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Chapter 8 Introduction to Number Theory

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2 Contents Prime Numbers Fermats and Eulers Theorems

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3 Prime Numbers Primes numbers An integer p > 1 is a prime number if and only if it is divisible by only 1 and p.

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4 Prime Numbers Integer factorization Any integer a > 1 can be factored in a unique way as where p 1 < p 2 < … < p t are prime numbers and each a i is a positive integer. 91 = 7 × 13; 11101 = 7 × 11 2 ×13

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5 Prime Numbers Another integer factorization If P is the set of all prime numbers, then any positive integer can be written uniquely in the following form: The right side is the product over all possible prime numbers p. Most of the exponents a p will be 0. 3600 = 2 4 ×3 2 ×5 2 ×7 0 ×11 0 ×….

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6 Prime Numbers Another integer factorization The value of any given positive integer can be specified by listing all the nonzero exponents. The integer 12 =2 2 ×3 1 is represented by {a 2 =2, a 3 =1}. The integer 18 =2 1 ×3 2 is represented by {a 2 =1, a 3 =2}. The integer 91= 7 2 ×13 1 is represented by {a 7 = 2, a 13 = 1}.

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7 Prime Numbers Multiplication Multiplication of two numbers is adding the corresponding exponents. k = 12 × 18 = 216 12 = 2 2 × 3 1 18 = 2 1 × 3 2 ------------------ 216 = 2 3 × 3 3

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8 Prime Numbers Divisibility a|ba p b p for all p a = 12;b= 36;12|36 12 = 2 2 ×3; 36 = 2 2 ×3 2 a 2 = 2 = b 2 a 3 = 1 2 = b 3

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9 Prime Numbers GCD k = gcd (a, b) k p = min(a p, b p ) for all p 300 = 2 2 ×3 1 ×5 2 18 = 2 1 ×3 2 ×5 0 gcd (18, 300) = 2 1 ×3 1 ×5 0 = 6

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10 Fermats and Eulers Theorems Fermats theorem If p is prime and a is a positive integer not divisible by p, then a p-1 1 (mod p)

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11 Fermats and Eulers Theorems Proof of Fermats theorem. Outline Show {1, 2, …, p-1}={a mod p, 2a mod p, …, (p-1)a mod p} Show. Since is relatively prime to p, we multiply to both sides to get.

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12 Fermats and Eulers Theorems Proof of Fermats theorem Show {1, 2, …, p-1}={a mod p, 2a mod p, …, (p-1)a mod p} Show ka mod p for any 1 k p-1 is in {1, 2, …, p-1} by showing that ka mod p k a mod p for k k. Show ka mod p k a mod p for 1 k k p-1. Proof by contradiction Assume that ka k a mod p for some 1 k k p-1. Since a is relatively prime to p, we multiply a -1 to get k k mod p, which contradiction the fact that k k.

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13 Fermats and Eulers Theorems Proof of Fermats theorem Show. {1, 2, …, p-1} = {a mod p, 2a mod p, …, (p-1)a mod p}

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14 Fermats and Eulers Theorems An alternative form of Fermats Theorem a p a mod p where p is prime and a is any positive integer. Proof If a and p are relatively prime, we get a p a mod p by multiplying a to each side of a p-1 1 mod p. If a and p are not relatively prime, a = cp for some positive integer c. So a p (cp) p 0 mod p and a 0 mod p, which means a p a mod p.

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15 Fermats and Eulers Theorems An alternative form of Fermats Theorem a p a mod p where p is prime and a is any positive integer. p = 5, a = 33 5 = 243 3 mod 5 p = 5, a = 1010 5 = 100000 10 mod 5 0 mod 5

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16 Fermats and Eulers Theorems Eulers Totient Function The number of positive integers less than n and relatively prime to n. = 36 37 is prime, so all the positive number from 1 to 36 are relatively prime to 37. = 24 35 = 5×7 1, 2, 3, 4, 6, 8, 9,11, 12, 13, 16, 17, 18, 19, 22, 23, 24, 26, 27, 29, 31, 32, 33, 34

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17 Fermats and Eulers Theorems How to compute In general, For a prime n, (Z n = {1,2,…, n-1}) For n = pq, p and q are prime numbers and p q

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18 Fermats and Eulers Theorems Proof of is the number of positive integers less than pq that are relatively prime to pq. can be computed by subtract from pq – 1 the number of positive integers in {1, …, pq – 1} that are not relatively prime to pq. The positive integers that are not relatively prime to pq are a multiple of either p or q. { p, 2p,…,(q – 1)p}, {q, 2q, …,(p – 1)q} There is no same elements in the two sets. So, there are p + q – 2 elements that are not relatively prime to pq. Hence, = pq – 1– (p + q – 2) = pq – p – q +1 = (p – 1)(q – 1)

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19 Fermats and Eulers Theorems Φ(21) = Φ(3)×Φ(7) = (3-1)×(7-1) = 2 ×6 = 12 Z 21 ={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20} Φ(3)={3,6,9,12,15,18} Φ(7)={7,14} where the 12 integers are {1,2,4,5,8,10,11,13,16,17,19,20}

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20 Fermats and Eulers Theorems Eulers theorem For every a and n that are relatively prime: a = 3;n = 10;Φ(10) = 4;3 4 = 81 1 mod 10 a = 2;n = 11;Φ(11) = 10;2 10 = 1024 1 mod 11

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21 Fermats and Eulers Theorems Proof of Eulers theorem If n is prime, it holds due to Fermats theorem. Otherwise (If n is not prime), define two sets R and S. show the sets R and S are the same. then, show

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22 Fermats and Eulers Theorems Proof of Eulers theorem Set R The elements are positive integers less than n and relatively prime to n. The number of elements is R={x 1, x 2,…, x Φ(n) } where x 1 < x 2 <…< x Φ(n) Set S Multiplying each element of R by a R modulo n S ={(ax 1 mod n), (ax 2 mod n),…(ax Φ(n) mod n)}.

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23 Fermats and Eulers Theorems Proof of Eulers theorem The sets R and S are the same. We show S has all integers less than n and relatively prime to n. S ={(ax 1 mod n), (ax 2 mod n),…(ax Φ(n) mod n)} 1. All the elements of S are integers less than n that are relatively prime to n because a is relatively prime to n and x i is relatively prime to n, ax i must also be relatively prime to n. 2. There are no duplicates in S. If ax i mod n = ax j mod n, then x i = x j. by cancellation law.

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24 Fermats and Eulers Theorems Proof of Eulers theorem Since R and S are the same sets,

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25 Fermats and Eulers Theorems Alternative form of the theorem If a and n are relatively prime, it is true due to Eulers theorem. Otherwise, ….

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26 Fermats and Eulers Theorem The validity of RSA algorithm Given 2 prime numbers p and q, and integers n = pq and m, with 0

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27 Fermats and Eulers Theorem Case 1: m is a multiple of p m=cp for some positive integer c. gcd(m, q)=1, otherwise, m is a multiple of p and q and yet m

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28 Fermats and Eulers Theorem Case 2: m is a multiple of q prove similarly. Thus, the following equation is proved.

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29 Fermats and Eulers Theorem An alternative form of this corollary is directly relevant to RSA.

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