Download Numbers Strand Lecture 5 (Week 8)

Section 4.
4.1.
Number Theory
Introduction
This section demonstrates some different techniques of
proving some general statements.
Examples:
•
Prove that the sum of any two odd numbers is even.
Firstly you should check a few examples to confirm (for
yourself) that the statement is true.
3 + 5 = 8 ; 9 + 7 = 16 ; 7 + 11 = 18
Write down what you are asked to prove in “mathematical
language” (if it is not already).
∀a , b ∈ , a , b are odd ⇒ a + b is even
Next, write down everything you know [related to the
problem] and what they mean.
Note that this also includes what you can assume – such as
Claim(k) in Mathematical Induction.
WUCT121
a is odd ⇒ ∃k ∈ , a = 2k + 1
(1)
b is odd ⇒ ∃l ∈ , b = 2l + 1
( 2)
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Write down what you are trying to prove, and what that
means (i.e. what you have to “get to”).
Prove that a + b is even, that is ∃s ∈ , a + b = 2 s
What you do at this point varies from proof to proof.
Whenever you have an equality, it is often best to start with
the LHS and work to the RHS.
LHS = a + b
= (2k + 1) + (2l + 1)
by (1), (2)
= 2l + 2k + 2
= 2(l + k + 1)
Distributivity
= 2s
s = l + k + 1∈ 
= RHS
∴ a + b is even
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•
Prove that 2 is the only even prime.
Clearly, 2 is an even number and a prime number.
To prove that 2 is the only even prime requires a proof by
contradiction.
We must suppose the opposite of what we are proving, then
follow the logical arguments to arrive at a contradictory
statement.
Suppose there exists another even prime, n, such that n > 2 .
Then, ∃k ∈ , n = 2k .
Now, 2 > 0 and 2 ≠ 1 .
Also, k > 0 (as n > 2 ) and k ≠ 1.
[If k were equal to 1, then n = 2 . However, we know that
n > 2 .]
Therefore, we have positive numbers 2 and k where n = 2k
and k ≠ 1, 2 ≠ 1 .
Thus, by definition, n is composite.
This is a contradiction, as we said n is prime.
Therefore, there does not exist an even prime n such that
n > 2 , and so 2 is the only even prime.
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Exercise:
Determine if the following statements are true or false. In
each case prove your answer
•
∀n ∈ , n is prime ⇒ n is odd
•
∀a , b ∈ , a 2 = b 2 ⇒ a = b
Note: You MUST give a counterexample. A “general
disproof” doesn’t work as the values a = b = 0 are a
counterexample to the “disproof”.
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4.2.
Divisibility
4.2.1.
Definition: Divisibility
If n and d are integers and d ≠ 0 , then n is divisible by d if
and only if n = d × k for some k ∈  .
We write d | n and say that “d is a divisor of n.”
Alternative expressions include:
“d is a factor of n.”
“d divides n.”
“n is a multiple of d.”
“n is divisible by d.”
When n is not divisible by d we write d /| n
Exercise:
Write the definition of divisibility using logic notation.
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Examples:
•
Is − 16 a divisor of 32?
Yes. 32 = −16 × (−2), − 2 ∈  .
•
If l ∈  and l ≠ 0 , does l | 0 ?
Yes. 0 = l × 0, 0 ∈  .
•
Find all values of a ∈  such that a | 1.
We need to show ∃k ∈ , 1 = ak ⇔ a =
Need
•
1
.
k
1
1
∈  ⇔ k = ±1 giving a =
= ±1.
k
±1
What is the relationship between a and b if a | b and
b | a , for a, b ∈  ?
b = ak ⇔ a =
b
k
a = bl
b
∴ = bl ⇒ 1 = kl
k
kl = 1 ⇒ (k = l = 1) ∨ (k = l = −1)
(k = l = 1) ⇒ (a = b )
(k = l = −1) ⇒ (a = −b )
∴ a = ±b
WUCT121
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Exercises:
•
If a, b ∈  , is 3a + 3b divisible by 3?
•
Let a , b, c, x, y ∈  . If b | a and b | c , does
b | (ax + cy )? Why?
•
a, b ∈  , is it true that a | b implies a ≤ b ?
•
Does 4 | 15 ?
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4.2.2.
Definition: Transitivity of Divisibility
For all integers a, b and c, if a | b and b | c , then a | c .
Exercise:
Write the definition of transitivity of divisibility using logic
notation.
Proof:
We know:
a | b ⇒ ∃k ∈ , b = ka
b | c ⇒ ∃l ∈ , c = bl
(1)
(2)
Show that a | c , that is, find m ∈  such that c = ma .
Now,
c = bl
= ( ka )l
= ( kl )a
= ma
∴a |c
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by( 2)
by(1)
by associativity and commutativity
where m = kl ∈ 
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4.2.3.
Theorem: Divisibility by a Prime
Every integer n > 1 is divisible by some prime number.
Exercise:
Write the theorem of divisibility by a prime using logic
notation.
Proof:
We need to show:
∀n ∈ , ( n > 1 ⇒ ∃ a prime number p, p | n )
Let n ∈ , n > 1.
Then there are two possibilities:
a. If n is prime, let p = n .
Then p | n and the result follows.
b. If n is not prime, then
∃r1 , s1 ∈ ,1 < r1 < n and 1 < s1 < n
n = r 1 s1 .
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Consider r 1 > 1, again there are two possibilities:
A. If r 1 is prime, let p = r 1 .
Then p | r 1 and r 1| n so p | n by transitivity of
divisibility
B. If r 1 is not prime, then
∃r 2 , s 2 ∈ ,1 < r 2 < r 1 and 1 < s 2 < r 1
r1 = r 2 s2 .
Consider r 2 > 1, again there are two possibilities
A. If r 2 is prime, let p = r 2 .
Then p | r 2 and r 2| r 1 and r 1| n so p | n by
transitivity of divisibility.
B. If r 2 is not prime, then we factorise as with r 1
and n.
We continue in this way until we find a prime factor.
This process will finish after a finite number of steps
because each new factor is less than the previous one and
greater than 1.
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Thus, we obtain the list r 1 r 2 r 3 ,K , r k where
n > r 1 > r 2 > r 3 > K > r k and each r i | n .
The list ends when r k is prime.
Therefore, let p = r k , then p | n and the result follows.
Example:
Find a prime factor of 693.
693 = 9 × 77
9 and 77 are not prime
Let r 1 = 9
9 = 3×3
∴ Let p = 3
3 is prime
Exercise:
Find a prime factor of 48.
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4.2.4.
Theorem: Infinite Number of Primes
The number of primes is infinite.
Proof:
Suppose there is a finite number of primes; that is, we can
list all prime numbers as follows: p1 = 2,
p 2 = 3,K , p n
Therefore, p n is the largest prime number.
Consider X = p1 p 2 p3 K p n + 1.
Clearly X > p n .
Since p n is the largest prime, X must be composite so, by
the Theorem of Divisibility by a Prime, X is divisible by a
prime.
Consider:
X
1 1
= p 2 p3 K p n +
.
∉  , so p1 /| X
p1
p1 p1
X
1
1
= p1 p3 K p n +
.
∉  , so p 2 /| X
p2
p2
p2
Similarly, p 3 /| X . p 4 /| X , …, p n /| X .
Therefore, no prime number divides X.
This is a contradiction, so our assumption must be
incorrect. Therefore, since the number of primes is not
finite, it must be infinite.
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4.3.
Quotient Remainder Theorem
Exercise:
Evaluate the following division without using decimals.
(Nor a calculator!)
4 11
We can express the previous division as 11 = 4 × 2 + 3 .
Note that the remainder, 3, is less than 4.
4.3.1.
Theorem: The Quotient-Remainder
Theorem.
If n and d > 0 are both integers, then there exist unique
integers q and r such that n = dq + r and 0 ≤ r < d .
Proof:
We know that n, d ∈ , and d > 0 .
The proof of this theorem involves proving two main
things:
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A.
Existence:
Prove that: ∃q, r ∈ , n = dq + r , 0 ≤ r < d .
Let S = {n − dk : n − dk ≥ 0, k ∈ }.
We want to show S is nonempty.
If n ≥ 0 , we can select k = 0 :
n −0⋅d = n ≥ 0
∴n −0⋅d ∈S
If n < 0 , we can select k = n :
n − n ⋅ d = n(1 − d ) ≥ 0, since n < 0 and 1 − d ≤ 0
∴n − n⋅d ∈S
Therefore S is nonempty.
Note that S ⊆ 0 and that 0 is well-ordered.
Therefore by the well-ordering principle, S contains a
least element r.
Then for some specific integer k = q
n − dq = r
∴ n = dq + r
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We prove r < d by contradiction.
Suppose r ≥ d .
n − d ( q + 1) = n − dq − d
= r−d
≥0
Since n − dq = r , n − d (q + 1) would be a nonnegative
integer in S that would be smaller than r.
But r is the smallest integer in S.
We have a contradiction.
Therefore our assumption, r ≥ d , was wrong, and so
r<d.
Uniqueness:
B.
Prove that for n = dq + r , r and q are unique.
Suppose that:
n = dq + r
n = dq1 + r 1
(1)
q1 ≠ q, r 1 ≠ r (2)
0 = d ( q − q1 ) + r − r 1
(1) − ( 2)
∴ d ( q − q1 ) = r 1 − r
Now q, q1 ∈  and q1 ≠ q .
∴ q − q1 ≥ 1.
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∴ r1 − r ≥ d since r1 ≠ r
(3)
0≤r<d
∴ −d < −r ≤ 0
and 0 ≤ r1 < d
∴ − d < r1 − r < d
i.e. r1 − r < d
This contradicts (3), so our assumption is wrong.
∴ q1 = q, r 1 = r
Therefore for n = dq + r , r and q are unique.
Summary of Proof:
The proof of this theorem involved proving two main
things:
Existence:
∃q, r ∈ , n = dq + r , 0 ≤ r < d
Let S = {n − dk : n − dk ≥ 0 and k ∈ }.
Apply well-ordering to get a least value r with
corresponding k = q , so that n − dq = r .
Prove that r < d by assuming r ≥ d and showing that
this gives a smaller element of S, which contradicts the
fact that r is the least element of S.
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Uniqueness:
For n = dq + r , r and q are unique. (i.e. the only ones).
Assume that there are two values each for q and r, and
prove by contradiction that the values are really the
same.
The Quotient-Remainder Theorem says that when we
divide any integer n by any positive integer d, there will
be a quotient q and a remainder r, where 0 ≤ r < d .
Example:
Find values for q, r ∈ , such that n = dq + r , 0 ≤ r < d , for
the following:
•
n = 54, d = 4
54 = 4 × 13 + 2
∴ q = 13, r = 2
•
n = −32, d = 9
− 32 = 9 × ( −4) + 4
∴ q = −4, r = 4
WUCT121
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Exercises:
Find values for q, r ∈ , such that n = dq + r , 0 ≤ r < d , for
the following:
•
n = −54, d = 4
•
n = 54, d = 70
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4.3.2.
Result: Even or Odd
Every integer is either even or odd.
Proof:
We must prove the statement
∀n ∈ , n is even ∨ n is odd
i.e.
∀n ∈ , (∃k ∈ , n = 2k ) ∨ (∃l ∈ , n = 2l + 1 )
Let n ∈  .
Now, n and 2 > 0 are integers, so we apply the QuotientRemainder Theorem to get:
∃q, r ∈ , n = 2q + r , 0 ≤ r < 2 .
For 0 ≤ r < 2 to be true, r = 0 or r = 1.
Therefore, n = 2q + 0 = 2q or n = 2q + 1 and the result
follows.
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4.4.
The Fundamental Theorem of Arithmetic
4.4.1.
Theorem: The Fundamental Theorem of
Arithmetic
If a ∈  and a > 1 then a can be factorised in a unique way
in the form
a = p1α1 p 2 α 2 p 3α 3 ... p k α k
where p1, p 2, K , p k are each prime numbers and α i ∈ 
for each i = 1, 2,K , k .
Example:
Write the following numbers in terms of their prime
factors.
•
32
32
2
16
2
8
2
4
2
2
32 = 25
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93
•
924
924
2
462
2
231
3
77
7
11
924 = 22 × 3 × 7 × 11
Exercises:
Write the following numbers in terms of their prime
factors.
•
1300
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94
•
2772
•
50193
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4.5.
Finding Prime Numbers
4.5.1.
Algorithm: The Sieve of Eratosthenes.
The Sieve of Eratosthenes is a method of finding primes up
to n as follows.
A. List all the primes up to
⎣ n⎦
B. Write down all integers from 1 to n, noting the listed
primes
C. Delete all multiples of the listed primes
D. The remaining values are the prime numbers up to n.
Note:
⎣ x ⎦ is the floor function and is defined as:
⎣ x ⎦ is the greatest integer that is less than or equal to x.
That is ⎣ x ⎦ = n, where n ∈ , n ≤ x < n + 1
Example:
⎣
117 ⎦ = ⎣10.82⎦ = 10
Exercise:
Find
⎣
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200 ⎦
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Example:
Find all primes between 1 and 100.
A.
⎣
100 ⎦ = 10 . Primes to 10 are: 2, 3, 5, 7
B. List the numbers 1 – 100 noting the primes 2, 3, 5, 7.
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15 16 17 18 19
21 22 23 24 25 26 27 28 29
20
30
31 32 33 34 35 36 37 38 39
40
41 42 43 44 45 46 47 48 49
50
51 52 53 54 55 56 57 58 59
60
61 62 63 64 65 66 67 68 69
70
71 72 73 74 75 76 77 78 79
80
81 82 83 84 85 86 87 88 89
90
91 92 93 94 95 96 97 98 99 100
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97
C. (i) Delete multiples of 2
1
2
3
9
10
11 12 13 14 15 16 17 18 19
21 22 23 24 25 26 27 28 29
20
30
31 32 33 34 35 36 37 38 39
40
41 42 43 44 45 46 47 48 49
50
51 52 53 54 55 56 57 58 59
60
61 62 63 64 65 66 67 68 69
70
71 72 73 74 75 76 77 78 79
80
81 82 83 84 85 86 87 88 89
90
4
5
6
7
8
91 92 93 94 95 96 97 98 99 100
(ii) Delete multiples of 3
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15 16 17 18 19
21 22 23 24 25 26 27 28 29
20
30
31 32 33 34 35 36 37 38 39
40
41 42 43 44 45 46 47 48 49
50
51 52 53 54 55 56 57 58 59
60
61 62 63 64 65 66 67 68 69
70
71 72 73 74 75 76 77 78 79
80
81 82 83 84 85 86 87 88 89
90
91 92 93 94 95 96 97 98 99 100
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98
(iii) Delete multiples of 5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15 16 17 18 19
21 22 23 24 25 26 27 28 29
20
30
31 32 33 34 35 36 37 38 39
40
41 42 43 44 45 46 47 48 49
50
51 52 53 54 55 56 57 58 59
60
61 62 63 64 65 66 67 68 69
70
71 72 73 74 75 76 77 78 79
80
81 82 83 84 85 86 87 88 89
90
91 92 93 94 95 96 97 98 99 100
(iv) Delete multiples of 7
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15 16 17 18 19
20
21 22 23 24 25 26 27 28 29
30
31 32 33 34 35 36 37 38 39
40
41 42 43 44 45 46 47 48 49
50
51 52 53 54 55 56 57 58 59
60
61 62 63 64 65 66 67 68 69
70
71 72 73 74 75 76 77 78 79
80
81 82 83 84 85 86 87 88 89
90
91 92 93 94 95 96 97 98 99 100
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D. Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Exercise:
Find all primes between 101 and 200.
A.
B.
C.
101 102 103 104 105 106 107 108 109 110
111 112 113 114 115 116 117 118 119 120
121 122 123 124 125 126 127 128 129 130
131 132 133 134 135 136 137 138 139 140
141 142 143 144 145 146 147 148 149 150
151 152 153 154 155 156 157 158 159 160
161 162 163 164 165 166 167 168 169 170
171 172 173 174 175 176 177 178 179 180
181 182 183 184 185 186 187 188 189 190
191 192 193 194 195 196 197 198 199 200
D.
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Numbers
100