4.5.2.
Definition: Twin Primes.
If p and p + 2 are both prime numbers, then p and p + 2
are called twin primes
Examples:
•
Find all twin primes between 15 and 50.
The primes 15 – 50 are: 17, 19, 23, 29, 31, 37, 41, 43, 47
Hence the twin primes between 1 and 50 are:
17, 19
•
29, 31
41, 43
Does 97 have a “twin” to form a pair of twin primes?
No.
Exercises:
•
Find all twin primes between 101 and 150.
•
Does 197 have a “twin” to form a pair of twin primes?
WUCT121
Numbers
101
4.6.
Greatest Common Divisor
4.6.1.
Definition: Greatest Common Divisor
Let a , b ∈ , with a ≠ 0 or b ≠ 0 .
The greatest common divisor of a and b, denoted gcd(a, b),
is a natural number c with the following properties:
A. c is a common divisor of both a and b.
That is c | a and c | b
B. For all natural numbers d, if d is a common divisor of
a and b, then d ≤ c .
That is ∀d ∈ , ( d | a ∧ d | b ) ⇒ d | c
Exercise:
Write the definition of greatest common divisor using logic
notation.
WUCT121
Numbers
102
Examples:
Evaluate:
•
gcd(18, 12)
Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 12: 1, 2, 3, 4, 6, 12
∴ gcd(18,12) = 6
•
gcd(18, –12)
Factors of 18: 1, 2, 3, 6, 9, 18
Factors of –12: 1, 2, 3, 4, 6, 12
∴ gcd(18,12) = 6
Note: need only consider positive factors since gcd is
a natural number
Exercises:
•
Evaluate gcd(45, 75)
•
Evaluate gcd(7, 11)
WUCT121
Numbers
103
If a = p1α1 p 2 α 2 p 3α 3 ... p k α k and
b = p1 β1 p 2 β 2 p 3 β3 ... p k β k , then
gcd( a , b ) = 1 × p1 min(α1 , β1 ) p 2 min(α 2 , β 2 ) ... p k min(α k , β k )
where α i ≠ 0 and β i ≠ 0
Example:
•
gcd(3220, 1155)
3220 = 2 2 × 5 × 7 × 23
1155 = 3 × 5 × 7 ×11
gcd(2772, 2310) = 1 × 5 × 7
= 35
•
gcd(35100, 6975)
35100 = 2 2 × 3 × 5 2 ×117
6975 = 3 2 × 5 2 × 31
gcd(35100, 6975) = 1 × 3 × 5 2
= 75
WUCT121
Numbers
104
Exercises:
•
Evaluate gcd(2772, 2310).
•
Evaluate gcd(6615, 1352).
•
If gcd(a , b ) = c , what are
gcd(a , − b ), gcd(−a , − b ), gcd(−a , b ) ?
•
For a , b ∈ , with a ≠ 0 or b ≠ 0 , does gcd(a, b)
always exist? Is it unique?
•
gcd(0, 0) is not defined. Why?
WUCT121
Numbers
105
4.7.
Least Common Multiple
4.7.1.
Definition: Least Common Multiple
Let a , b ∈ , with a ≠ 0 or b ≠ 0 .
The least common multiple of a and b, denoted lcm(a, b), is
a natural number c with the following properties:
A. c is a common multiple of both a and b.
That is a | c and b | c
B. For all natural numbers d, if d is a common multiple
of a and b, then c ≤ d .
That is ∀d ∈ , ( a | d ∧ b | d ) ⇒ c | d
Exercise:
Write the definition of least common multiple using logic
notation.
WUCT121
Numbers
106
Example:
Evaluate:
•
lcm(9, 15)
Multiples of 9: 9, 18, 27, 36, 45,….
Multiples of 15: 15, 30, 45….
∴ lcm(9,15) = 45
•
lcm(32, –24)
Multiples of 32: 32, 64, 96, 128,….
Multiples of –24: 24, 48, 72, 96,….
∴ lcm(32, − 24) = 96
Note: Need only consider positive multiples since lcm
is a natural number
Exercise:
•
Evaluate lcm(45, 75)
•
Evaluate lcm(7, –11)
WUCT121
Numbers
107
If a = p1α1 p 2 α 2 p 3α 3 ... p k α k and
b = p1 β1 p 2 β 2 p 3 β3 ... p k β k , then
lcm( a , b ) = p1max(α1 , β1 ) p2 max(α 2 , β 2 ) ... pk max(α k , β k ) .
Example:
Evaluate:
•
lcm(3220, 1155)
3220 = 22 × 5 × 7 × 23
1155 = 3 × 5 × 7 ×11
lcm( 2772, 2310) = 22 × 3 × 5 × 7 ×11 × 23
= 106260
•
lcm(35100, 6975)
35100 = 22 × 3 × 52 ×117
6975 = 32 × 52 × 31
lcm(35100, 6975) = 22 × 32 × 52 × 31 ×117
= 3264300
WUCT121
Numbers
108
Exercise:
•
Evaluate lcm(2772, 2310).
•
Evaluate lcm(6615, 1352).
•
If lcm( a , b ) = c , what are
lcm( a , − b ), lcm( −a , − b ), lcm( −a , b ) ?
•
For a , b ∈ , with a ≠ 0 or b ≠ 0 , does lcm(a, b)
always exist? Is it unique?
WUCT121
Numbers
109
4.8.
The Euclidean Algorithm
The Euclidean Algorithm is an efficient process for finding
the greatest common divisor of two integers. It uses the
Quotient-Remainder Theorem and the following lemmas:
Lemma 1: gcd(r , 0) = r
If r is a positive integer, then gcd(r , 0) = r .
Proof:
Let r ∈ , r > 0 , and let c = r.
Clearly, c | r and c | 0 .
Obviously, any other divisor of r (all of which also divide
0) must divide c as c = r.
That is ∀d ∈ , ( d | r ∧ d | 0) ⇒ d | c
∴ gcd(r , 0) = c = r .
WUCT121
Numbers
110
Lemma 2: a = bq + r ⇒ gcd(a , b ) = gcd(b, r )
If a , b ∈ , b ≠ 0 and q, r ∈  that a = b × q + r , then
gcd(a , b ) = gcd(b, r ) .
Proof:
The proof of this lemma has two parts:
a. Prove that gcd(a , b ) | gcd(b, r ) ; and
b. Prove that gcd(b, r ) | gcd(a , b ) .
Once we have shown these two results and noting that the
greatest common divisor is always positive, we can
conclude that gcd(a , b ) = gcd(b, r ) .
A.
Prove that gcd(a , b ) | gcd(b, r ) .
Let gcd(a , b ) = k ∈  .
Then k | a and k | b ; that is,
∃m, n ∈ , (a = km ∧ b = kn )
(1) .
We are given that a = b × q + r , r > 0
(2) ,
Substituting (1) into (2), we have
km = knq + r ⇒ km − knq = r
⇒ r = k (m − nq )
WUCT121
Numbers
111
Clearly, m − nq ∈  and so k | r .
Therefore, k is a common divisor of b and r.
So k | gcd (b, r ) , that is, gcd(a , b ) | gcd(b, r ) .
B.
Prove that gcd(b, r ) | gcd(a , b ) .
Let gcd(b, r ) = l ∈  .
Then l | b and l | r ; that is,
∃m, n ∈ , (b = lm ∧ r = ln )
(3) .
We are given that a = b × q + r , r > 0
(4) .
Substituting, (3) into (4), we have
a = lmq + ln ⇒ a = l (mq + n ) .
Clearly, mq + n ∈  and so l | a .
Therefore, l is a common divisor of a and
b.
So l | gcd(a , b ) , that is, gcd(b, r ) | gcd(a , b ) .
Therefore, by A. and B. gcd(a , b ) = gcd(b, r ) .
WUCT121
Numbers
112
4.8.1.
Process: The Euclidean Algorithm
The Euclidean Algorithm is used to find gcd(a, b) and can
be described as follows:
1.
Let a , b ∈ , with | a |>| b |≥ 0 .
2.
If b = 0, then gcd(a , b ) = a (by Lemma 1).
If b ≠ 0 , apply the quotient remainder theorem to get
a = bq + r , where 0 ≤ r < b , then, (by Lemma 2)
gcd(a , b ) = gcd(b, r ) .
Repeat the process in step 2, to find gcd(b, r )
(i.e. let a = b and b = r).
The process is guaranteed to terminate eventually with
r = 0 because each new remainder is less than the
preceding one and all are nonnegative.
Note if a < 0 or b < 0 then let a = | a |, b =| b | in the above
algorithm
WUCT121
Numbers
113
Examples:
Find the greatest common divisor of the following pairs of
numbers using the Euclidean Algorithm:
•
2772 and 2310
Divide 2772 by 2310 to find :
2772 = 2310 × 1 + 462
Hence gcd(2772,2310) = gcd(2310,462)
Divide 2310 by 462 to find :
2310 = 462 × 5 + 0
Hence gcd(2310,462) = gcd(462,0)
= 462
Therefore gcd(2772,2310) = gcd(2310,462)
= gcd(462,0)
= 462
WUCT121
Numbers
114
•
–243 and 223
gcd(−243,223) = gcd(243,223)
Divide 243 by 223 to find :
243 = 223 × 1 + 20
Hence gcd(243,223) = gcd(223,20)
Divide 223 by 20 to find :
223 = 20 × 11 + 3
Hence gcd(223,20) = gcd(20,3)
Divide 20 by 3 to find :
20 = 3 × 6 + 2
Hence gcd(20,3) = gcd(3,2)
Divide 3 by 2 to find :
3 = 2 ×1 + 1
Hence gcd(3,2) = gcd(2,1)
Divide 2 by 1 to find :
2 = 1× 2 + 0
Hence gcd(2,1) = gcd(1,0)
=1
Therefore gcd(−243,223) = gcd(243,223)
= gcd(223,20)
= gcd(20,3)
= gcd(3,2)
= gcd(2,1)
= gcd(1,0)
=1
WUCT121
Numbers
115
Exercises:
Find the greatest common divisor of the following pairs of
numbers using the Euclidean Algorithm:
•
15 and 10
•
5 and 9
WUCT121
Numbers
116
•
90 and –54
•
24 and 63
WUCT121
Numbers
117
4.8.2.
Definition: Relatively Prime
If a , b ∈  and gcd(a , b ) = 1 then a and b are relatively
prime.
Example:
gcd(17,9 ) = 1, therefore 17 and 9 are relatively prime
Exercises:
•
Which of these pairs are relatively prime?
∗ 3, 8
∗ 27, 32
∗ 81, 33
∗ 72, 139
•
True or False? If true, what is y? If false, give a
counterexample
∀x ∈ , ∃y ∈  , gcd( x, y ) = 1
•
Let p be prime, n ∈  . What are the two possibilities
for gcd( p, n ) ?
WUCT121
Numbers
118
4.8.3.
Theorem: Linear Combination of gcd
If a , b ∈  and not both equal zero, then gcd(a , b ) exists
and there exist m, n ∈  such that gcd(a , b ) = ma + nb .
Note.
The theorem says two things:
•
gcd(a , b ) always exists (as long as a and b are not both
zero)
and
•
gcd(a, b ) can be written as a linear combination of a
and b.
Corollary 1:
If a and b are relatively prime, gcd (a , b ) = 1.
Thus ∃m, n ∈  such that gcd(a , b ) = 1 = ma + nb .
Corollary 2:
∀a , b, c ∈  , if gcd (a , b ) = 1 and a | bc , then a | c .
WUCT121
Numbers
119
Proof:
We must prove that:
∀a , b, c ∈  , ((gcd(a , b ) = 1 ∧ a | bc ) ⇒ a | c ) .
We know:
gcd(a , b ) = 1 that is ∃m, n ∈  such that
1 = ma + nb
(1)
and
a | bc , that is ∃k ∈  such that bc = ka
(2) .
We must prove that a | c , that is, we must find l Π٠such
that c = la .
Multiplying (1) by c we have c = mac + nbc (3) .
Substituting (2) into (3)
c = mac + nqa
= (mc + nq )a
= la where l = mc + nq ∈ 
∴a | c
WUCT121
Numbers
120