Download In this lecture we will start with Number Theory. We will start

In this lecture we will start with Number Theory. We will start by studying the integers. These are the
most basic numbers there are, Z = {0..., −2, −1, 0, 1, 2, 3, ...}. The symbol for the integers Z, a capital stylized
Z, comes from the german Zahl, which means number.
A very important property of the non-negative is the Well-ordering principle: Any non-empty set of nonnegative integers has a smallest element. On this principle, a lot of theory rests. It leads to so-called mathematical induction (which we will study later on) and it is therefore the basis of why the integers are discrete,
that is, that we can deal with them one at at a time and then move on to the next. The real numbers R and
the rational numbers Q do not have this property.
The division algorithm. What does it mean to divide? When we study integers sometimes we can divide and
the division works out evenly and sometimes it does not work out evenly. We can divide 6 evenly by 3, yielding
the quotient 3, but we cannot divide 13 evenly by 3, this leaves a so-called remainder and then we have to
write 13 = 4 · 3 + 1. We can formulate this in a theorem.
Theorem: (The Division Algorithm) Given any two integers a, b, where b 6= 0, there exists uniquely determined
integers q, r such that
a = qb + r
where 0 ≤ r < |b|.
Definition: In the relation above, the number q is called the quotient and the number r is called the remainder.
Proof: We prove this theorem when b > 0, the proof for b < 0 is similar. The set of numbers {kb; k ∈ Z}
contains numbers that are both larger and smaller than a. This means that the set S = {a−kb; kb ≤ a∧k ∈ Z}
is a non-empty of positive numbers.
Pause: Let’s see what this set S is when we have a = 13 and b = 3, from the example above, then
S = {a − kb; kb ≤ a ∧ k ∈ Z} = {13 − k · 3; k · 3 ≤ 13 ∧ k ∈ Z} = {1, 4, 7, ...}
we must have all k ≤ 4 and for these k we get 13 − k · 3 ≥ 0 assuming the values 1, 4, 7, ....
Since the set S is non-empty and has only non-negative integers in it, there is a smallest element in the set.
We can write this smallest element a − qb for some qZ. That this is the smallest element in the set means that
a − (q + 1)b is not in the set, that is we have
a − qb ≥ 0 and a − (q + 1)b < 0 ⇔ a − qb < b.
But if we put r = a − bq we have just shown that there are numbers q, r such that
a = bq + r
where 0 ≤ r < b = |b| so that the numbers q, r really exist. Now we have to show that they are unique and
therefore we assume that there are more numbers of this kind, that is we assume that
a = bq1 + r1 = bq2 + r2
for q1 , r1 and q2 , r2 both having 0 ≤ r1 < b and 0 ≤ r2 < b. But subtracting we get
0 = a − a = bq1 + r1 − (bq2 + r2 ) ⇔ b(q1 − q2 ) + r1 − r2 ⇔ b(q1 − q2 ) = r2 − r1 .
Since r1 , r2 are both in the interval 0 ≤ r < b then so must the difference between them, this means that
the number r2 − r1 is less than b but greater than −b. But it is also a multiple of b since it can be written
b(q1 − q2 ). Since the only multiple of b in the interval between −b and b is 0, we have condluded that we must
have q1 − q2 = 0 ⇔ q1 = q2 from which also follows r1 = r2 .
This is a proof of existence and unqiueness, we prove that there exist numbers with a certain property (in this
case q, r so that a = qb + r, with 0 ≤ r < |b|). And it is only after the proof is done that we really can make
the definition above which layed down the meaning of the terms quotient and remainder.
Example: What are the quotient and remainders when (a, b) are (13, 3), (13, −3), (−13, 3), and (−13, −3)
respectively?
Solution: We have seen above that 13 = 4 · 3 + 1 so that in the first case we have (q, r) = (4, 1). From this
we can deduce that 13 = −4 · (−3) + 1 so that (q, r) becomes (−4, 1) when (a, b) = (13, −3). Further we get
−13 = −4 · 3 − 1 and here we have to rearrange the right hand side to obtain a positive remainder (remainders
are always positive), we rewrite this as −13 = −5 · 3 + 2 so that for (a, b) = (−13, 3) we obtain (q, r) = (−5, 2).
And the equality −13 = −5 · 3 + 2 can also be written −13 = 5 · −3 + 2 so that for (a, b) = (−13, −3) we obtain
(q, r) = (5, 2).
The division algorithm lies behind the representation of integers in various bases, study this independently.
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Divisibility and the Euclidean Algorithm. We begin this with a definition straight away:
Definition: Given integers a, b with b 6= 0 we say that
(a)
(b)
(c)
(d)
b is a divisor of a, or that
b is a factor of a, or that
a is divisible by b, or that
”b divides a”
if and only if there is an integer q such that a = qb. The notation for this relationship between a and b is
written b|a and is read as (d) above (”b divides a”.) If b does not divide a we write this as b - a
Example: 3|15 since 15 = 5 · 3, that is, the q in the definition is 5. 3 - 13 since we cannot find any integer
q such that 13 = q · 3.
We make the following important observation: Given any two integers a, b 6= 0, if q, r are given by the
division algorithm, then
b|a ⇔ r = 0.
This is illustrated by the two examples above, 3|15 and 15 = 5 · 3 + 0, that is r = 0 (and q = 5). And 3 - 13
and 13 = 4 · 3 + 1, that is r = 1 6= 0 (and q = 4).
We will make another observation about the division. If we take any set of positive integers, call it U , then
the division relation defined on U becomes a partial order or U . The book proves this specifically for the set
of natural numbers (that is the set {1, 2, 3, ...}), but we can prove this for any set of positive integers. We
formulate it as a proposition:
Proposition: Let A be any set of positive numbers, then the relation b|a defined on A is a partial order of A
Proof: To prove that the relation is a partial order we must show that three things hold: reflexivity, antisymmetry and transitivity.
Reflexivity: We must show that a|a for all a ∈ A, but this is true since a = 1 · a for all a ∈ A, hence
q = 1 works in the defintion of divisibility. Reflexitivity is shown.
Antisymmetry: We must show that b|a∧a|b ⇒ a = b. What will be crucial here is that we are dealing
with positive integers. The fact that b|a ∧ a|b can also be expressed ∃q1 , q2 : a = q1 b ∧ b = q2 a. But then
we can write a = q1 b = q1 q2 a and as a is a positive number we have a = q1 q2 a ⇒ 1 = q1 q2 . As also b
is positive both q1 , q2 must be positive. The only posibility of having 1 = q1 q2 for two positive integers
q1 , q2 is if both are 1. But if q1 = q2 = 1 then we have proved that a = b which shows antisymmetry.
Transitivity: We must show that a|b ∧ b|c ⇒ a|c, but this is very easy, since a|b ∧ b|c ⇔ ∃q, q2 : b =
q1 a ∧ c = q2 b which allows us to write c = q2 b = q2 q1 a = qa for q = q2 q1 . But this means exactly that
a|c. Transitivity is shown.
Since we have shown that all three properties hold for the divisibility relation, we have shown that it sis a
partial order of A.
This proposition means that we can think of the divisibility relation as a ”less-than” relation. (Indeed, if the
numbers a, b are positive with a|b we certainly also have a ≤ b. This need not hold if we consider the number
0, why?)
One very useful fact is the following proposition:
Proposition: If a, b are two integers with c|a and c|b for an integer c, then the integer c also divides every
linear combination of a and b, that is, for every integers x, y we have c|xa + yb. Particularly c|a + b and c|a − b
(this is just the proposition restated with (x, y) = (1, ±1)).
Proof: We have c|a and c|b so that there exists integers q1 , q2 such that a = q1 c and b = q2 c. But then, when
we form xa + yb we see that this is
xa + yb = xq1 c + yq2 c = (xq1 + yq2 )c = qc for q = xq1 + yq2
and this precisely means that c|xa + yb which is what we wanted to prove.
The Greatest Common Divisor. Again we start with a definition:
Definition: If a, b are two integers, then c is called a common divisor of a and b if c|a and c|b. The greatest
common divisor of a, b is denoted gcd(a, b), or shorter here, g, and is the unique number with the two properties
that
* g|a and g|b (that is it is a common divisor of a and b).
* No greater number than g is a common divisor of a and b.
Examples: The greatest common divisor of 63 and 42 is gcd(63, 42) = 21. If a|b then, is gcd(a, b) = a? Are
there any nonnegative integers a, b such that gcd(a, b) > b?
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Finding The Greatest Common Divisor of two numbers. It is a very important procedure to find the greatest
common divisor of two numbers. We will see that it has important implications in many areas of the theory
of numbers. In fact it is very easy, we do it through a certain lemma:
Lemma: If a = qb + r for integers a, b, q, r, then gcd(a, b) = gcd(b, r).
Proof: It may not be apparent, but sometimes the greatest common divisor may not exist. This is only the
case when we are studying the number 0. So if a = b = 0, then r = 0 and, similarly, if b = r = 0 then also
a = 0 so then the greatest common divisor of a, b is not defined preceisely when the greatest common divisor
of b, r is not defined. Then we consider them to be equal. (Some authors set gcd(0, 0) = ∞, and here we would
have gcd(a, b) = gcd(b, r) = ∞. Thus we can always consider gcd(a, b) and gcd(b, r) to be well-defined positive
integers. We now introduce the notation g1 = gcd(a, b) and g2 = gcd(b, r). Since g2 |b and g2 |r we must have
g2 |qb + r, by the proposition above about divisiors of linear combinations, this means that g2 |a = qb + r. Since
g2 apparently is divisor of both b and a, g2 is a common divisor of a, b. But g1 is the greatest common divisor
of a, b which means we will have g2 ≤ g1 . But a very similar argument can be used to also show that g1 ≤ g2 .
Hence we have g1 ≤ g2 and g2 ≤ g1 . The only possibility is that g1 = g2 which is what we wanted to prove.
Now to find the greatest common divisor of two numbers, we simply repeat the division algorithm. We
study this in several examples. This way of finding the gcd of two numbers is called The Euclidean Algorithm
in honor of Euclid of Alexandria, one of the giants of mathematics and geometry.
Example: Find the greatest common divisor of 234 and 567
Solution: Using the division algorithm repeatedly, as prescribed by The Euclidean Algorithm, we get
567 = 2 · 234 + 99
so that, by the lemma we just proved, gcd(567, 234) = gcd(234, 99). Now we repeat the procedure with 234
and 99, we have
234 = 2 · 99 + 36
so that gcd(567, 234) = gcd(234, 99) = gcd(99, 36). The problem gets smaller and smaller! We proceed,
99 = 2 · 36 + 27
36 = 1 · 27 + 9
so that gcd(99, 36) = gcd(36, 27) = gcd(27, 9) = 9. In conclusion we get
gcd(567, 234) = gcd(234, 99) = gcd(99, 36) = gcd(36, 27) = gcd(27, 9) = 9.
Example: Find the greatest common divisor of 360 and 735.
Solution: We again utilize The Euclidean Algorithm and we have
735 = 2 · 360 + 15
so that gcd(735, 360) = gcd(360, 15), and plugging in 360 and 15 into the algorithm yields
360 = 24 · 15 + 0
from which we get gcd(735, 360) = gcd(360, 15) = gcd(15, 0) = 15. Only two steps!
The Euclidean algorithm will always work for non-negative integers. It will always produce a sequence of
remainders that are descreasing (why?), (above 15,0, and above this example we saw the sequence 99,36,27,9,0).
We may or may not emphasize that it ends with 0 but anyhow the last non-vanishing remainder must be our
greatest common divisor. Why? We can see this in the following way. As mentioned above the sequence of
remainders must be decreasing. All the remainders must be non-negative so if we continue on and do the
process again and again, we must eventually get 0. If we would not get 0, then we would eventually get the
same remainder twice, but then if we get the same remainder twice in a row, then, the next remainder must
be 0. So one of the remainders must be 0. If we denote the last non-vanishing remainder with r, then, the
lemma above still holds which states that gcd(a, b) = gcd(r, 0), but gcd(r, 0) must of course be r so that we
indeed have gcd(a, b) = r.
Two numbers a, b has a common divisor c, as is defined above, if this number c divides both of the a, b.
This means also that c is a factor in both a, b and in a way a, b both then contain c. This means that a, b are
related via c, we could say that c connects a and b. Since the number 1 is always a common divisor of any
two numbers, we can then express the fact that two numbers are not connected by stating that the greatest
common divisor is 1. This is the basis of this definition:
Definition: Two integers a, b are called relatively prime if the greatest common divisor of a and b is 1.
Example: Since gcd(9, 49) = 1, the integers 9 and 49 are relatively prime. Since gcd(15, 21) = 3, the integers
15 and 21 are not relatively prime. We will shortly see that numbers that are relatively prime have interesting
properties. But first e will now study an extremely useful property of the greatest common divisor.
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Theorem: The greatest common divisor of integers a, b is an integral linear combination of them; that is, if
g = gcd(a, b), then there are integers m, n such that g = ma + nb.
”Proof”: We do not do the proof in it’s full detail, we only give a few illustrating examples and leave to the
reader to convince him/her that it always works.
Example: The greatest common divisor of a = 735 and b = 360 has shown to be 15, above. Now the Euclidean
algorithm gave the equalities:
735 = 2 · 360 + 15
360 = 24 · 15 + 0
and the first of these can be rewritten 15 = gcd(735, 360) = 1 · 735 − 2 · 360 which means that m, n are the
numbers 2, −1.
Example: We can extend the procedure to a more complicated example, above we saw that gcd(a, b) = 9 if
a = 567, b = 234. We now show how to find m, n such that gcd(a, b) = ma + nb. We proceed to work backwards
in the Euclidean algorithm, firt we list all equalities that were obtained when we established that gcd(a, b) = 9.
Those were:
567 = 2 · 234 + 99,
234 = 2 · 99 + 36,
99 = 2 · 36 + 27,
36 = 1 · 27 + 9
(⇒ gcd(567, 234) = 9).
Starting with the last equality we get 9 = gcd(a, b) = 36 − 1 · 27. Using the third equality we replace 27 by
99 − 2 · 36 giving 9 = gcd(a, b) = 36 − 1 · 27 = 36 − 1 · (99 − 2 · 36) = −1 · 99 + 3 · 36. We now continue in the
same fashion and replace 36 by the number 234 − 2 · 99 giving 9 = −1 · 99 + 3 · 36 = −1 · 99 + 3 · (234 − 2 · 99).
Rearranging and gathering all the 99’s and the 234’s together we get 9 = gcd(a, b) = −7 · 99 + 3 · 234. Now we
only need to take one final step and use the firt equality and replace 99 by 567 − 2 · 234 finally giving
9 = gcd(a, b) = −7 · 99 + 3 · 234 = −7 · (567 − 2 · 234) + 3 · 234 = −7 · 567 + 17 · 234
hence m, n = −7, 17 gives gcd(a, b) = ma + nb for (a, b) = (567, 234). It is evident that this works in all cases,
for all numbers a, b so we can regard the theorem as being proven.
From this theorem we have two immediate corollaries:
Corollary: Suppose that a|bx and a, b are relatively prime, then a|x.
Proof: Since gcd(a, b) = 1 we have ma + nb = 1 for some choice of m, n. Multiplying this identity by x gives
us max + nbx = x. Since a|bx we have bx = qa for some q, therefore we have
max + nbx = x ⇒ mxa + nqa = x ⇔ (mx + nq)a = x
which shows that a|x.
This is an important property and we can formulate it in words like this: ”if a product of two numbers bx
has a as a factor, but does not have anything of a in b, then all of a must be in x”.
Another corollary also follows:
Corollary: The greatest common divisor of nonzero integers a, b is divisible aby every common divisor of a, b.
Proof: Let c be any common divisor of a, b. Then there is q1 , q2 such that a = q1 c and b = q2 c. As there also
are integers m, n such that gcd(a, b) = ma + nb we can write
gcd(a, b) = ma + nb = mq1 c + nq2 c = (mq1 + nq2 )c
but this shows precisely that c|gcd(a, b) and since c was an arbitrary common divisor of a, b the proof is
complete.
Lattices will be studied separately when we have develop the theory of prime numbers.
Prime numbers. These are very important numbers, it is why they are called ”prime”, they are truly the first
numbers. We will see why this is true in someting called the Fundamental Theorem of Arithmetic. But first
we will state what a prime number is:
Definition: A natural number (a positive integer) p > 1 is called prime if and only if the only natural numbers
that divides p are 1 and p itself. A natural number n > 1 that is not prime is called a composite.
Let us express this in terms of quantifiers:
p ∈ N prime ⇔ ∀a, b ∈ N : p = ab ⇒ a = p ∨ b = p,
n ∈ N composite ⇔ ∃a, b ∈ N : n = ab ⇒ a > 1 ∧ b > 1.
We can write a list of the first 6 prime numbers:
2, 3, 5, 7, 11, 13
and all numbers in between these are composite numbers
4 = 2 · 2, 6 = 2 · 3, 8 = 4 · 2, 9 = 3 · 3, 10 = 2 · 5, 12 = 3 · 4.
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Any number always has a prime divisor, in fact we shall also see the Fundamental Theorem of Arithmetic
which states something stronger than this, but we will start small. We formulate this as a lemma:
Lemma: Given any natural number n > 1, there exists a prime p such that p|n.
Proof: This proof is by contradiction so we assume that there are natural numbers that do not have prime
divisors, call them n1 , n2 , n3 , .... By the well-ordering principle of the integers, one of these numbers must be
the smallest, we can assume that it is n1 (it is only a matter of naming). We have now then a number n1 with
the following two properties:
(a) n1 has no prime divisors, particularly, n1 itself cannot be prime so n1 = ab where a > 1 and b > 1
(b) All natural numbers smaller than n1 has prime divisors
But the two numbers a, b must both be smaller than n1 , this means that they must both have prime divisors,
since n1 was the smallest number without prime divisors. Assume for example that p|a for some prime number
p. But then a = qp ⇒ n1 = bqp which shows that p|n1 and we have found a prime factor in n1 which is a
contradiction. What was wrong? It was the assumption about the existence of any number without prime
factors, hence there cannot be any such numbers, hence all numbers must have prime factors in them and the
proof is complete.
Now we will show the great theorem on the infinitude of the primes. It is an extremely beautiful theorem with
a most elegant proof. The argument is attributed to Euclid of Alexandria.
Theorem: There is an infinite number of primes.
Proof: Assume the opposite, that there is only a finite number of primes. The first prime number we know
are 2, 3 and 5, etc, but let us call all of them p1 (= 2), p2 (= 3), ..., pN where N is some (big) number. Then
form the special number
n = 2 · 3 · 5 · · · pN + 1 = p1 · p2 · p3 · · · pN + 1.
Then, by the lemma above, this number must have a prime divisor. But which one can it be? We have
a complete list of primes in p1 , p2 , . . . , pN so we can test everyone of them. But as n gives remainder 1
when we divide with any of the numbers p1 , p2 , . . . , pN , not a single one of them can divide n. But this is a
contradiction because each number must be divisible by some prime number. We have reached a contradiction.
The assumption that there is only a finite number of prime numbers must be wrong, hence the number of prime
numbers must be infinite which was what we wanted to prove.
We will conclude this section with stating and proving theorems that lead to the Fundamental Theorem of
Arithmetic that states that every natural number can be written as the product of prime numbers in essentially
only one way.
Theorem: Every natural number n > 1 can be written as the product of prime numbers.
Proof: Assume that there are numbers that cannot be written as a product of prime numbers. Then, according
to the well-ordering principle of the integers, we have a smallest such numbers. This means that we have a
smallest number n which is not a product of primes. Particularly n itself is composite so that n = ab for
n > a > 1 and n > b > 1. But since both a, b are smaller than n, each of these numbers is the product of
prime numbers. The equality n = ab then gives
n = ab = {a product of primes} · {a product of primes} = a product of primes
which is a contradiction. Hence there cannot be any numbers which cannot be written as a product of primes
so that all natural numbers can be written as a product of primes which completes the proof.
The next proposition illustrated the most important property of prime numbers. They must always be factors
of one number:
Proposition: Suppose a, b are integers and p is a prime number with p|ab then either pa or p|b.
Proof: We want to prove that p|a ∨ p|b. One equivalent way of proving this is to prove that p - a ⇒ p|b.
Assume therefore that p - a. Then p and a are relatively prime, otherwise we would have p|a. But since p|ab
this proves that p|b. (The earlier corollary.)
Corollary: If a prime number p divides a product of integers a1 · a2 · · · ak , then p divides one of the ai .
The Fundamental Theorem of Arithmentic: Every integer n > 2 can be written in the form
n = pα1 1 pα2 2 · · · pαr r
and if the prime factors are sorted in ascending order, the prime factors and the numbers α1 , . . . , αr are uniquely
determined.
Proof: We will not go through the proof here, we will merely study some examples, the proof is better studied
after one has gone over the examples.
Example: 36 = 4 · 9 = 22 · 32 , this is one way of factoring 36, but we can also write 36 = 6 · 6 and then factor
each 6, giving 2 · 3 · 2 · 3 and again collecting the prime factors in ascending order again gives the factorization
6 = 2 · 2 · 3 · 3 = 22 · 32 . This shows that 36 can only be written in one way as a product of prime factors. Try
some more complicated examples, factor 567, 123, and 789.