Number Theory: Divisibility, Congruences, and Representations of Integers, Lecture notes of Engineering

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Chapter Motivation

 Number theory is the part of mathematics devoted to the study of the integers and their properties.  Key ideas in number theory include divisibility and the primality of integers.  Representations of integers, including binary and hexadecimal representations, are part of number theory.  Number theory has long been studied because of the beauty of its ideas, its accessibility, and its wealth of open questions.  We’ll use many ideas about proof methods and proof strategy in our exploration of number theory.  Mathematicians have long considered number theory to be pure mathematics, but it has important applications to computer science and cryptography.

Division

Definition : If a and b are integers with a ≠ 0 , then

a divides b if there exists an integer c such that b = ac.

 When a divides b we say that a is a factor or divisor of b and that b is a multiple of a.  The notation a | b denotes that a divides b.  If a | b , then b / a is an integer.  If a does not divide b , we write a ∤ b.

Example : Determine whether 3 | 7 and whether 3 | 12.

Properties of Divisibility

Theorem 1 : Let a , b , and c be integers, where a ≠0. i. If a | b and a | c , then a | ( b + c ); ii. If a | b, then a | b c for all integers c ; iii. If a | b and b | c , then a | c. Proof : (i) Suppose a | b and a | c , then it follows that there are integers s and t with b = as and c = at. Hence, b + c = as + at = a ( s + t ). Hence, a | ( b + c )

Corollary : If a , b , and c be integers, where a ≠0, such that a | b and a | c, then a | mb + nc whenever m and n are integers.

Congruence Relation

Definition : If a and b are integers and m is a positive integer, then a is congruent to b modulo m if m divides a – b.  The notation a ≡ b (mod m ) says that a is congruent to b modulo m.  We say that a ≡ b (mod m ) is a congruence and that m is its modulus.  Two integers are congruent mod m if and only if they have the same remainder when divided by m.  If a is not congruent to b modulo m , we write a ≢ b (mod m ) Example : Determine whether 17 is congruent to 5 modulo 6 and whether 24 and 14 are congruent modulo 6. Solution :  17 ≡ 5 (mod 6) because 6 divides 17 − 5 = 12.  24 ≢ 14 (mod 6) since 24 − 14 = 10 is not divisible by 6.

The Relationship between

(mod m ) and mod m Notations

 Theorem 3 : Let a and b be integers, and let m be a

positive integer.

Then a ≡ b (mod m ) if and only if a mod m = b mod m.

Congruences of Sums and Products

Theorem 5 : Let m be a positive integer. If a ≡ b (mod m ) and c ≡ d (mod m ), then a + c ≡ b + d (mod m ) and ac ≡ bd (mod m ) Proof :  Because a ≡ b (mod m ) and c ≡ d (mod m ), by Theorem 4 there are integers s and t with b = a + sm and d = c + tm.  Therefore,  b + d = ( a + sm ) + ( c + tm ) = ( a + c ) + m ( s + t ) and  b d = ( a + sm ) ( c + tm ) = ac + m ( at + cs + stm ).  Hence, a + c ≡ b + d (mod m ) and ac ≡ bd (mod m ). Example : Because 7 ≡ 2 (mod 5 ) and 11 ≡ 1 (mod 5 ) , it follows from Theorem 5 that 18 = 7 + 11 ≡ 2 + 1 = 3 (mod 5 ) 77 = 7 ∙ 11 ≡ 2 ∙ 1 = 2 (mod 5 )

Algebraic Manipulation of Congruences

 Multiplying both sides of a valid congruence by an integer preserves validity. If a ≡ b (mod m ) holds then c ∙ a ≡ c ∙ b (mod m ), where c is any integer, holds by Theorem 5 with d = c.  Adding an integer to both sides of a valid congruence preserves validity. If a ≡ b (mod m ) holds then c + a ≡ c + b (mod m ), where c is any integer, holds by Theorem 5 with d = c.  Dividing a congruence by an integer does not always produce a valid congruence. Example : The congruence 14 ≡ 8 (mod 6 ) holds. But dividing both sides by 2 does not produce a valid congruence since 14/2 = 7 and 8/2 = 4, but 7 ≢4 (mod 6).

Arithmetic Modulo m

Definitions : Let Z m be the set of nonnegative integers less than m : { 0 , 1 , …., m −1}  The operation + m is defined as a + m b = ( a + b ) mod m. This is addition modulo m.  The operation ∙ m is defined as a ∙ m b = ( a ∙ b ) mod m. This is multiplication modulo m.  Using these operations is said to be doing arithmetic modulo m. Example : Find 7 + 11 9 and 7 ∙ 11 9. Solution : Using the definitions above:  7 + 11 9 = (7 + 9) mod 11 = 16 mod 11 = 5  7 ∙ 11 9 = (7 ∙ 9) mod 11 = 63 mod 11 = 8

• Chapter
• Section 4.
• Section 4.

Representations of Integers

 In the modern world, we use decimal, or base 10,

notation to represent integers. For example when we

write 965, we mean 9 ∙10^2 + 6∙10^1 + 5∙10^0.

 We can represent numbers using any base b , where b

is a positive integer greater than 1.

 The bases b = 2 ( binary ), b = 8 ( octal ) , and b = 16

( hexadecimal ) are important for computing and

communications

 The ancient Mayans used base 20 and the ancient

Babylonians used base 60.

Binary Expansions

Most computers represent integers and do arithmetic with binary (base 2 ) expansions of integers. In these expansions, the only digits used are 0 and 1.

Example : What is the decimal expansion of the integer that has (1 0101 1111) 2 as its binary expansion?

Solution :

(1 0101 1111) 2 = 1∙2^8 + 0∙2^7 + 1∙2^6 + 0∙2^5 + 1∙2^4 + 1∙2^3

• 1∙2^2 + 1∙2^1 + 1∙2^0 =351.

Example : What is the decimal expansion of the integer that has ( 11011 ) 2 as its binary expansion?

Solution : ( 11011 ) 2 = 1 ∙2^4 + 1∙2^3 + 0∙2^2 + 1∙2^1 + 1∙2^0 =27.

Hexadecimal Expansions

The hexadecimal expansion needs 16 digits, but our decimal system provides only 10. So letters are used for the additional symbols. The hexadecimal system uses the digits {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}. The letters A through F represent the decimal numbers 10 through 15. Example : What is the decimal expansion of the number with hexadecimal expansion (2AE0B) 16? Solution : 2 ∙16^4 + 10∙16^3 + 14∙16^2 + 0∙16^1 + 11∙16^0 =

Example : What is the decimal expansion of the number with hexadecimal expansion (E 5 ) 16? Solution : 14 ∙16^1 + 5∙16^0 = 224 + 5 = 229

Base Conversion

To construct the base b expansion of an integer n :

 Divide n by b to obtain a quotient and remainder. n = bq 0 + a 0 0 ≤ a 0 ≤ b  The remainder, a 0 , is the rightmost digit in the base b expansion of n. Next, divide q 0 by b. q 0 = bq 1 + a 1 0 ≤ a 1 ≤ b  The remainder, a 1 , is the second digit from the right in the base b expansion of n.  Continue by successively dividing the quotients by b , obtaining the additional base b digits as the remainder. The process terminates when the quotient is 0.

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