Tải bản đầy đủ - 49 (trang)
GCD & Linear combination (contd.)

GCD & Linear combination (contd.)

Tải bản đầy đủ - 49trang

GCD & Linear combination

(contd.)







Second part of proof





Any other divisor is smaller than d

Let c | a, c | b, c > 0

a = cm, b = cn

d = ax1 + by1 = c(mx1 + ny1)



⇒ c|d

⇒ d is the gcd

Sriram Srinivasan



15/47



Summary 1













All numbers are expressible as unique

products of prime numbers

GCD calculated using Euclid’s algorithm

gcd(a,b) = 1 ⇒ a & b are mutually prime

gcd(a,b) equals the minimum positive

ax+by linear combination



Sriram Srinivasan



16/47



Modular/Clock Arithmetic











1:00 and 13:00 hours are the same

 1:00 and 25:00 hours are the same

1 ≡ 13 (mod 12)

a ≡ b (mod n)

 n is the modulus

 a is “congruent” to b, modulo n

 a - b is divisible by n



a%n=b%n

Sriram Srinivasan



17/47



Modular Arithmetic







a ≡ b (mod n), c ≡ d (mod n)

Addition

a - b = jn

c - d = kn









aa++c c- (b

+ k) nn)

≡ +

b d)

+=

d (j(mod



Multiplication

 ac ≡ bd (mod n)

Sriram Srinivasan



18/47



Modular Arithmetic (contd.)





Power



a ≡ b (mod n) ⇒ ak ≡ bk (mod n)

Using induction,

If ak ≡ bk (mod n),

a . ak ≡ b . bk (mod n), by multiplication rule







∴ ak+1 ≡ bk+1 (mod n)

Going n times around the clock

 a + kn ≡ b (mod n)

Sriram Srinivasan



19/47



Chinese Remainder Theorem





m ≡ a (mod p), m ≡ a (mod q)

⇒ m ≡ a (mod pq) (p,q are primes)

m-a = cp.

Now, m-a is expressible as p1. p2 .p3 . . .

If m - a is divisible by both p and q,

p and q must be one of p1 , p2 , p3

⇒ m - a is divisible by pq

Sriram Srinivasan



20/47



GCD and modulus





If gcd(a,n) = 1, and a = b (mod n),

then gcd(b,n) = 1

a ≡ b (mod n) ⇒ a = b + kn

gcd(a,n) = 1

ax1 + ny1 = 1, for some x1 and y1

(b + kn)x1 + ny1 = 1

bx1 + n(kx1 + y1) = bx1 + ny2 = 1

gcd(b,n) = 1

Sriram Srinivasan



21/47



Multiplicative Inverse





If a, b have no common factors, there

exists ai such that a.ai ≡ 1 (mod b)

 ai is called the “multiplicative inverse”

gcd(a,b) = 1 = ax1+ by1, for some x1 and y1

ax1 = 1 – by1

ax1 = 1 + by2



(making y2 = -y1)



ax1 - 1 = by2

ax1 ≡ 1 (mod b) (x1 is the multiplicative inverse)

Sriram Srinivasan

22/47



Summary 2













Modular arithmetic

 Addition, multiplication, power, inverse

Chinese Remainder Theorem

 If m ≡ a (mod p) and m ≡ a (mod q),

then m ≡ a (mod pq)

Relationship between gcd and modular

arithmetic

 gcd(a,b) = 1

⇒ aai ≡ 1 (mod b)

Sriram Srinivasan



23/47



Euler’s Totient function









φ(n) = Totient(n)

= Count of integers ≤ n coprime to n

 φ (10) = 4

(1, 3, 7, 9 are coprime to 10)

 φ (7) = 6 (1, 2, 3, 4, 5, 6 coprime to 10)

φ(p) = p - 1, if p is a prime



Sriram Srinivasan



24/47



Totient lemma #2: product





φ(pq) = (p - 1)(q - 1) = φ(p) . φ(q)

 if p and q are prime

Which numbers ≤ pq share factors with pq?

1.p, 2.p, 3.p, … (q-1)p and

1.q, 2.q, 3.q, … (p-1)q and

pq

The rest are coprime to pq. Count them.

φ(pq) = pq - (p - 1) - (q - 1) - 1 = (p - 1)(q - 1)

Sriram Srinivasan



25/47



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