the multiplication of all numbers between 1 and n.
Formula: n! = n × (n-1) × (n-2) × (n-3) × ….× 3 × 2 × 1
or
n! = n. (n-1)!
For example, the factorial of 10 is written as
10! = 10. 9!
10! = 10 (9 × 8 × 7 × 6 × 5× 4 × 3 × 2 × 1)
10! = 10 (362,880)
10! = 3,628,800
Therefore, the value of 10 factorial is 3,628,800

Fibonacci Sequence: The Fibonacci sequence is the series of
numbers where each number is the sum of the two preceding
numbers.
The Rule is Fn = Fn-1 + Fn-2
Where,
Fn is the term number "n“ where n ≥ 2
Fn-1 is the previous term (n−1)
Fn-2 the term before that (n−2)
For example: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, …

Example: Find the Fibonacci number between 7 to 10.
Solution: For Fibonacci series-
F(0) = 0 & F(1) = 1
And Fn = Fn-1 + Fn-2
For n=2, F2 = F2-1 + F2-2 = F1 + F0 = 1 + 0 = 1
For n=3, F3 = F3-1 + F3-2 = F2 + F1 = 1 + 1 = 2
For n=4, F4 = F4-1 + F4-2 = F3 + F2 = 2 + 1 = 3
For n=5, F5 = F5-1 + F5-2 = F4 + F3 = 3 + 2 = 5
Similarly,
F6 = F5 + F4 = 5 + 3 = 8
F7 = F6 + F5 = 8 + 5 = 13
F8 = F7 + F6 = 13 + 8 = 21
F9 = F8 + F7 = 21 + 13 = 34
F10 = F9 + F8 = 34 + 21 = 55
So, the Fibonacci number between 7 to 10 are 13, 21, 34, 55.

Perfect Numbers: A Perfect Number N is defined as any positive integer where the sum of
its divisors minus the number itself equals the number.
N = 2p-1(2p -1) where p is a prime for which 2p -1 is a Mersenne prime.
The first 5 perfect numbers are 6, 28, 496, 8128, and 33550336.

Example-1: Find all the perfect numbers from 1 to 500.
Solution: We know that every perfect number can be expressed as
2p-1(2p -1) where p is a prime number.
Using the above formula let us find the perfect numbers from 1 to
500.
For n = 2, 22-1(22 – 1) = 21 (4 –1) = 2 × 3 = 6.
For n = 3, 23-1(23 – 1) = 22 (8 – 1) = 4 × 7 = 28
For n = 5, 25-1 (25 – 1) = 24 (32 – 1) = 16 × 31 = 496
So, the perfect numbers between 1 to 500 are 6, 28 and 496.

Example-2: Check whether the following numbers are perfect numbers or not.
(i) 282
(ii) 8128
Solution:
(i) Factors of 282 are 1, 2, 3, 6, 47, 94, 141, 282.
The proper divisors of 282 = 1, 2, 3, 6, 47, 94, 141.
Now,
1 + 2 + 3 + 6 + 47 + 94 + 141 = 294 ≠ 282.
Thus, 282 is not a perfect number.
(ii) Factors of 8128 are 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064 and 8128
Proper divisors of 282 = 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064
Now,
1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064 = 8128
Thus, 8128 is a perfect number.

Deficient Number: The sum of the factors (excluding itself) is less
than the number.
The first few Deficient Numbers are: 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13,
14, 15, 16, 17, 19 …..
Example: Find out, 8 is deficient number or not.
Solution: The factor of 8 are 1,2,4.
So, 8: 1+2+4=7
Here 7 is less than 8.
So, this number is deficient number.

Abundant: The sum of the factors (excluding itself) is greater than
the number.
The first few Abundant Numbers are: 12, 18, 20, 24, 30, 36, 40, 42, 48,
54, 56, 60, 66 …..
Example: Find out 18 is abundant number or not.
Solution: The factor of 18 are 1,2,3,6,9.
So, 18: 1+2+3+6+9 = 21
Here, 21 is greater than 18.
So, this number is abundant number.

Lucas numbers: Lucas numbers are similar to Fibonacci numbers.
Lucas numbers are also defined as the sum of its two
immediately previous terms. But here the first two terms are 2 and
1 whereas in Fibonacci numbers the first two terms are 0 and 1
respectively.
The Lucas numbers are in the following integer sequence:
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123 …………..
For Lucas sequence-
The Rule is Fn = Fn-1 + Fn-2
Where,
Fn is the term number "n“ where n ≥ 2
Fn-1 is the previous term (n−1)
Fn-2 the term before that (n−2)

Example: Find the Lucas number between 7 to 10.
Solution: For Lucas series-
L(0) = 2 & L(1) = 1
And Ln = Ln-1 + Ln-2
For n=2, L2 = L2-1 + L2-2 = L1 + L0 = 2 + 1 = 3
For n=3, L3 = L3-1 + L3-2 = L2 + L1 = 3 + 1 = 4
For n=4, L4 = L4-1 + L4-2 = L3 + L2 = 4 + 3 = 7
For n=5, L5 = L5-1 + L5-2 = L4 + L3 = 7 + 4 = 11
Similarly,
L6 = L5 + L4 = 11 + 7 = 18
L7 = L6 + L5 = 18 + 11 = 29
L8 = L7 + L6 = 29 + 18 = 47
L9 = L8 + L7 = 47 + 29 = 76
L10 = L9 + L8 = 76 + 47 = 123
So, the Lucas number between 7 to 10 are 29, 47, 76, 123.

Modular arithmetic
Modular arithmetic is a system of arithmetic for integers, where the
numbers are reduced within a certain range, defined by the
modulus.
For two integers a and b, and a positive integer n, we say that a is
congruent to b modulo n if their difference is an integer multiple of n.
This is denoted as:
a ≡ b (mod n)

❑ Quotient Remainder Theorem:
It states that, for any pair of integers a and b (b is positive), there exist
two unique integers q and r such that:
a = b x q + r where 0 <= r < b
Example: If a = 20, b = 6 then q = 3, r = 2;
So, 20 = 6 x 3 + 2
❑ Modular Addition: The rule for modular addition is-
(a + b) mod m = ((a mod m) + (b mod m)) mod m
Example:
(15 + 17) % 7
= ((15 % 7) + (17 % 7)) % 7
= (1 + 3) % 7
= 4 % 7
= 4

Modular Subtraction: The rule for modular subtraction is-
(a - b) mod m = ((a mod m) - (b mod m)) mod m
Example:
(17 - 15) % 7
= ((17 % 7) - (15 % 7)) % 7
= (3 - 1) % 7
= 2 % 7
= 2
Modular Multiplication: The Rule for modular multiplication is-
(a x b) mod m = ((a mod m) x (b mod m)) mod m
Example:
(12 x 13) % 5
= ((12 % 5) x (13 % 5)) % 5
= (2 x 3) % 5
= 6 % 5
= 1

❑ Applications of Modular Arithmetic-
✓ Cryptography
✓ Computer Science
✓ Number Theory
✓ Digital Signal Processing
✓ Clock Arithmetic

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