lect16 - From the last lecture: Definition. Let a, b Z. A...

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1 Definition. Let a , b Z. A positive integer d is called a common divisor of a and b if d | a and d | b . For any two integers a and b the greatest common divisor is always defined (except the case a = b =0) and is denoted by gcd ( a , b ). Note, that gcd ( a , b ) 1. gcd ( a , b ) = gcd ( b , a ) = gcd ( - a , b ) = gcd ( a , - b ) = gcd ( - a , - b ) if gcd ( a , b ) =1, integers a and b are called relatively prime . Theorem 1. The gcd( a , b ) is the least positive value of ax + by , where x and y range over all integers. From the last lecture:
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2 If an integer c is expressible in the form c = ax + by , it is not necessary that c is the gcd( a , b ). But it does follow from such an equation that gcd( a , b ) is a divisor of c . Why? From the fact that ax + by= 1 for some integers x and y we can imply that a and b are relatively prime. Why? Because ax + by= 1 is divisible by gcd( a , b ), that means that gcd( a , b )=1. If d is any common divisor of two integers, i. e. d | a and d | b , then d | gcd( a , b ). Why? Some consequences of the Theorem 1: gcd( na , nb )= n gcd( a , b ) for any integers a , b , n .
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3 How to find gcd of two integers? a =10, divisors: 1, 2, 5, 10 b =24, divisors: 1, 2, 3, 4, 6, 8, 12, 24 gcd(10, 24)=2 By the last Theorem 1, there are some integers x and y , such that 10 x +24 y =2. 10 5+24 ( - 2) = 2 (found by inspection). So, we found an integer solution of equation 10 x +24 y =2. What if we consider equation 10 x +24 y =4? 5 x +12 y =1? 10 x +24 y =1? Is it unique? 10 (5+12 k )+24 ( - 2+5 k ) = 2
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4 Theorem 2. An integer solution ( x , y ) of equation ax + by = c exists if and only if c is divisible by gcd( a , b ). Proof. Let d = gcd( a , b ). We need to prove: 1). d | c 5 x , y Z such that ax + by = c 2) 5 x , y Z such that ax + by = c d | c 1) Assume d | c c = n d , n Z c = n ( ax 0 + by 0 ) = a ( n x 0 )+ b ( n y 0 ) 2) Assume ax + by = c d | a and d | b d | ax + by d | c x y
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Given two integers a and b how can we find gcd ( a, b )? Euclidian Algorithm. Consider an example: a =963, b = 637. We have 963 = 1 657+306 657 = 2 306+45 306 = 6 45+36 45 = 36+9 36 = 4 9 a=b q 1 + r 1 b=r 1 q 2 + r 2 r 1 = r 2 q 3 + r 3 r 2 = r 3 q 4 + r 4 r 3 = r 4 q 5 0< r 1 < b 0< r 2 < r 1 0< r 3 < r 2 0< r 4 < r 3 We are going to show that the last nonzero remainder ( r 4 =9) is the gcd(963, 657). Let d is any common divisor of a and b: What can be implied from here? d
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lect16 - From the last lecture: Definition. Let a, b Z. A...

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