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Lecture 4 The Euclidean Algorithm (long division) Recall 4th grade If A and B are integers (whole numbers) then we say that B divides A if there is an integer Q such that A = BQ Examples: • 2 divides 6 since there is an integer (3) such that 6 = 2*3 •1 divides any 291 since there is a number (291) such that 291 = 219*1 •If B is any number then B divides 0 since there is a number (0) such that 0 = B*0 •Another way to say that B divides A is to say that B is a factor of A More examples • 3 does not divide 5 since there is no whole number Q such that 5 = 3*Q • 9 does not divide 10 since there is no whole number Q such that 10 = 9*Q Visualizing division 6 6 = 2*3 2 divides 6 7 7 = 2*3 + 1 2 does not divide 7 Division Algorithm A and B are whole numbers and B is not 0. To determine if B divides A do the following: i. If A = 0 then B divides A ii. If 0 < A < B then B does not divide A ii. If B <= A then replace A by B – A and repeat i. 7 – 2 - 2 – 2 = 7 – 3*2 = 1 < 2 (step 4) 7 – 2 - 2 = 7 – 2*2 = 3 > 2 (step 3) 7 – 1*2 = 5 > 2 7 >2 (step2) (step 1) The process subtracts B from A as many times as it can. At some point A is reduced to 0 or B cannot be subtracted from what remains. The effect is to write A = BQ + R with R = 0 or 0 < R < B R is called the remainder when A is divided by B. Q is called the quotient. “Long Division” __3 _ 7 | 220 210 10 220 = 7*30 + 10 (does 30 subtractions) 10 > 7 so can subtract more 31 7 | 220 210 10 _7 3 220 - 7*30 = 1*7 + 3 0 < 3 < 7 process terminates Quotient = 31 remainder = 3 220 = 7*31 + 3 Division Algorithm If A and B are integers with B not zero then there are unique Q and R such that A = BQ + R with 0 <= R < |B| Note this allows A and B to be negative 5 = (-2)(-2) + 1 0 < 1 <|-2| -7 = (-2)(4) + 1 0 < 1<|-2| Works Exactly the Same for Polynomials • B(x) divides A(x) if there is a Q(x) so that A(x) = B(x)*Q(x) • Every polynomial divides 0 0 = B(x)*0 • Any non-zero number1 divides any polynomial (e.g. A(x) = 7*( 7 A(x) ) • x – 1 divides 2 x 1 since 2 x 1 ( x 1 ) ( x 1 ) Recall that if f(x) and g(x) are not 0 then degree f(x)*g(x) = degree f(x) + degree g(x) This says that if B(x) is a factor of A(x) then degree B(x) degree A(x) Example: x does not divide 1 since degree (x) = 1 which is strictly less than the degree of 1 (which is 0). 3 5 x 3 x 1 does not divide 2 x 1 Division Algorithm for Polynomials If A(x) and B(x) are polynomials with B not zero then there are unique polynomials Q(x) and R(x) such that A(x) = B(x)Q(x) + R(x) with R(x)=0 or degree R(x) < degree B(x) Note that R(x) = 0 is another way to say that B(x) divides A(x). The Long Division of Polynomials is a Way to calculate Q(x) and R(x) Observations: To calculate Q(x) and R(x) it suffices to find R(x) since we can divide A(x)- R(x) by B(x) to get R(x) The uniqueness of the remainder says if in any way you arrange to write A(x) = B(x)K(x) + P(x) where P(x) is zero or of smaller degree than B(x) then it must be that P(x) is the R(x) you would get by long division. 2 2 x For instance : x x ( x 1 )x so the remainder when is divided by x-1 will be x. 3 2 2 3 x ( x 1 ) ( 1x x )1 Also ( x 1 ) ( 1x x )x 1 so which says that the remainder when x 2 is divided by x -1 is 1. Algebra of Remainders (modular arithmetic) Principle: When calculating the remainder when an algebraic expression of polynomials is divided by a polynomial B, one can replace any factor or summand by its remainder upon division by B. The remainder upon dividing the sum of two polynomials by B is the same as the remainder if either (or both) terms is first replaced by its remainder (or any polynomial that has the same remainder). The remainder upon dividing the product of two polynomials by B is the same as the remainder if either (or both) terms is first replaced by its remainder (or any polynomial that has the same remainder). Calculate the remainder upon division of 2 4 3 x 3 x x 1 by x x 1 First note that since x 2( x 2x 1 ) 1( 1x ) the remainder of x 2 is 1x 2 4 3 2 2 x 3 x x 1( x ) 3 x x x 1 We can replace x 2 by its remainder 1-x 4 3 2 x 3 x x 1( 1x ) 3 x ( 1x )x 1 = 12 x x 23 x 3 x 2x 1 = 4 x 4 x 2 Now we can replace x 2 again to get 8 x 4 Similarly Calculate the remainder when x 4 is divided by As with the previous example the remainder of the remainder of x 2 4 x 4 x 1 This says 1 2 Replace x by 4 x 2 is = is a root of 2x+1 ( 2 x 1 ) is the same as the remainder of 2 or 4 ( 2 x 1 )4 x 112 x 5 4 2 x ( x 2 x 1 ) Q( x )12 x 5 2 x 2 x 1 1 2 Since it is a root of 2 x 2 x 1 2 x 2 x 1 in . for some polynomial Q(x) What is ( 1 2 ) 4 4 2 x ( x 2 x 1 ) Q( x )12 x 5 we have 4 ( 1 2 ) 12 ( 1 2 )5 or 1712 2 The Greatest Common Divisor (GCD) (also called GCFactor) • The GCD of two integers A1 and A2 is the largest integer that divides both. – Examples: gcd(n,1)=1 for any n – gcd(n,0) = |n| if n is not 0 • (gcd(0,0) does not exist • GCD 12 and 20 – – – – Factors of 12 = {-12,-6,-4,-3,-2,-1,1,2,3,4,6,12} Factors of 20 = {-20,-10,-5,-4,-2,-1,1,2, 4, 5,10,20} Common factors = {-4, -2,-1,1,2,4} Greatest common factor = 4 • Not a practical for large numbers • Factors of 5280 2 3 4 5 6 1 10 11 12 15 16 8 22 24 30 32 33 20 44 48 55 60 66 40 88 96 110 120 132 80 165 176 220 240 264 160 352 440 480 528 660 330 880 1056 1320 1760 2640 5280 • Factors of 4680 2 1 9 8 18 20 39 40 72 78 130 156 312 360 780 936 GCD(5280,4680) = ? 3 10 24 45 90 180 390 1170 6 12 13 15 26 30 36 52 60 65 104 117 120 195 234 260 468 520 585 1560 2340 4680 4 5 • 146057167872 has 1056 positive factors • 5228296875 has 120 positive factors • There is no known way to find a single factor (other than itself and 1) of a randomly chosen number in a “small” number of steps. • We can find the GCD of pairs of HUGE numbers in a small number of steps. Theorem: If A, B, d and n are numbers and d divides both A and B then it also divides both A and B+/-n*A Proof: A = d*r B = d*s n*A = d*(n*r) B + n*A = d*s +/- d*(n*r) = d*(s +/- n*r) Theorem: GCD(A,B) = GCD(A,B+/- n*A) Using this over and over … GCD(146057167872, 5228296875) = GCD(146057167872 – 27* 5228296875, 5228296875) = GCD(4893152247, 5228296875) = GCD(5228296875- 1*4893152247, 4893152247) = GCD(335144628, 4893152247 ) = GCD(4893152247-14* 335144628,335144628 ) = GCD(201127455, 335144628) = GCD( 201127455, 335144628- 1*201127455 ) = GCD( 201127455, 134017173) = GCD( 201127455 – 1*134017173, 134017173) = GCD( 67110282, 134017173) = GCD( 67110282, 134017173 – 1* 67110282) =GCD(67110282,66906891)= GCD(67110282-328* 66906891,66906891)=GCD(203391, 66906891)= GCD(203391, 66906891-328*203391) =GCD(203391,194634) = GCD(203391-1*194634, 194634) = GCD(8748, 194634) = GCD(8748, 194634-22*8748) = GCD(8748,2187) = GCD(8748-4*2187, 2187) = GCD(2187,0) So we calculated the GCD without factoring. It has to stop because each time we subtracted a multiple of the smaller from the larger so that the resulting number is even smaller. Process has to eventually get to 0. Euclid Algorithm To find the gcd of numbers A1 and A2 with A1 > A2 >= 0 a. If A2 = 0 then gcd = A1 b. If A2 > 0 then A1 = A2 q2 + A3 with A2>A3 >=0 c. Replace A1 by A2, A2 by A3 and go to step a. This is exactly what we did in the previous example. Example: gcd(120,85) 120 = 85*1 + 35 85 = 35*2 + 15 35 = 15*2 + 5 15 = 5*3 + 0 gcd = 5 (gcd is the last non-zero remainder) Tabulate: note 120 = 85*1 + 35 85 = 35*2 + 15 35 = 15*2 + 5 15 = 5*3 + 0 pattern 120 85 35 15 5 0 " " 1 2 2 3 " " Find numbers a, n so we can write GCD(120,85) = a*120- b*85 Idea: find a, b for the last two then modify them to serve for the previous pair. 120 " " * 120 " " Last pair = 5,0 85 1 85 1 * GCD=5 obviously 35 2 5*1 – 0*0 = 5 35 2 * Add column to left 15 5 0 2 3 " " * 0 1 15 +5 - 0 2 3 " " 5*1 – 0*0 = 5 (gcd) 1*5 – 0*0 = 5 ( the gcd) Now want a and b so that a*15+b*5 = 5 Have which says 15=5*3+0 or 15 – 5*3 = 0 Substituting in the first equation 1*5 – 0*(15 – 5*3) = 5 1*5 +(0*3)*5 -0*15 = 5 (1+0*3)*5 - 0*15 = 5 * * * * 0 1 120 85 35 15 5 0 " " 1 2 2 3 " " Find the next number (1+0*3)*5 - 0*15 =5 * 120 85 * 35 * * - 15 0 + 5 0 1 " " 1 2 2 3 " " Note the pattern Pattern * 120 85 * 35 *? 15 1 + 5 0 0 1 " " 1 2 2 3 " " * * * 1 0 1 120 = * 85 35 15 5 0 " " 1 2 2 3 " " + - + = + * = * "Answers" "Integers" "Divisions" - 7 120 "Begin" 5 85 1 + 2 35 2 - 1 15 2 + 0 5 3 - 1 0 "done" + 5*120-7*85 = 5 2*85-5*35 = -5 1*35-2*15 = 5 0*15 -1* 5 = -5 1*5 - 0*0=5 Same thing works with polynomials Note we are differing from the book slightly. The book would ask that the gcd be monic. That is it wants the gcd in this problem to be x-2 so it would put a ¼ at the lower left instead of 1. "Polynomials" "Answers" 2 3 1x 5 x 2 4 x x 2 1 x 5 x 6 0 4 x 8 1 0 "Divisions" "Begin" 1x x 3 4 4 "done" 2 3 2 ( 5 x 24 x x ) 1(1+x) ( x 5 x 6 )4 x 8 Multiply through by ¼ to write x – 2 as a combination of 2 3 5 x 24 x x and 2 x 5 x 6 "Polynomials" "Answers" 3x 5 2 3 3 8 x 8 x 3 x 2 2 2 1 2 x 2 x 0 33 x 1 0 1*( 2 3 ) - ( 3 x 5 38 x 8 x 3 x 2 2 "Divisions" "Begin" 3 x 5 2 2 2x 3 "done" )*( 2 2 x 2 x ) = -3 - 3x Lead coefficient of the gcd is -3. Monic gcd is (-3 -3x)/(-3) = 1+x. Divide both sides by -3 to get monic gcd as a linear combination.