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CompSci 230
Discrete Math for Computer Science
Announcements
• Homework 4 out
October 22, 2013
Prof. Rodger
Slide ideas from Rosen and Benjamin
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2
Chap. 4.3 - Greatest Common Divisor
Chap. 4.3 - Greatest Common Divisor
Definition: Let a and b be integers, not both zero.
The largest integer d such that d | a and also d | b
is called the greatest common divisor of a and b.
The greatest common divisor of a and b is
denoted by gcd(a,b).
Definition: Let a and b be integers, not both zero.
The largest integer d such that d | a and also d | b
is called the greatest common divisor of a and b.
The greatest common divisor of a and b is
denoted by gcd(a,b).
Examples:
gcd(12, 30) = 6
gcd(12, -30) = 6
gcd(16, 9) = 1
gcd(7, 63) = 7
gcd(7, 0) = 7
relatively prime
3
Examples:
gcd(12, 30) = 6
gcd(12, -30) = 6
gcd(16, 9) = 1
gcd(7, 63) = 7
gcd(7, 0) = 7
relatively prime
4
Greatest Common Divisor
Greatest Common Divisor
Definition: The integers a and b are relatively prime if
their greatest common divisor is .
Example:
Definition: The integers a and b are relatively prime if
their greatest common divisor is .
Example:
Definition: The integers a1, a2, …, an are pairwise
relatively prime if gcd(ai, aj)= whenever
i j n.
Definition: The integers a1, a2, …, an are pairwise
relatively prime if gcd(ai, aj)= whenever
i j n.
Example: Are
Solution: Yes
and
Example: Are
Yes
Example: Are
Solution: No,
and
Example: Are
No,
5
6
What numbers can be created using
12 and 30 as building blocks?
What numbers can be created using
12 and 30 as building blocks?
• That means 12s + 30t
• Can you create any positive integer using
this formula?
• Can you create 4?
• That means 12s + 30t
• Can you create any positive integer using
this formula?
• Can you create 4?
– 4 = 12x + 30 y
– = 6(2x + 5y)
– NO, must be a multiple of 6
4 = 12s + 30 t
= 6(2s + 5t)
NO, must be a multiple of 6
7
8
Can you create 6 using 12s + 30t?
Can you create 6 using 12s + 30t?
6 = 12s + 30 t
= 12(3) + 30(-1) = 36 – 30 = 6, yes!
Can you create any multiple of 6 using 12 and 30?
6 = 12s + 30 t
= 12(3) + 30(-1) = 36 – 30 = 6, yes!
Can you create any multiple of 6 using 12 and 30?
Yes
12(3m) + 30(-m) = 6m
Yes
12(3m) + 30(-m) = 6m
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10
Can you create any number using
16 and 9 as building blocks?
Can you create any number using
16 and 9 as building blocks?
16s + 9t
16s + 9t
Can you create the number 1?
16(4) + 9(-7) = 64 – 63 = 1
Yes, can create any number with 16 and 9
Can you create the number 1?
16(4) + 9(-7) = 64 – 63 = 1
Yes, can create any number with 16 and 9
To create any number m:
To create any number m:
16(4m) + 9(-7m) = m
16(4m) + 9(-7m) = m
B zout showed back in 18th century
B zout showed back in 18th century
• If a and b are relatively prime,
then there exists integers x and y such that ax+by=1
• If a and b are relatively prime,
then there exists integers s and t such that as+bt=1
11
12
gcds as Linear
Combinations
gcds as Linear
Combinations
Étienne Bézout
(1730‐1783)
Étienne Bézout
(1730‐1783)
B zout’s Theorem: If a and b are positive integers,
then there exist integers s and t such that
gcd(a,b) = sa + tb.
)
(proof in exercises of Section
Idea:
gcd(a,b) = g
a/g and b/g are relatively prime
g is the largest that divides both of them
Thus, there exists s and t such that (a/g)s + (b/g)t = 1
B zout’s Theorem: If a and b are positive integers,
then there exist integers s and t such that
gcd(a,b) = sa + tb.
)
(proof in exercises of Section
Idea:
gcd(a,b) = g
a/g and b/g are relatively prime
g is the largest that divides both of them
Thus, there exists s and t such that (a/g)s + (b/g)t = 1
Definition: If a and b are positive integers, then
integers s and t such that gcd(a,b) = sa + tb are called
B zout coefficients of a and b. The equation gcd(a,b) 13=
sa + tb is called B zout’s identity.
Definition: If a and b are positive integers, then
integers s and t such that gcd(a,b) = sa + tb are called
B zout coefficients of a and b. The equation gcd(a,b) 14=
sa + tb is called B zout’s identity.
Euclidean Algorithm
Euclidean Algorithm
Euclid
(325 B.C.E. – 265 B.C.E.)
• The Euclidian algorithm is an efficient method for
computing the greatest common divisor of two integers. It
is based on the idea that gcd(a,b) is equal to gcd(a,c) when
a b and c is the remainder when a is divided by b.
Example: Find gcd(
,
• 287 91∙3 14
• 91 14∙6 7
• 14 7∙2 0
):
Divide 287 by 91
Divide 91 by 14
Divide 14 by 7
Stopping
condition
gcd(
,
) = gcd(
,
) = gcd(
, ) =
15
• The Euclidean algorithm expressed in
pseudocode is:
procedure gcd(a, b: positive integers)
x := a
y := b
while y
r := x mod y
x := y
y := r
return x {gcd(a,b) is x}
16
Correctness of Euclidean Algorithm
Correctness of Euclidean Algorithm
Lemma : Let a = bq + r, where a, b, q, and r
are integers. Then gcd(a,b) = gcd(b,r).
Proof:
Lemma : Let a = bq + r, where a, b, q, and r
are integers. Then gcd(a,b) = gcd(b,r).
Proof:
– Suppose that d divides both a and b.
Then d also divides a bq = r (by Theorem of
Section
). Hence, any common divisor of a and b
must also be any common divisor of b and r.
– Suppose that d divides both b and r.
Then d also divides bq + r = a. Hence, any common
divisor of a and b must also be a common divisor of
b and r.
– Therefore, gcd(a,b) = gcd(b,r).
– Suppose that d divides both a and b.
Then d also divides a bq = r (by Theorem of
Section
). Hence, any common divisor of a and b
must also be any common divisor of b and r.
– Suppose that d divides both b and r.
Then d also divides bq + r = a. Hence, any common
divisor of a and b must also be a common divisor of
b and r.
– Therefore, gcd(a,b) = gcd(b,r).
17
18
Correctness of Euclidean Algorithm
•
Suppose that a and b are positive
r0
integers with a b.
r1
Let r0 = a and r1 = b.
Successive applications of the division
algorithm yields:
= r 1 q1 + r 2
= r 2 q2 + r 3
∙
∙
∙
rn‐2 = rn‐1qn‐1 + r2
rn‐1 = rnqn .
0
0
0
r 2 < r 1,
r 3 < r 2,
Correctness of Euclidean Algorithm
•
rn < rn‐1,
•
Suppose that a and b are positive
r0
integers with a b.
r1
Let r0 = a and r1 = b.
Successive applications of the division
algorithm yields:
= r 1 q1 + r 2
= r 2 q2 + r 3
∙
∙
∙
rn‐2 = rn‐1qn‐1 + r2
rn‐1 = rnqn .
0
0
0
r 2 < r 1,
r 3 < r 2,
rn < rn‐1,
Eventually, a remainder of zero occurs in the sequence of terms: a = r0> r1
> r2 ∙∙∙ 0.Thesequencecan’tcontainmorethana terms.
• ByLemma1
gcd(a,b) = gcd(r0,r1) = ∙∙∙ gcd rn‐1,rn gcd rn ,0 rn.
• Hencethegreatestcommondivisoristhelastnonzeroremainderin
thesequenceofdivisions.
•
Eventually, a remainder of zero occurs in the sequence of terms: a = r0> r1
> r2 ∙∙∙ 0.Thesequencecan’tcontainmorethana terms.
• ByLemma1
gcd(a,b) = gcd(r0,r1) = ∙∙∙ gcd rn‐1,rn gcd rn ,0 rn.
• Hencethegreatestcommondivisoristhelastnonzeroremainderin
thesequenceofdivisions.
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20
Finding gcds as Linear Combinations
Finding gcds as Linear Combinations
Example: Express gcd(252,198) = 18asalinearcombination
of252and198.
Solution: First use the Euclidean algorithm to show
gcd(252,198) = 18
Example: Express gcd(252,198) = 18asalinearcombination
of252and198.
Solution: First use the Euclidean algorithm to show
gcd(252,198) = 18
i.
ii.
iii.
iv.
252 1∙198 54
198 3 ∙54 36
54 1 ∙36 18
36 2 ∙18
i.
ii.
iii.
iv.
– Nowworkingbackwards,fromiii andi above
– Nowworkingbackwards,fromiii andi above
• 18 54 1 ∙36
• 36 198 3 ∙54
• 18 54 1 ∙36
• 36 198 3 ∙54
– Substitutingthe2nd equationintothe1st yields:
• 18 54 1 ∙ 198 3 ∙54
252 1∙198 54
198 3 ∙54 36
54 1 ∙36 18
36 2 ∙18
– Substitutingthe2nd equationintothe1st yields:
4 ∙54 1 ∙198
• 18 54 1 ∙ 198 3 ∙54
– Substituting54 252 1 ∙198 fromi yields:
4 ∙54 1 ∙198
– Substituting54 252 1 ∙198 fromi yields:
• 18 4 ∙ 252 1 ∙198 1 ∙198 4 ∙252 5 ∙198
• 18 4 ∙ 252 1 ∙198 1 ∙198 4 ∙252 5 ∙198
• This method illustrated above is a two pass method. It first uses
the Euclidian algorithm to find the gcd and then works
backwards to express the gcd as a linear combination of the
original two integers.
• This method illustrated above is a two pass method. It first uses
the Euclidian algorithm to find the gcd and then works
backwards to express the gcd as a linear combination of the
original two integers.
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Euclid’s algorithm
22
Euclid’s algorithm
• It is smart!
• It is smart!
– It not only finds the gcd, it also finds the s and t
that gives you the gcd!
• It is fast!
– It not only finds the gcd, it also finds the s and t
that gives you the gcd!
• It is fast!
– If a>b, then finds gcd(a,b) in
steps
– Slightly less than the number of digits in b
– If a and b are 100 digit numbers, finds the gcd
in under 500 steps
– If a>b, then finds gcd(a,b) in
steps
– Slightly less than the number of digits in b
– If a and b are 100 digit numbers, finds the gcd
in under 500 steps
23
24
What are the worse numbers to try?
• How about gcd(99, 98)?
What are the worse numbers to try?
• How about gcd(99, 98)?
gcd(99,98) = 1
gcd(99,98) = 1
• How about gcd(99,50)?
• How about gcd(99,50)?
gcd(99,50) = gcd(50,49) = gcd(49,1) = 1
gcd(99,50) = gcd(50,49) = gcd(49,1) = 1
• How about gcd(89,55)?
• How about gcd(89,55)?
gcd(89,55) = gcd(55,34) = gcd(34,21) =
gcd(21,13) = gcd(13,8) = gcd(8,5) = gcd(5,3) =
gcd(3,2) = gcd(2,1) = 1
Recognize Fibonnaci numbers?
gcd(89,55) = gcd(55,34) = gcd(34,21) =
gcd(21,13) = gcd(13,8) = gcd(8,5) = gcd(5,3) =
gcd(3,2) = gcd(2,1) = 1
25
Recognize Fibonnaci numbers?
26
Primes
Primes
Definition: A positive integer p greater than
is called prime if the only positive factors of p
are and p. A positive integer that is greater
than and is not prime is called composite.
Definition: A positive integer p greater than
is called prime if the only positive factors of p
are and p. A positive integer that is greater
than and is not prime is called composite.
Example: is prime only factors are and ,
is composite because it is divisible by .
What about 1?
Neither prime or composite
It is a unit! It divides everything
27
Example: is prime only factors are and ,
is composite because it is divisible by .
What about 1?
Neither prime or composite
It is a unit! It divides everything
28
The Fundamental Theorem of
Arithmetic
The Fundamental Theorem of
Arithmetic
Theorem: Every positive integer greater than
can be written uniquely as a prime or as the
product of two or more primes where the
prime factors are written in order of
nondecreasing size.
Examples:
2
Theorem: Every positive integer greater than
can be written uniquely as a prime or as the
product of two or more primes where the
prime factors are written in order of
nondecreasing size.
Examples:
2
2910
10
The Sieve of Erastosthenes
Erastothenes
(276‐194 B.C.)
• The Sieve of Erastosthenes can be used to find all
primes not exceeding a specified positive integer.
For example, begin with the list of integers between
and
.
a. Delete all the integers, other than 2, divisible by 2.
b. Delete all the integers, other than 3, divisible by 3.
c. Next, delete all the integers, other than 5, divisible by
5.
d. Next, delete all the integers, other than 7, divisible by
7.
e. Since all the remaining integers are not divisible by
any of the previous integers, other than 1, the primes
are:
continued → 31
32
Infinitude of Primes
The Sieve of Erastosthenes
Euclid
(325 B.C.E. – 265 B.C.E.)
Theorem: There are infinitely many primes. (Euclid)
If an integer n is a composite integer, then it has a prime
divisor less than or equal to √n.
To see this, note that if n = ab, then a √n or b
√n.
Trial division, a very inefficient method of determining if a
number n is prime, is to try every integer i √n and see if
n is divisible by i.
Proof: Assume finitely many primes: p1, p2, ….., pn
– Let q = p1p2··· pn + 1
– Either q is prime or by the fundamental theorem of
arithmetic it is a product of primes.
• But none of the primes pj divides q since if pj | q, then pj
divides
q p1p2··· pn = 1 .
• Hence, there is a prime not on the list p1, p2, ….., pn. It is
either q, or if q is composite, it is a prime factor of q.
This contradicts the assumption that p1, p2, ….., pn are
all the primes.
– Consequently, there are infinitely many primes.
In previous example, why did we use only 2, 3, 5 and 7?
33
Mersenne Primes
Marin Mersenne
(1588-1648)
Definition: Prime numbers of the form 2p 1 ,where p
is prime, are called Mersenne primes.
– 22 1 3,23 1 7, 25 1 37,and 27 1 127
areMersenne primes.
– 211 1 2047 is not a Mersenne prime since2047 23∙89.
– Thereisanefficienttestfordeterminingif 2p 1 is
prime.
– ThelargestknownprimenumbersareMersenne primes.
– Asofmid2011,47Mersenne primeswereknown,the
largestis243,112,609 1,whichhasnearly13million
decimaldigits.
– TheGreat Internet Mersenne Prime Search GIMPS isa
distributedcomputingprojecttosearchfornew
Mersenne Primes.
http://www.mersenne.org/
35
This proof was given by Euclid The Elements. The proof is considered to be one of the
most beautiful in all mathematics. It is the first proof in The Book, inspired by the
famous mathematician Paul Erdős’ imagined collection of perfect proofs maintained by
God.
Paul Erdős
(1913-1996)
34
Distribution of Primes
• Mathematicians have been interested in the
distribution of prime numbers among the positive
integers. In the nineteenth century, the prime
number theorem was proved which gives an
asymptotic estimate for the number of primes not
exceeding x.
Prime Number Theorem: The ratio of the number
of primes not exceeding x and x/ln x approaches as
x grows without bound. (ln x is the natural logarithm
of x)
– The theorem tells us that the number of primes not
exceeding x, can be approximated by x/ln x.
– The odds that a randomly selected positive integer less
than n is prime are approximately (n/ln n)/n = 1/ln n.
36
Conjectures about Primes
Generating Primes
Many conjectures about them are unresolved, including:
• Goldbach’s Conjecture:Everyevenintegern,n 2,isthe
sumoftwoprimes.Ithasbeenverifiedbycomputerforall
positiveevenintegersupto1.6∙1018.Theconjectureis
believedtobetruebymostmathematicians.
• Finding large primes with hundreds of digits is
important in cryptography.
• There is no simple function f(n) such that f(n) is
prime for all positive integers n.
• Consider
– f(n) = n2 n +
f( ) = 2
is prime for all integers
• Thereareinfinitelymanyprimesoftheformn2 1,wheren
isapositiveinteger.Butithasbeenshownthatthereare
infinitelymanyprimesoftheformn2 1,wheren isa
positiveintegerortheproductofatmosttwoprimes.
.
• Fortunately, we can generate large integers which
are almost certainly primes. See Chapter .
37
• The Twin Prime Conjecture:Thetwinprimeconjectureisthat
thereareinfinitelymanypairsoftwinprimes.Twinprimes
arepairsofprimesthatdifferby2.Examplesare3and5,5
and7,11and13,etc.Thecurrentworld’srecordfortwin
primes asofmid2011 consistsofnumbers
65,516,468,355∙2333,333 1,whichhave100,355decimal
digits.
38
Finding the Greatest Common
Divisor Using Prime Factorizations
Finding the Greatest Common
Divisor Using Prime Factorizations
• Suppose the prime factorizations of a and b are:
• Suppose the prime factorizations of a and b are:
where each exponent is a nonnegative integer, and where all primes
occurring in either prime factorization are included in both. Then:
•
This formula is valid since the integer on the right (of the equals
sign) divides both a and b. No larger integer can divide both a and b.
•
where each exponent is a nonnegative integer, and where all primes
occurring in either prime factorization are included in both. Then:
•
This formula is valid since the integer on the right (of the equals
sign) divides both a and b. No larger integer can divide both a and b.
•
23
Example: 120 =
∙3∙5500 =
gcd(120,500) = 2min 3,2 ∙3min 1,0 ∙5min 1,3 22 ∙30 ∙51 20
• Finding the gcd of two positive integers using their prime
factorizations is not efficient because there is no efficient algorithm
for finding the prime factorization of a positive integer.
22
∙53
39
Example: 120 = 23 ∙3∙5500 = 22 ∙53
gcd(120,500) = 2min 3,2 ∙3min 1,0 ∙5min 1,3 22 ∙30 ∙51 20
• Finding the gcd of two positive integers using their prime
factorizations is not efficient because there is no efficient algorithm
for finding the prime factorization of a positive integer.
40
Least Common Multiple
Least Common Multiple
Definition: The least common multiple of the positive
integers a and b is the smallest positive integer that is
divisible by both a and b. It is denoted by lcm(a,b).
• The least common multiple can also be computed
from the prime factorizations.
This number is divided by both a and b and no
smaller number is divided by a and b.
Example: lcm(30, 35) =
5*2*3, 7*5
5*2*3*7 = 210
Example: lcm(
3 5 2,
max 3,4
4 3)
max 5,3
=
max 2,0
4
41
LCM and GCD relation
Theorem : Let a and b be positive integers. Then
ab = gcd(a,b)
a,b
Example: lcm(30, 35) =
5*2*3, 7*5
5*2*3*7 = 210
3 5 2,
max 3,4
2
42
Least Common Multiple
Example: lcm(
5
4 3)
max 5,3
=
max 2,0
4
5
2
43
Proof:
Notethatmin x,y max x,y x y
oneusesthelargerexponentandtheotheronethe
smallerexponent,butyougetallfactorsback.
44
LCM and GCD relation
Consequences of Bézout’s Theorem
Theorem : Let a and b be positive integers. Then
ab = gcd(a,b)
a,b
Lemma 2: If a, b, and c are positive integers such that gcd(a, b)
= 1 and a | bc, then a | c.
Proof: Assume gcd(a, b) = 1 and a | bc
– Since gcd(a, b) = 1, by Bézout’s Theorem there are integers s and
t such that
sa + tb = 1.
– Multiplying both sides of the equation by c, yields sac + tbc = c.
– From Theorem 1 of Section 4.1:
a | tbc
part ii) and a divides sac + tbc since a | sac and a|tbc (part i)
– We conclude a | c, since sac + tbc = c.
Lemma 3: If p is prime and p | a1a2∙∙∙an, then p | ai for some i.
(proof uses mathematical induction; see Exercise 64 of Section
5.1)
Proof:
Notethatmin x,y max x,y x y
oneusesthelargerexponentandtheotheronethe
smallerexponent,butyougetallfactorsback.
• Lemma 3 is crucial in the proof of the uniqueness of prime
factorizations.
45
46
Consequences of Bézout’s Theorem
Uniqueness of Prime Factorization
Lemma 2: If a, b, and c are positive integers such that gcd(a, b)
= 1 and a | bc, then a | c.
Proof: Assume gcd(a, b) = 1 and a | bc
• We will prove that a prime factorization of a positive integer
where the primes are in nondecreasing order is unique. (This is
part of the fundamental theorem of arithmetic. The other part,
which asserts that every positive integer has a prime
factorization into primes, will be proved in Section 5.2.)
Proof: (by contradiction) Suppose that the positive integer n can
be written as a product of primes in two distinct ways:
n = p1p2 ∙∙∙ ps and n = q1q2 ∙∙∙ pt.
– Since gcd(a, b) = 1, by Bézout’s Theorem there are integers s and
t such that
sa + tb = 1.
– Multiplying both sides of the equation by c, yields sac + tbc = c.
– From Theorem 1 of Section 4.1:
a | tbc
– Remove all common primes from the factorizations to get
part ii) and a divides sac + tbc since a | sac and a|tbc (part i)
– We conclude a | c, since sac + tbc = c.
– By Lemma 3, it follows that
divides
contradicting the assumption that
and
primes.
Lemma 3: If p is prime and p | a1a2∙∙∙an, then p | ai for some i.
(proof uses mathematical induction; see Exercise 64 of Section
5.1)
• Lemma 3 is crucial in the proof of the uniqueness of prime
factorizations.
, for some k,
are distinct
– Hence, there can be at most one factorization of n into primes in
nondecreasing order.
47
48
Dividing Congruences by an Integer
Dividing Congruences by an Integer
• Dividing both sides of a valid congruence by an integer
does not always produce a valid congruence (see
).
Section
• Dividing both sides of a valid congruence by an integer
does not always produce a valid congruence (see
).
Section
• But
• But
Theorem 7
c
ac
a b
bc (mod m) and gcd(c,m) =
Proof: Since ac bc (mod m), m | ac bc = c(a b)
by Lemma and the fact that gcd(c,m) = , it follows
that m | a b.
m
Hence, a b
49
Theorem 7
c
ac
a b
bc (mod m) and gcd(c,m) =
Proof: Since ac bc (mod m), m | ac bc = c(a b)
by Lemma and the fact that gcd(c,m) = , it follows
that m | a b.
m
Hence, a b
50