Transcript 109_lecture4_fall05

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 ) ( 1x x )1
Also ( x 1 ) ( 1x 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 2x 1 ) 1( 1x ) the remainder of x 2 is
1x
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( 1x ) 3 x ( 1x )x 1 = 12 x x 23 x 3 x 2x 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 112 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
1712 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




220 240 264
160 165 176




480 528 660
330 352 440




880 1056 1320 1760 2640 5280
• Factors of 4680
2
3
 1


9
10
 8


24
 18 20


45
 39 40


90
 72 78


180
130 156


390
312 360


780 936 1170
GCD(5280,4680) = ?
4
12
26
52
104
195
468
1560
6


13
15


30
36


60
65


117 120


234 260


520 585


2340 4680
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
 1x
5 x 24 x x



2

1
x
5 x 6




0
4 x 8




1
0

"Divisions"


"Begin" 


1x 

x 3 
 
4 4 


"done" 
2
3
2
( 5 x 24 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 24 x x
and
2
x 5 x 6
"Polynomials"
"Divisions"
"Answers"




3
x
5


2
3

38 x 8 x 3 x
"Begin" 


 2

2






3
x
5
2

1
2 x 2 x
 

2
2 





2x 



0
33 x


3




1
0
"done" 

1*(
2
3 ) - ( 3 x  5
38 x 8 x 3 x
2
2
)*(
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.