Quiz 2 key

1
Quiz 2 key

2
The Euclidean Algorithm

• (long division)

3
First The Division algorithm

• If a and b are integers with b ltgt 0, then there
  are unique integers q and r so that a q b r
  and 0 lt r lt b
• Example 3745 __q__ 45 __r___
• Long division
• Calculator

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Divisor, common divisor, greatest common divisor

• b is a divisor of a if a bq for some integer
  q
• b is common divisor of a and c if _____
• b is the greatest common divisor of a and c if
  ____

5
Arayabhata-Euclids algorithm How to find
gcd(a,b), the greatest common divisor of a and b

• based on a single observation if a b q
  r, then
• any divisor of a and b is also a divisor of r,
  and any divisor
• of b and r is also a divisor of a, so gcd(a,b)
  gcd(b,r)
• Euclid algorithm use the division algorithm
  repeatedly
• To reduce the problem to one you can solve.
• Example gcd(55,35)
• 55 351 20 so gcd(55,35) gcd(35,20)
• 35 201 15 so gcd(35,20) gcd(20,15)
• 20 151 5 done gcd(55,35) 5

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2 columns of the arayabhata table the columns of
quotients and remainders
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Writing gcd(a,b) as a linear combination of a and
b

• gcd(55,35) 5 55x 35y Solve for x
  and y.

8
The 1st column of the arayabhata table

9
An application

• Problem Given a 3 pint can, a 5 pint can, a
  large tank of water, and a large empty tank, how
  can we get exactly 1 pint of water into the empty
  tank?

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We will talk about this on WednesdayDivision
algorithm for polynomials in x.

• If a(x) and b(x) are polynomials in x then there
  are unique polynomials q(x) and r(x) so that
  a(x) q(x)b(x) r(x) where
• r(x) 0 or deg(r(x)) lt deg(b(x))
• Example x4 __q(x)___ (x2 -1)
  __r(x)___
• Long division

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Divisor, common divisor, and gcd for polynomials
in x

• b(x) is a divisor of a(x) if ____
• c(x) is a common divisor of a(x) and b(x) if ___
• c(x) is a greatest common divisor

12
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 23
• 1 divides any 291 since there is a number (291)
  such that 291 2191
• If B is any number then B divides 0 since there
  is a number (0) such that 0 B0
• Another way to say that B divides A is to say
  that B is a factor of A

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More examples

• 3 does not divide 5 since there is no whole
  number Q such that 5 3Q
• 9 does not divide 10 since there is no whole
  number Q such that 10 9Q

14
Visualizing division
6
2 divides 6
6 23
7
7 23 1
2 does not divide 7
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Division Algorithm
A and B are whole numbers and B is not 0. To
determine if B divides A do the followingi.
If A 0 then B divides A ii. If 0 lt A lt B
then B does not divide Aii. If B lt A then
replace A by B A and repeat i.
(step 4)
(step 3)
(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 lt
R lt BR is called the remainder when A is divided
by B. Q is called the quotient.
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Long Division
__3 _ 7 220 210 10
(does 30 subtractions)
220 730 10
10 gt 7 so can subtract more
3 1 7 220 210
10 _7 3
220 - 730 17 3 0 lt 3 lt 7 process
terminates Quotient 31 remainder 3
220 731 3
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Division Algorithm

• If A and B are integers with B not zero then
  there are unique integers Q and R such that
  A BQ R with 0 lt R lt B
• Note this allows A and B to be negative
  5 (-2)(-2) 1 0 lt 1 lt-2
• -7 (-2)(4) 1 0 lt 1lt-2
  

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Works Exactly the Same for Polynomials

• B(x) divides A(x) if there is a Q(x) so
  thatA(x) B(x)Q(x)
• Every polynomial divides 0 0 B(x)0
• Any non-zero number divides any polynomial
  (e.g. A(x) 7( A(x) )
• x 1 divides since

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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) thendegree 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).
does not divide
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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 ordegree R(x) lt degree B(x)
• Note that R(x) 0 is another way to say that
  B(x) divides A(x).

21
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. For instance
so the remainder when is
divided by x-1 will be x. Also
so
which
says that the remainder when is divided by
x -1 is 1.

22
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).
23
Calculate the remainder upon division of
by
First note that since
the remainder of is

We can replace by its remainder 1-x


Now we can replace again to get
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Similarly
Calculate the remainder when is divided
by
As with the previous example the remainder of
is 2x1 the remainder of
is the same as the remainder of
or


This says
for some polynomial Q(x)
is a root of
. What is
Replace x by in

Since it is a root of we
have
or
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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

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• Factors of 5280
• Factors of 4680
• GCD(5280,4680) ?

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• 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.

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• Theorem If A, B, d and n are numbers and d
  divides both A and B then it also divides both
  A and B/-nA
• Proof A dr B ds
• nA d(nr)
• B nA ds /- d(nr) d(s /- nr)
• Theorem GCD(A,B) GCD(A,B/- nA)

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Using this over and over

• GCD(146057167872, 5228296875)
  GCD(146057167872 27 5228296875, 5228296875)
  GCD(4893152247, 5228296875)
• GCD(5228296875- 14893152247, 4893152247)
• GCD(335144628, 4893152247 )
• GCD(4893152247-14 335144628,335144628 )
• GCD(201127455, 335144628)
• GCD( 201127455, 335144628- 1201127455 )
• GCD( 201127455, 134017173)
• GCD( 201127455 1134017173, 134017173)
• GCD( 67110282, 134017173)
• GCD( 67110282, 134017173 1 67110282)
  GCD(67110282,66906891)
• GCD(67110282-328 66906891,66906891)GCD(203391,
  66906891)
• GCD(203391, 66906891-328203391)
  GCD(203391,194634)
• GCD(203391-1194634, 194634) GCD(8748, 194634)
  
• GCD(8748, 194634-228748) GCD(8748,2187)
  GCD(8748-42187, 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.

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Euclid Algorithm

• To find the gcd of numbers A1 and A2 with A1 gt
  A2 gt 0
• If A2 0 then gcd A1
• If A2 gt 0 then A1 A2 q2 A3 with A2gtA3 gt0
• 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)
• 120 851 35
• 85 352 15
• 35 152 5
• 15 53 0 gcd 5 (gcd is the last
  non-zero remainder)

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Tabulate note pattern

• 120 851 35

85 352 15
35 152 5
15 53 0
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Find numbers a, n so we can write GCD(120,85)
a120- b85

• Idea find a, b for the last two then modify
  them to serve for the previous pair.
• Last pair 5,0
• GCD5 obviously
• 51 00 5
• Add column to left

-
51 00 5 (gcd)
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First Fill in columns 2 and 3
15 00 5 ( the gcd) Now want a and b so
that a15b5 5Have which
says 15530 or 15 53 0 Substituting
in the first equation 15 0(15 53)
515 (03)5 -015 5 (103)5 - 015
5
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Then begin filling in column 1
35
Note the pattern
Pattern



36
-
5120-785 5

285-535 -5
-
135-215 5

015 -1 5 -5
-
15 - 005

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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.
Multiply through by ¼ to write x 2 as a
combination of
and
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1( )
- ( )(
) -3 - 3x
Lead coefficient of the gcd is -3. Monic gcd is
(-3 -3x)/(-3) 1x. Divide both sides by -3
to get monic gcd as a linear combination.