Download Notes for Numbers

Contents
Numbers Strand
Section
Topic
Page
6
The Natural Numbers
1
7
Mathematical Induction
5
8
The Integers
11
9
Elementary Number Theory
14
10
Congruence Arithmetic
28
1
Section 6.
The Natural Numbers
6.1 The Natural Numbers
The Natural Numbers, denoted N , consist of all positive \whole" numbers.
N = f: : : ; 2; 3; 4; : : :g
Discussion:
1. Is 0 an element of N (notation: 0 2 N )?
2. Is 175964 2 N ? Is 1010
10
2 N?
Is there a largest element? If there is, what is it? If there isn't,
can you prove it?
3. What's the smallest element of N ?
4. Let
E
= f2; 4; 6; 8; : : :g that is, the even numbers which are elements of N . What is the largest
element? What is the smallest element?
5. What are the four main \things" we can do with the elements of N ? [Think about what you
could do by Year 4!!]
Denition 6.1.1. A closed
operation
is a rule for combining any two elements of a set that always
produces another element in the same set. We say the set is
Exercise:
Which of these are closed operations on N ?
Addition [4 + 2 = 6]
Subtraction [4
2 = 2]
Multiplication [4 2 = 8]
Division [4 2 = 2]
Why or why not?
closed
under the operation.
2
For now we are only interested in the closed operations (that is, + and
).
Exercise:
Carefully write down every small step you take when simplifying the following expressions without
the aid of a calculator. Can you give reasons for each step?
(i) 3x + 4y + 2x + y
(ii) (157 + 25) + 75
(iii) (x + 3)(x + 9)
6.2 Properties of Addition and Multiplication
6.2.1 Commutativity
Let
a; b
2 N . Then
(i)
a+b=b+a
(ii) a
b=ba
We say that the natural numbers (N ) are commutative under addition and multiplication.
6.2.2 Associativity
Let
a; b; c
2 N . Then
(i)
(a + b) + c = a + (b + c)
(ii) (a
b) c = a (b c)
We say N is associative under addition and multiplication. This property allows us to omit parenthesis in some expressions.
6.2.3 Distributivity
Let
Eg.
a; b; c
(3 + 4) + 2 is written 3 + 4 + 2.
2 N . Then
(i)
a (b + c) = a b + a c
(ii) (a + b)
c = ac+bc
3
Exercise:
Let x; y 2 N . Simplify the following expression giving reasons for your steps.
[8(x + y )] + 2x
6.3 Order Properties of the Natural Numbers
Discussion:
1. Given two natural numbers a; b, what are the three possible orderings of the two numbers?
2. Write down a denition for \a < b".
3. Use your denition in question 2 to prove the following.
(i) 5 < 20
(ii) If
x; y; s; t
2 N with
x < y
and
, then
s < t
x
+ s < y + t.
The Natural Numbers, N , is a set of numbers with order properties{these are more formally expressed as
the
Law of Trichotomy.
Property 6.3.1 The Law of Trichotomy.
relationships hold:
(i)
(ii)
(iii)
a < b
a
=b
a > b
If
a; b
2 N , then one and only one of the following
4
Discussion:
1. Let's turn our attention for a moment to the real numbers, represented by R . Consider the
following sets of real numbers.
1 3 5
A =
0; 2 ; 1; 2 ; 2; 2
B
C
D
= fx 2 R : 0 < x 1g
= 12 ; 31 ; 14 ; 15 ; : : :
= f: : : ; 10; 8; 6; 4; 2; 2; 4; 6; 8; 10; : : :g
Write down the smallest and largest elements in each set.
2. Consider these sets of natural numbers.
E
O
P
= f2; 4; 6; 8; : : :g
= f1; 3; 5; 7; 9; : : :g
= f2; 3; 5; 7; 11; : : :g
What is the smallest element in each set?
3. Write down any non-empty subset of N . Does it have a smallest element? Can you give a subset
of N that does not have a smallest element?
The last question above demonstrates an important order property of the natural numbers.
Property 6.3.2 The Well-Ordering Property for N .
If
A
is a non-empty subset of N , then
A
has a least element.
We say that N is a
well-ordered set.
Denition 6.3.3. We say a set is well-ordered if every non-empty subset has a least element.
Exercise:
Are the following sets well-ordered?
(i) [0; 1) = fx 2 R : 0 x < 1g, the interval of real numbers between 0 and 1, including 0
(ii) R , the set of all real numbers
(iii)
E
= f2; 4; 6; 8; : : :g, the set of even natural numbers
The well-ordering property is an important statement about N . It can sometimes be used to prove an
innite number of claims as we shall see in the next section.
5
Section 7.
Mathematical Induction
7.1 The Basic Principle of Mathematical Induction
Example.
Prove n3 > 2n
2 for all natural numbers, n.
Let CLAIM(n) be \n3 > 2n
For n = 1:
2".
CLAIM(1) is 13 > 2 1
For n = 2:
CLAIM(2) is 23 > 2 2
For n = 3:
CLAIM(3) is ...
2.
LHS = 1; RHS = 0
2.
LHS = 8; RHS = 2
=)
=)
LHS
>
RHS.
LHS
>
RHS.
How do we prove CLAIM(n) for all other values of n? The process above will NEVER prove the CLAIM
for ALL natural numbers n.
Therefore, we need to use the following theorem that is a result of the well-ordering property on N .
Theorem 7.1.1
The Principle of Mathematical Induction.
For all natural numbers n, let
CLAIM(n) be a statement. If
1. CLAIM(1) is true, and
2. for all natural numbers k 1, if CLAIM(k ) is true, then CLAIM(k + 1) is also true,
then CLAIM(n) is true for all natural numbers n, that is, for all n 2 N .
Proof: Let T = ft 2 N : CLAIM(t) is falseg.
If we can show that T is empty, then CLAIM(n) is true for all n 2 N . Note that T is a subset of N . Also,
we know the following:
1. CLAIM(1) is true.
2. For all natural numbers k 1, if CLAIM(k ) is true, then CLAIM(k + 1) is also true.
Assume
is not empty. Then
T
T
has a least element. [Why?] Let this least element be t0 , so
CLAIM(t0 ) is false.
By statement 1 above, t0 =
6 1, so t0
CLAIM(t0
(*)
1 is a natural number. Also, t0
1 is not in
T
. Therefore,
1) is true. However, by statement 2 above,
CLAIM(t0
1) true implies that CLAIM(t0
1 + 1) = CLAIM(t0 ) is true.
This is a contradiction with statement (*). So, our assumption that T is not empty must be
Hence,
T
must be empty. Therefore, CLAIM(n) is true for all n 2 N .
Example.
( + 1)
for n 2 N .
2
n(n + 1) "
Let CLAIM(n) be \1 + 2 + : : : + n =
.
2
Prove 1 + 2 + : : : + n =
n n
Does the claim satisfy the conditions of the Principle of Mathematical Induction?
false.
6
1. Is CLAIM(1) true?
1(1 + 1)
.
LHS = 1; RHS = 22 = 1.
CLAIM(1) is 1 =
2
Therefore, LHS = RHS and so CLAIM(1) is true.
2. We must show that for k 1, CLAIM(k ) true =) CLAIM(k + 1) true.
k (k + 1)
Assume CLAIM(k ) is true, that is, 1 + 2 + : : : + k =
.
2
(k + 1)(k + 1 + 1)
Prove CLAIM(k + 1) is true, that is, 1 + 2 + : : : + k + (k + 1) =
.
2
(k + 1)(k + 2)
LHS = 1 + 2 + : : : + k + (k + 1)
RHS =
2 2
k (k + 1)
k + 3k + 2
=
+ (k + 1)
=
2
2
2
k + k + 2k + 2
=
2
2
k + 3k + 2
Therefore, LHS = RHS.
=
2
Therefore, CLAIM(k + 1) is true.
So, CLAIM(n) satises all the conditions of the Principle of Mathematical Induction.
Therefore, by the Principle of Mathematical Induction, CLAIM(n) is true for all n 2 N .
In summary:
To establish an innite family of claims, (CLAIM(1), CLAIM(2), CLAIM(3),
:::
) using the Principle of
Mathematical Induction, it is suÆcient to carry out the following two steps:
Step 1: Prove that the rst claim, CLAIM(1) is true.
Step 2: Give a general proof that for each k 2 N , if CLAIM(k) is true, then CLAIM(k + 1) is also
true.
Example.
Returning to our previous example (on the previous page), we had CLAIM(n) :
Step 1:
From earlier work, CLAIM(1) is true.
Step 2:
Assume CLAIM(k ) is true for k 1, that is, k 3 > 2k
2 ... (1)
Prove CLAIM(k + 1) is true, that is, prove that (k + 1)3
Note: RHS = 2k + 2
3 > 2n 2.
n
>
2(k + 1)
2
2 = 2k .
LHS = (k + 1)3
= k 3 + 3k 2 + 3k + 1
>
2k
2
k
2 +3k2 + 3k + 1
2+ 3 + 3 + 1
(by hypothesis (1))
(since
k
1)
= 2k + 5
>
2k
= RHS
Therefore, (k + 1)3 > 2(k + 1)
2 and so CLAIM(k + 1) is true.
Therefore, by the Principle of Mathematical Induction, CLAIM(n) is true for all n 2 N .
7
Aside:
The expression \n!" reads \n factorial" and is dened by n! = n(n
2):::3 2 1.
1)(n
For example, 5! = 5 4 3 2 1 = 120.
Example.
Prove (n + 1)! 2n for all n 2 N .
Let CLAIM(n) be (n + 1)! 2n .
CLAIM(1) is (1 + 1)! 21 .
Step 1:
RHS = 21 = 2.
LHS = (1 + 1)! = 2! = 2;
RHS and so CLAIM(1) is true.
Assume CLAIM( ) is true for 2 N , that is, ( + 1)! 2
Therefore, LHS
Step 2:
k
k
k
k
... (1)
Prove CLAIM(k + 1) is true, that is, prove that [(k + 1) + 1]! 2k+1 .
LHS = (k + 2)!
= (k + 1)!(k + 2)
2
k
>
(k + 2)
2k 2
= 2k+1
(by hypothesis (1))
(as
k
1)
= RHS
Therefore, CLAIM(k + 1) is true.
Therefore, by the Principle of Mathematical Induction, CLAIM(n) is true for all n 2 N .
Aside: Notes on Divisibility
The expression \a
j
b
" reads \a divides b", or, \a is a factor of b", or, \a is a divisor of b". It means \b
is divisible by a", or, \b is a multiple of a", or more formally, there is an l 2 N so that b = al.
Examples.
(i)
6 j 12
(ii) 7 j (a2
2)
()
there is l 2 N ; a2
2 = 7l.
(iii) 6 6 j 25 reads \6 does not divide 25".
When a proof involves showing a j b, you need to express b as a multiple of a, that is,
b
nd l 2 N so that
= al.
Example.
To prove 3 j (4k+1
1), you need to nd an expression for l 2 N so that 4k+1
Divisibility will be covered in more detail in Section 9.
1 = 3l.
8
Example.
Prove that 6 j (3n2 + 3n) for all n 2 N .
Let CLAIM(n) be 6 j (3n2 + 3n).
Step 1:
CLAIM(1) is 6 j (3(1)2 + 3(1)), or 6 j 6 after simplifying.
This is true, therefore CLAIM(1) is true.
Step 2:
Assume CLAIM(k ) is true for k 2 N , that is, 6 j (3k 2 + 3k ),
or there is
l
2N
;
3k2 + 3k = 6l ... (1)
Prove CLAIM(k + 1) is true, that is, prove that 6 j (3(k + 1)2 + 3(k + 1)).
We need to nd an expression for m 2 N (using l) so that
3(k + 1)2 + 3(k + 1) = 6m.
Now, 3(k + 1)2 + 3(k + 1) = 3k 2 + 6k + 3 + 3k + 3
= (3k 2 + 3k ) + (3k + 3k + 6)
= 6l + 6k + 6
(by hypothesis (1))
= 6(l + k + 1)
Let
m
=l+k+12N
= 6m
Therefore, 6 j (3(k + 1)2 + 3(k + 1)) and so CLAIM(k + 1)is true.
Therefore, by the Principle of Mathematical Induction, CLAIM(n) is true for all n 2 N .
7.2 The General Principle of Mathematical Induction
Example.
Prove that 2n > 2n + 1.
Let CLAIM(n) be 2n > 2n + 1 for n 2 N .
Step 1:
CLAIM(1) is 21 > 2(1) + 1.
LHS
<
LHS = 2; RHS = 3
RHS!!!
The fact is that 2n > 2n + 1 is true for n 3, n 2 N .
Sometimes we would like to prove a CLAIM that only works for values greater than 4, say. The Principle
of Mathematical Induction can be generalised to \accommodate" this type of problem.
Theorem 7.2.1 The Generalised Principle of Mathematical Induction. For all natural numbers
, let CLAIM(n) be a statement and let
n
q
2 N . If
1. CLAIM(q ) is true, and
2. for all natural numbers k q , if CLAIM(k ) is true, then CLAIM(k + 1) is also true,
then CLAIM(n) is true for all natural numbers n q .
Proof: Left as an exercise as it is almost identical to the proof to Theorem 1.4.1.
9
This general principle is applied in the same way as before, except that at step 1, we prove CLAIM(q ) is
true and at step 2, we must note that
k
q
.
Exercise:
Prove that 2n > 2n + 1 for n 3, n 2 N .
Recall that the inequality is not true for
n
= 1 nor
n
= 2!
7.3 The Strong Principle of Mathematical Induction
This section is covered in the Discovery Class. However, you will need the following two ideas.
X
7.3.1 Sigma Notation
The notation
X4
n
i=1
ai
denotes the
=1
X
n
i=1
Examples.
(i)
sum of the numbers a1 , a2 , a3 , : : :, an ; that is,
ai
= a1 + a2 + a3 + : : : + an.
2i = (2 1) + (2 2) + (2 3) + (2 4)
i
(ii)
X12
=4
1=1+1+1+1+1+1+1+1+1
i
=2+4+6+8
= 20
Note. The expression
X1
=1
ai
=9
will NOT be used in MATH121|it has a special meaning.
i
X
n
We will use sigma notation to \shorten" expressions like 2 + 7 + 12 + : : : + (5n
3) =
=1
i
(5i
3):
10
It is important that you satisfy yourself that the following equalities hold.
(i)
(ii)
(iii)
X = +X
X =X +
X =X
n
n
i=1
n
i=1
n
i=1
a1
ai
1
n
ai
ai
i=1
n
ai
X(
X
i=2
j =1
ai
an
aj
X
X
Also, it is easy to prove the following statements.
(iv)
(v)
n
ai
i=m
n
i=m
Discussion:
True or False?
kai
X3
i
=1
n
+ bi ) =
=k
2i
X
ai
i=m
n
i=m
ai
,
n
+
n
i=m
m
bi
,
n
m
.
1 = 9.
7.3.2 Recursive Denitions
Sequences of numbers are dened
recursively if each new number depends on previous values.
Example.
Let u1 ; u2 ; : : : ; un ; : : : be real numbers dened by
un
= 5un 1
6un 2 , for n 3
and u1 = 2, u2 = 4.
Find u4 :
u
3 = 5u2
=54
6u1
(n = 3)
4 = 5u3
u
62
=58
=8
The numbers can often be dened directly. For example,
6u2
(n = 4)
64
= 16:
un
= 2n in the previous example (check for
yourself!). The Discovery Class introduces a \strong" mathematical induction to prove this type of direct
denition for un .
11
Section 8.
The Integers
The set of Integers, denoted Z, consists of all \whole" numbers, positive, negative and zero.
Z = f: : : ; 3; 2; 1; 0; 1; 2; 3; : : :g
Discussion:
1. Represent Z geometrically.
2. What special element do we now have that was lacking in N ?
3. How can subtraction be redened in terms of addition? Why would we want to do this?
4. What property of N do we lose?
5. What relationship is there between N and Z?
Notes.
1. When graphing the integers on the number line, there are \gaps".
every point on the number line.
The integers
do not
comprise
[Which set of numbers does cover the entire number line?]
2. Under addition, zero satises an important property:
For all a 2 Z, a + 0 = 0 + a = a.
0 is called the identity under addition. This property is the reason why division by 0 is undened!
For similar reasons, the symbol 0 is used to \ll" places in numbers such as 1; 029. And, how would
you perform long multiplication without 0? For example, multiplying XC by CMLXI [90 961] is
diÆcult using only Roman numerals!
3. The fact that subtraction of integers can be redened in terms of addition leads to an interesting
property of Z:
For all a 2 Z, a
We say that
a
= a + ( a) = ( a) + a = 0.
a is the inverse of a under addition.
12
4. Z has the properties of commutativity under addition and multiplication, associativity under addition
and multiplication, and
distributivity.
Z also has the order properties described by the
ordered. Even so, it's not as bad as it sounds.
Law of Trichotomy.
Any non-empty subset of
However, Z is not well-
Z that
is bounded below is
well-ordered.
For example, W = f 100; 99; 98; : : :g is well ordered. Can we say the same for real numbers?
8.1 Even and Odd Integers
Discussion:
1. Write down a denition for an even integer.
Write down a denition for an odd integer.
2. Is 0 an even integer?
3. Is
4. If
301 an odd integer? If so, demonstrate this using the denition.
a; b
2 Z, explain from the denition why 6
2
a b
is even.
5. Is every integer either even or odd?
Recall that \p
()
q
" says two things,
p
=)
q
and
q
=)
.
p
Since our denition for evens and odds uses a \ () ", we can use it in two ways.
(i) Knowing that
n
is even (odd), deduce there is a k 2 Z such that
(ii) Knowing that there is a k 2 Z such that
n
n
= 2k (n = 2k + 1).
= 2k (n = 2k + 1), deduce that
n
is even (odd).
13
8.2 Prime and Composite Numbers
Denition 8.2.1. An integer n > 1 is prime if and only if for all positive integers r and s, if n = r s,
then r = 1 or s = 1.
Exercise:
1. Write the expression \if n = r s, then r = 1 or s = 1" using logic.
2. Write down the negation of the logic expression in question 1.
3. What kind of integer do you think this statement about
n
describes?
Denition 8.2.2. An integer n > 1 is composite if and only if there exist positive integers r and s so
that
n
= r s and r 6= 1 and s 6= 1.
Discussion:
1. Is 1 prime or composite?
2. Is it true that every integer greater than 1 is either prime or composite?
3. Write down the rst six prime numbers.
4. Write down the rst six composite numbers.
5. Is 2 the only even prime number? Can you prove this?
14
Section 9.
Elementary Number Theory
Challenge:
If all the odd numbers between 1 and 100 are multiplied together, what is the second last digit?
Outline how to nd this digit without calculating the product.
9.1 Introductory Examples
Within the context of Number Theory, this section demonstrates the dierent techniques of proving
general statements that were introduced in Section 3.
Example 9.1.1
. Prove that the sum of any two odd numbers is even.
Firstly, you should check a few examples to conrm (for yourself) that the statement is true.
3 + 5 = 8;
9 = 17 = 26;
7 + 11 = 18
Write down what you are asked to prove in \mathematical language" (if it is not already).
8a; b 2 Z; (a; b odd =) a + b 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.
a is odd =) 9k 2 Z; a = 2k + 1
b is odd =) 9l 2 Z; b = 2l + 1
Write down what you are trying to
prove,
and what that means (ie, what you have to \get to").
Prove that a + b is even, that is, 9s 2 Z; a + b = 2s.
What you do at this point varies from proof to proof. Whenever you have an equality, it is best to start
with the LHS and work to the RHS.
a + b = (2k + 1) + (2l + 1)
(hypothesis)
= 2l + 2k + 2
= 2(l + k + 1)
(Distributivity)
Let s = l + k + 1 and s 2 Z (as addition is a closed operation on Z).
Then, a + b = 2s and so by denition, a + b is even.
The following example gives a formal proof of the result discussed at the end of Section 8.
Example 9.1.2
. 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.
15
Suppose there exists another even prime, n, such that n > 2. Then, 9k 2 Z; n = 2k .
Now, 2 is positive and 2 6= 1. Also, k is positive (as n > 2) and k 6= 1.
[If k were equal to 1, then n = 2. However, we know that n > 2.]
Therefore, we have positive integers 2 and k where n = 2 k and k 6= 1, 2 6= 1.
Thus, by denition, 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.
Discussion:
Determine if the following statements are true or false. In each case, prove your answer.
(i)
8n 2 N ; (n is prime =) n is odd)
(ii)
8a; b 2 Z; (a2 = b2 =) a = b)
9.2 Divisibility
Denition 9.2.1. If n and d are integers and d 6= 0, then n is
for some k 2 Z. We write d j n and say that d is
divisible by
a divisor of n
d if and only if n = d k
.
In logic notation, the denition of divisibility is written
djn
() 9k 2 Z; n = dk.
Discussion:
1. Is -16 a divisor of 32?
2. If l 2 Z and l 6= 0, does l j 0?
3. Find all values of a 2 Z such that a j 1.
4. What is the relationship between a and b if a j b and b j a, a; b 2 Z?
continued over ...
16
5. If a; b 2 Z, is 3a + 3b divisible by 3?
6. Let a; b; c; x; y 2 Z. If b j a and b j c, does b j (ax + cy )? Why?
7. If a; b 2 N , is it true that a j b implies a b?
8. What do you make of 0 j 0?
Result 9.2.2 Transitivity of Divisibility. For all integers a, b and c, if a j b and b j c, then a j c.
Proof:
a j b =) 9k 2 Z; b = ka
We know that
b j c =) 9l 2 Z; c = bl
(1)
(2)
Show that a j c, that is, nd m 2 Z such that c = ma.
Now,
c = bl
by (2)
= (ka)l
by (1)
= (kl)a
by Associativity and Commutativity
Let m = kl, then m 2 Z (since is a closed operation on Z). Therefore, c = ma and so a j c.
Theorem 9.2.3. Every integer n > 1 is divisible by some prime number.
Proof: In logic notation, we must prove
8n 2 Z; (n > 1 =) 9 a prime number p; p j n).
Let n > 1. Then there are two possibilities.
A. If n is prime, let p = n. Then p j n and the result follows.
B. If n is not prime, then 9r1 ; s1 2 N ; n = r1 s1
^ r1 < n ^ r1 6= 1 ^ s1 6= 1.
Consider r1 > 1. Again, there are two possibilities.
a. If r1 is prime, let p = r1 . Then p j n and the result follows.
b. If r1 is not prime, then 9r2 ; s2 2 N ; r1 = r2 s2
^ r2 < r1 ^ r2 6= 1 ^ s2 6= 1.
Consider r2 > 1. Once again, there are two possibilities.
i. If r2 is prime, let p = r2 . Then p j r1 and r1 j n giving p j n by transitivity.
ii. If r2 is not prime, then we factorise as with r1 and n.
17
We continue in this way until we nd a prime factor. This process will nish after a nite number of
steps because each new factor is less than the previous one and greater than 1. Thus, we obtain the list
r1 ; r2 ; r3 ; : : : ; rk
where n > r1 > r2 > r3 > : : : > rk and each ri j n. The list ends when rk is prime.
Therefore, let p = rk , then p j n and the result follows.
Theorem 9.2.4. The number of primes is innite.
Proof: Suppose there is a nite number of primes; that is, we can list all prime numbers as follows.
p1 = 2; p2 = 3; : : :; pn
Therefore, pn is the largest prime number. Consider
X = p1 p2 p3 : : : pn + 1.
Clearly, X > pn . Since pn is the largest prime, X must be composite. By Theorem 3.2.3, X is divisible
by a prime. Consider
X
1
= p2 p3 : : : pn +
So, p1 6 j X
p1
p1
1
X
= p1 p3 : : : pn +
So, p2 6 j X
p2
p2
Similarly, p3 6 j X , p4 6 j X , : : :, pn 6 j 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 nite, it must be innite.
9.3 The 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.
Theorem 9.3.1 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: The proof of this theorem involves proving two main things:
1.
Existence: There exist q; r 2 Z such that n = dq + r and 0 r < d.
2.
Uniqueness: The values of q and r are unique (ie, the only ones).
18
We know that n; d 2 Z and d > 0.
1.
Existence. Prove that 9q; r 2 Z; n = dq + r; 0 r < d.
Let S = fn
dk : n dk 0 and k 2 Zg.
If n 0, then n
d 0 = n 0 and so n 2 S .
If n < 0, then consider n d n = n(1 d) 0 as both n < 0 and 1 d 0. So, n d n 2 S .
Note that S N [ f0g and that N [ f0g is well-ordered.
Therefore, as S is non-empty, we can apply the well-ordering principle to conclude that S has a
least element r. Further, this value of r has a corresponding integer k = q , such that
r = n dq
=)
n = dq + r.
Now, to prove that r < d, suppose that r d (ie, r
d 0) and consider
n d(q + 1) = n dq d = r d 0.
Since r
d is of the form n dk with k = q + 1, we have r d 2 S with r d < r.
However, r was said to be the least element in S . So r
d 62 S . Thus, we have a contradiction.
Therefore, our assumption must be wrong, giving r < d.
2.
Uniqueness. Prove that for n = dq + r, r and q are unique.
Suppose that
and
n = dq + r
: : : (1)
n = dq1 + r1
: : : (2),
Subtracting (1)
(2), we have 0 = dq
where q 6= q1 and r 6= r1 .
dq1 + r r1
=) d(q
=) jr1
q1 ) = r1
r
(note: jq
q1 j 1)
rj d since q 6= q1 and r 6= r1
Now, 0 r < d, giving 0 r > d. Further, 0 r1 < d. Therefore,
d < r1
: : : (3).
r < d, that is,
jr1 rj < d. However, this contradicts (3). So, our assumption must be wrong.
Therefore, q1 = q and r1 = r.
Although the above proof to the Quotient-Remainder Theorem is not examinable, the following summary
is part of the syllabus.
Summary of Proof: The proof of this theorem involves proving two main things:
1.
9q; r 2 Z; n = dq + r; 0 r < d.
fn dk : n dk 0 and k 2 Zg.
Existence:
Let S =
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 .
2.
Uniqueness: For n = dq + r, r and q are unique. (ie, 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.
19
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.
Let n = 11, d = 4. Then, we have 11 = 4 2 + 3; that is, q = 2 and r = 4.
Exercise:
Find values for q; r 2 Z such that n = d q + r, 0 r < d, for the following.
(i) n = 54, d = 4
(ii) n = 54, d = 4
(iii) n = 54, d = 70
Can you nd a second set of q and r (r < d)?
Note. Theorem 9.3.1 said that r and q are unique, that is, there is only one solution for r and q in the
previous exercise.
Recall earlier we stated|without proof|that every integer is either odd or even. The proof of the
statement uses the Quotient-Remainder theorem as follows.
Result 9.3.2. Every integer is either odd or even.
Proof: We must prove the statement 8n 2 Z; (n is odd
Let n 2 Z. Prove that (9k 2 Z; n = 2k + 1)
_ n is even).
_ (9l 2 Z; n = 2l).
Now, n and 2 > 0 are integers, so we apply the Quotient-Remainder Theorem to get
9q; r 2 Z; 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.
9.4 Greatest Common Divisor
Denition 9.4.1. Let a; b 2 Z with a 6= 0 or b 6= 0. A natural number
common divisor
(gcd) of a and b if and only if
(i) c j a and c j b; and
(ii) if d j a and d j b for some d 2 N , then d j c.
We write gcd(a; b) = c.
c
is said to be the
greatest
20
Exercise:
1. Evaluate gcd(18; 12), gcd(18; 12), gcd( 18; 12) and gcd( 18; 12).
2. What can you say about gcd(a; b); gcd(a; b); gcd( a; b); gcd( a; b)?
3. Evaluate gcd(7; 11).
4. For a; b 2 Z where not both are zero, does gcd(a; b) always exist? Is it unique?
5. gcd(0; 0) is not allowed. Why? What would it be if it were found in the same way as for other
pairs of numbers?
6. Evaluate gcd(2772; 2310).
Lemma 9.4.2. If r is a positive integer, then gcd(r; 0) = r.
Proof: Let r > 0, r 2 Z, and let c = r. Clearly, c j r and c j 0. Obviously, any other divisor of r (all of
which also divide 0) must divide c as c = r. Therefore, gcd(r; 0) = c = r.
Lemma 9.4.3.
If a; b
2
Z with b =
6 0 and q, r > 0 are integers such 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) j gcd(b; r); and
(b) prove that gcd(b; r) j 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) Let gcd(a; b) = k 2 N . Then k j a and k j b; that is, 9m; n 2 Z; (a = km
Now, we are given that a = b q + r, r > 0. Substituting, we have
km = knq + r
Clearly, m
=)
km knq = r
nq 2 Z and so k j r.
=)
^ b = kn).
r = k(m nq)
Therefore, k is a common divisor of b and r, so k j gcd(b; r), that is, gcd(a; b) j gcd(b; r).
21
(b) Let gcd(b; r) = l 2 N . Then l j b and l j r; that is, 9m; n 2 Z; (b = lm
^ r = ln).
Therefore, l is a common divisor of a and b, so l j gcd(a; b), that is, gcd(b; r) j gcd(a; b).
Therefore, by (a) and (b), gcd(a; b) = gcd(b; r).
9.5 The Euclidean Algorithm
The Euclidean Algorithm is a process for nding the greatest common divisor between two integers. It
is presented in the Discovery Class.
The EÆciency of the Euclidean Algorithm.
If the Euclidean Algorithm is applied to two integers in the range 0 to m, m 8 (not both zero), then
2m
at most log3=2
divisions are required (more details and a proof can be found in Discrete Mathematics
3
by Richard Johnsonbaugh). The logarithm function grows slowly, so the Euclidean Algorithm is quite
eÆcient, even for large values. For example,
2(1; 000; 000)
log3=2
= 33:07:::
3
This indicates that the Euclidean Algorithm requires at most 34 divisions to calculate the greatest
common divisor of any pair of integers in the range 0 to 1; 000; 000!
Exercise:
For interested students.
Find out the maximum number of divisions you would need to nd gcd(486; 446), gcd(5; 9) and
gcd(7684; 4148). How many were used in each case (see the Discovery Class)?
Consider gcd(17; 9) =
We say that 17 and 9 are relatively prime.
Denition 9.5.1. If a; b 2 Z and gcd(a; b) = 1, then a and b are relatively
.
prime
22
Discussion:
1. Which of these pairs are relatively prime?
3, 8
27, 32
81, 33
72, 139
2. True or False?
8x 2 N ; 9y 2 N ; gcd(x; y) = 1
If true, what is y ? If false, give a counterexample.
3. Let p be prime, n 2 Z. What are the two possibilities for gcd(p; n)?
In the Discovery Class on the Euclidean Algorithm, you were presented with the following theorem.
Theorem 9.5.2 Theorem on Linear Combination of gcd. If a; b 2 Z and not both equal zero,
then gcd(a; b) exists and there exist m; n 2 Z such that gcd(a; b) = ma + nb.
Notes.
1. The theorem says two things.
(i) gcd(a; b) always exists (as long as both a and b are not both zero; and
(ii) gcd(a; b) can be written as a linear combination of a and b.
2. The proof to Theorem 3.5.2 is similar to that of the Quotient-Remainder Theorem. The proof can
be found on the MATH121 web site and will NOT be examined.
Corollary 9.5.3. If a and b are relatively prime, then 9m; n 2 Z such that 1 = ma + nb.
Proof: Exercise!
Corollary 9.5.4. Let a; b; c 2 Z, a; b 6= 0. If gcd(a; b) = 1 and a j bc, then a j c.
Proof: We must prove that 8a; b 2 Z; (gcd(a; b) = 1
^ a j bc) =) a j c .
We know that gcd(a; b) = 1 and a j bc, that is, 9q 2= Z; bc = qa.
We must prove that a j c, that is, we must nd k 2 Z such that c = ka.
By Corollary 3.5.3, gcd(a; b) = 1 implies that 1 = ma + nb for some m; n 2 Z.
Multiplying this equation by c we have c = mac + nbc.
However, bc = qa, so we have
c = mac + nqa = a(mc + nq)
Let k = mc + nq 2 Z, and so c = ak , that is, a j c.
23
9.6 The Fundamental Theorem of Arithmetic
Exercise:
1. Write the following numbers in terms of their prime factors.
32
924
2. Are these factorizations unique? How do you know?
3. Can you express EVERY natural number as a product of primes?
At this stage, the most we can say is that every integer greater than 1 is divisible by a prime. The
Fundamental Theorem of Arithmetic takes the idea further.
Theorem 9.6.1 The Fundamental Theorem of Arithmetic. If a 2 Z and a > 1, then a can be
factorised in a unique way in the form
a = p1 1 p2 2 p3 3 : : : pk k
where p1 , p2 , : : :, pk are each prime numbers and i 2 N for each i = 1; 2; : : : ; k .
The proof will not be covered in MATH121. This theorem answers questions in the last exercise: Each
integer greater than
is
the only one
1
(ie, not every one) can be decomposed into prime factors|and this decomposition
.
9.7 Finding Prime Numbers
The easiest method to nd prime numbers up to n is to use the Sieve of Eratosthenes.
Algorithm 9.7.1.
The Sieve of Eratosthenes
is a method of nding primes up to n as follows.
(i) Write down all integers from 1 to n.
p
(ii) Delete all multiples (not the number) of primes up to ( n + 1), that is, all multiples of 2; 3; 5; 7; : : :.
(iii) The remaining values are the prime numbers up to n.
24
Exercise:
Find all primes between 1 and 100.
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
Denition 9.7.2. If p and p + 2 are both prime numbers, then p and p + 2 are called twin
.
primes
Example.
5 and 7 are twin primes.
Exercise:
1. Write down all the sets of twin primes between 1 and 50.
2. Does 97 have a \twin" to form a pair of twin primes?
Conjecture 9.7.3. The number of twin primes is innite.
At this stage, we don't know whether this conjecture is true or not. It seems likely, but a proof is yet to
be found!
25
9.8 The Pigeonhole Principle
Principle 9.8.1 The Pigeonhole Principle. If n pigeons y into k pigeonholes and k < n, then
some pigeonhole contains at least two pigeons.
Proof: (By contradiction) Suppose each hole contains at most one pigeon. Since there are k pigeonholes
then we have at most k pigeons in holes. This contradicts that we know that n pigeons have own into
the holes and k is strictly bigger than n.
Example.
For any 8 people there is a day of the week when at least two people have birthdays.
In this case the people are pigeons so n = 8 and the days of the week are the pigeonholes (so k = 7).
Since n > k there is a day (pigeonhole) with at least two pigeons (people).
Exercise:
Four cards are selected out of a standard pack of 52 cards. Must there be at least two from the same
suit? What if 5 cards are selected?
Some Equivalent Mathematical Statements of the Pigeonhole Principle:
1. Let A be a set containing n elements. If A is partitioned into k pairwise disjoint subsets, where
k < n, then at least one of the subsets of the partition contains more than one element.
2. A function from one nite set to a smaller nite set cannot be one-to-one. That is there must be at
least 2 elements mapped to the same point.
Discussion:
In a group of 700 people must there be two whose family names have the same rst and last letter?
Why?
Example.
Five dierent numbers are selected from the set A = f1; 2; 3; 4; 5; 6; 7; 8g. Show that two of the
selected numbers sum to 9.
26
Note that there are 4 pairs that sum to 9, namely, 1 + 8 = 2 + 7 = 3 + 6 = 4 + 5 = 9. Each number
in A belongs to exactly one of these pairs. Dene a function on A by f (i) = pair containing i which
sums to 9. Now when we make our selection of 5 numbers we have a function from the 5 selections
to the 4 pairs. By the principle above two of the selections must be mapped to the same pair. That
is both elements of the pair are amongst our selections.
Principle 9.8.2 The Generalised Pigeonhole Principle. If n pigeons y into k pigeonholes and
for some number m, k:m < n, then some pigeonhole contains at least m + 1 pigeons.
Example.
In a group of 85 people the family name of at least 4 must start with the same letter.
Exercise:
1. We number 151 CSCI subjects using the numbers 1 to 300. Show that at least two of the
subjects must have consecutive numbers.
2. How many integers must you pick in order to be sure that at two of them have the same
remainder when divided by 15.
3. How many integers between 100 and 999 must you pick in order to sure that at least two of
them have a digit in common. (For example 256 and 530 have the common digit 5.)
continued over...
27
4. We want to assign 70 students to 11 classes so that no class has more than 15 people. Show
that there must be at least 3 classes with 5 or more. (Hint: try contradiction.)
5. Suppose a1 ; a2 ; : : : ; an are real numbers and
ai b=n.
n
X
i=1
ai b. Show that there is at least one ai with
28
Section 10.
Congruence Arithmetic
10.1 Introduction
A number of computer languages have built-in functions that compute the quotient and remainder of
division.
Denition 10.1.1. Let n 2 Z, d 2 N . We dene
(i)
n
(ii)
n
jdivj
jmodj
d
= the integer quotient obtained when n is divided by d.
= the integer remainder obtained when n is divided by d.
d
Exercise:
1. Calculate 32 jdivj 9 and 32 jmodj 9.
2. For n 2 N , write down all the possible answers for n jmodj 5.
Congruence Arithmetic centres around a relation based on the idea of \ jmodj ".
Denition 10.1.2. Let n 2 N . We will dene a relation on Z called congruence modulo
) by 8
a; b
2Z
;
a
(mod n)
b
()
n
j(
a
b
n
(denoted
).
Notes.
1. \a b(mod n)" reads \a is congruent to b modulo n."
2. The denition says that a b(mod n) if and only if n divides the dierence between a and b.
Example. 38 2(mod 6) because 38 2 = 36 and 6 j 36.
3. Another way to think about congruence modulo n is in terms of remainders:
a
b
(mod n) if and only if a jmodj n = b jmodj n, that is, if a and b have the same remainder after
being divided by n.
Example. 38 2(mod 6) because 38 jmodj 6 = 2 and 2 jmodj 6 = 2.
29
Exercise:
1. Find a value for m 2 N in each case.
(i) 14 m(mod 8)
3 m(mod 8)
(ii)
(iii) (n + 1) m(mod n);
n >
1
2. Determine whether each of the following is true or false.
(i) 7 9(mod 8)
(ii) 2 8(mod 11)
(iii) 11 1(mod 5)
3. Find a value for x 2 N in each case.
(i)
x
(ii)
x
(iii)
x
1(mod 7)
3(mod 5)
4(mod 9)
4. If m 0(mod 2), what can you say about m?
If n 1(mod 2), what can you say about n?
5. Fill in the spaces.
(i) 21 (mod 4)
(ii) 18 (mod 4)
(iii) 18 + 21 (mod 4)
Do you notice anything interesting?
10.2 Arithmetic of Congruences
Lemma 10.2.1. Let n 2 N , and a; b; c; d 2 Z.
(i) If a b(mod n) and c d(mod n), then (a + c) (b + d)(mod n) and ab cd(mod n).
(ii) If gcd(a; n) = 1 and ab cd(mod n), then b c(mod n).
Proof:
(i) We know that
a
c
b
d
(mod n)
(mod n)
=)
=)
n
n
j(
j(
9 2Z
=) 9 2 Z
), that is, j [(
)
a
b
c
d
)
We must prove that (a + c) (b + d)(mod n
nd l such that (a + c)
=)
k
;a
b
= nk
:::
(1)
q
;c
d
= nq
:::
(2)
n
a
+ c)
(b + d)]. To do this, we must
(b + d) = ln.
Adding equations (1) and (2), we have
a
+c
b
d
= nk + nq
=)
(a + c)
(b + d) = n(k + q ).
By letting l = k + q 2 Z, we have (a + c) (b + d)(mod n).
30
We now must prove that ac bd(mod n), that is, n j (ac
ac
bd
). To do this, we must nd s such that
bd
= sn. Consider
(1) c:
(2) b:
ac
bc
= nkc
:::
(3)
cb
db
= nqb
:::
(4)
Adding equations (3) and (4), we have
ac
bc
+ cb
db
=)
= nkc + nqb
ac
bd
= n(kc + qb).
By letting s = kc + qb 2 Z, we have ac bd(mod n).
(ii) We know that gcd(a; n) = 1 and
ab
(mod n). Now, by denition of congruence modulo n, we
ac
have
n
j(
ab
ac
)
=)
n
j
(
a b
c
)
using Corollary 3.5.4. Therefore, b c(mod n).
=)
n
j(
b
)
c
Examples of (ii).
(i) gcd(5; 4) = 1 and 5 6 5 2(mod 4) [that is, 30 10(mod 4)].
Also, 6 2(mod 4).
(ii) gcd(6; 3) = 3 and 6 1 6 2(mod 3) [that is, 6 12(mod 3)].
However, 1 6 2(mod 3).
Note that the condition that gcd(a; n) = 1 is necessary!
Lemma 10.2.1allows us to calculate remainders without actually performing a long division.
Example.
Find x such that 39 x(mod 5), 0 x < 5.
3 3(mod 5)
32 = 9 and 9 4(mod 5)
33 = 3 9; so 33 3 4(mod 5) 12(mod 5)
But 12 2(mod 5); so 33 2(mod 5)
36 = 33 33 ; so 36 2 2(mod 5) 4(mod 5)
39 = 33 36 ; so 39 2 4(mod 5) 8(mod 5)
But 8 3(mod 5); so 39 3(mod 5):
Exercise:
Find the remainder when 77 is divided by 16.
31
10.3 Equivalence Classes of Congruence Modulo n
Lemma 10.3.1.
f0 1 2
;
;
Let
n
1g.
;:::;n
2 N.
If
x
2
Z, then
x
is congruent (modulo n) to exactly one element in
Proof: Exercise! [Hint: Use the Quotient-Remainder Theorem.]
This lemma is important as it allows us to group integers according to their remainder after dividing by
a given number n 2 N .
Denition 10.3.2. Let n 2 N . We will dene the equivalence class determined by s 2 Z, denoted
[s] by [s] = fx 2 Z : x s(mod n)g.
Exercise:
1. Write down 10 elements in the following equivalence classes if n = 4.
[0] = f
[1] = f
[2] = f
[3] = f
[4] = f
[5] = f
2. How many distinct (ie, dierent) equivalence classes (mod 4) do you expect there to be?
Lemma 10.3.3. If n 2 N , then there are exactly n distinct equivalence classes determined by n, namely
[0]; [1]; [2]; : : : ; [n
1].
Proof: By Lemma 4.3.1, every
x
2
Z is congruent to exactly one of the numbers 0; 1; 2; : : : ; n
Therefore, x to one of [0]; [1]; [2]; : : : ; [n
1.
1]. Hence, we have n classes.
To prove the equivalence classes are distinct (disjoint), we must show that for i 6= j , [i] \ [j ] = .
Let 0 i < j n 1. Suppose that [i] \ [j ] 6= ; that is, there exists x 2 Z such that x 2 [i] and x 2 [j ].
Then, we have
x
and
x
2 [ ] =)
2 [ ] =)
i
x
j
x
i
(mod n)
j
(mod n):
However, this contradicts Lemma 4.3.1, so our assumption that [i] \ [j ] 6= is false, so [i] \ [j ] = and
the result follows.
32
10.4 Set of Residues, Zn
Denition 10.4.1. For all n 2 N , we will let
Zn = f[0]; [1]; [2]; : : : ; [n
1]g
and say that Zn is the complete set of residues modulo n.
Exercise:
1. What are Z3 , Z218 and Z1 ?
2. In Z3 , what are the sets [4], [ 2], [7] and [40] usually expressed as?
3. How many \names" are there for [3] in Z10 ? List three.
4. In Zn , [0] [ [1] [ [2] [ : : : [ [n
[0] \ [1] \ [2] \ : : : \ [n
1] =
1] =
As seen in question 3, Zn is a set of elements, each of which has an innite number of names.
Aside:
Consider Q , the set of all rational numbers. For all x 2 Q , x has an innite number of names.
1 2 3 100
= :::
Eg. = = =
2 4 6 200
1 3 7 + 6 13
How have we dened addition of rational numbers? Eg. + =
= .
2 7
14
14
1 3 1+3 4
+ =
= :
Why couldn't we dene addition as follows?
2 7 2+7 9
When we dene + and on Zn , we must make sure our denitions do NOT depend on the name of the
equivalence class. We can then say that + and on Zn are well-dened or consistent.
Denition 10.4.2. We dene + and on Zn as follows.
8[
] [ ] 2 Zn , [a] + [b] = [a + b] and [a] [b] = [ab].
a ; b
To prove that + and are consistent, we will use a property that you will prove in tutorial exercises:
[a] = [b]
()
a
(mod n)
b
The idea of the proof is to show that if dierent representations of the same elements in Zn are used, the
same result is obtained.
33
Example.
In Z3 , we know [1] = [4] and [2] = [5].
We want [1] + [2] = [4] + [5]. Now, [1] + [2] =
and [4] + [5] =
Therefore, [1] + [2] = [4] + [5].
Proposition 10.4.3. Let n 2 N . Addition and multiplication on Zn are consistent.
Proof: Let [a] = [c] and [b] = [d] in Zn . We must prove that [a + b] = [c + d].
[a] = [c] implies
a
(mod n) and [b] = [d] implies
c
b
(mod n). Using Lemma 4.2.1, we can deduce
d
that (a + b) (c + d)(mod n) and so [a + b] = [c + d]. Therefore, addition on Zn is consistent.
The proof for multiplication is identical, and is an exercise.
Exercise:
1. Write out the addition and multiplication tables for Z3 . Are + and closed operations?
+
[0]
[1]
[2]
[0] [1] [2]
[0]
[0]
[1]
[1]
[2]
[2]
2. Solve these equations for x in Z3 .
x
x
x
x
+ [2] = [0]
[2] = [1]
[0] = [1]
+ [0] = [2]
3. Is Z3 commutative under + or ? Is it associative under + or ? Does it have the distributive
property? How would you prove or disprove your answers?
4. Does Z3 have an identity under + or ?
Recall: 0 is the identity of Z under + because a + 0 = 0 + a = a.
5. Does each element of Z3 have an inverse under + or ? What are they?
34
Exercise:
1. Write out the multiplication table for Z4 . Is a closed operation on Z4 ?
+
[0]
[1]
[2]
[3]
[0]
[1]
[2]
[3]
2. Solve these equations for x in Z4 .
[2]x = [2]
[2]x = [0]
[2]x = [1]
[3]x = [1]
3. Is Z4 commutative or associative under ?
4. Does Z4 have an identity under ?
5. Does each element of Z4 (except [0]) have an inverse under ?
Recall the properties of Z:
Properties of Zn are similar:
+ and are closed operations
+ and are closed operations
associative under + and associative under + and commutative under + and commutative under + and distributive
distributive
identities: 0 under +; 1 under identities: [0] under +; [1] under inverses:
inverses:
+:
:
a
is the inverse of a
only 1 have inverses
+: In Zn , each element has an additive inverse: [
:
] (or [n
a
] ... why?)
a
For certain values of n, each element other
than [0] has a multiplicative inverse.
Example. In Z3 , [1] [1] = [1]
and [2] [2] = [1].
Discussion:
It was just noted that all non-zero elements in Zn have multiplicative inverses for certain values of n.
We saw in Z4 that [2] has no inverse. What do you think might be the condition on n for all non-zero
elements in Zn to have multiplicative inverses?