Download Number systems. - Elad Aigner

Numbers systems
Elad Aigner-Horev
Abstract. In this lecture we shall define the main number systems that we shall be using throughout
the course.
§1. Naive definitions
N - The natural numbers {0, 1, 2, . . .}. A naive way
to think of the natural numbers is as dots along
an infinite straight line where consecutive dots are
separated by one unit of length.
N
Z - The integer numbers {. . . , −2, −1, 0, 1, 2, . . .}.
The integers can be thought of as two copies of the
naturals numbers that are disjoint apart from a single common point at zero.
Z
Q
R
C
Q - the rational numbers { ab : a, b ∈ Z, b 6= 0}. The
rationals arose from everyday life necessities of expressing fractions of units of length and amounts.
The word rational comes from ratio and does not
mean that these numbers make sense. Another definition is that the rationals are those numbers whose
decimal expansion is periodic.
√
R \ Q - the irrational numbers are numbers like 2
and π. Often these are defined as those numbers
whose decimal expansion is non-periodic. The irrational arise naturally.
1
√
2
N⊆Z⊆Q⊆R
1
1
1
On the right we have some of the more commonly
known definitions for the above number systems. R - the union of the rational and irrational numbers.
The main goal of this lecture is to provide a more
systematic definition for these number systems.
C - the complex numbers.
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Number Theory ArielU 2016
Elad Aigner-Horev
§2. Fundamental set theory
The construction of the natural numbers requires some terminology from set theory.
Definition 2.1. An unordered collection of elements with no repetitions is called a set.
Example 2.2. We use {·} to denote sets.
{a} - a set containing a single element is called a singleton.
{a, b, 1, 4} = {1, b, a, 4} - both these sets are equal.
{{a}, {a, b, 3}, 1, {1, {2{3}}}} - sets can contain other sets.
{a, a} = {a}, {a, 1, 2, a} = {2, 1, a} - ignore repetitions.
We write a ∈ A do denote that the element A belongs to the set A. For instance,
1
1 1
∈ {x ∈ R : x − < }.
4
2
2
We write a ∈
/ A to denote that a is not in A. For instance
1 1
10 ∈
/ {x ∈ R : x − < }.
2
2
Example 2.3.
{2k : k ∈ Z} - the even numbers.
{2k : k ∈ N} - the positive even numbers.
{n ∈ Z+ : 3 n and n ≤ 106 } - all positive integers that are divisible by 3 not exceeding 106 .
Definition 2.4. For a finite set A we write |A| to denote the cardinality (i.e., size) of A.
Definition 2.5. Let A and B be two sets.
1. We write A ⊆ B to denote that every a ∈ A satisfies a ∈ B. We say that A is contained in B and
that A is a subset of B.
2. We write A ⊂ B to denote that every a ∈ A satisfies a ∈ B yet ∃b ∈ B such that b ∈
/ A. We say that
A is properly contained in B and that A is a proper subset of B.
Trivially, for any set A we have A ⊆ A and A 6⊂ A.
Example 2.6.
N ⊂ Z ⊂ Q ⊂ R ⊂ C.
{x ∈ Z : x < 100} 6⊆ {x ∈ R : x < 99}.
Example 2.7. Let A = {1, {1}, {2}}.
1 ∈ A - true
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Number Theory ArielU 2016
Elad Aigner-Horev
2 ∈ A - false
1 ⊆ A - No 1 is not a set
{1} ⊆ A - true
{2} ⊆ A - false
{2} ∈ A - true
{{2}} ∈ A - false
{{2}} ⊆ A -true
Lemma 2.8. Let A and B be two sets. Then
A ⊆ B and B ⊆ A ⇐⇒ A = B.
Example 2.9. Let us use Lemma 2.8. Let
A = {2n + 1 : n ∈ Z} B = {2n + 3 : n ∈ Z}.
Determine whether A = B.
The set A consists of the odd numbers in Z. As every member of B is odd we have that B ⊆ A. In
order to show that A ⊆ B fix an element 2m + 1 ∈ A. We show that 2m + 1 ∈ B. That is we show that
there exists an n ∈ Z such that 2m + 1 = 2n + 3. Let us make this clear. Fixing 2m + 1 in particularly
fixes m. Finding n then means that we express n in terms of m. That is n is a function of m: n = n(m).
2m + 1 = 2n + 3
2(m − n) = 2
n = m − 1.
We have just seen that n = n(m) = m − 1. This completes the proof that A ⊆ B.
Example 2.10. Continuing the theme of the previous example let k1 , k2 ∈ Z be odd and set
A0 = {2n + k1 : n ∈ Z} B 0 = {2n + k2 : n ∈ Z}.
Is it true that A0 = B 0 ? The question reduces to whether the following has a solution.
2m + k1 = 2n + k2
2(m − n) = k2 − k1 .
We see that 2(m − n) is an even number and k2 − k1 is an even number as k1 , k2 are odd, by assumption.
More explicitly, we have that A0 ⊆ B 0 because given an arbitrary element 2m + k1 ∈ A0 this element lies
0
0
2
in B 0 as 2m + k1 = 2n(m) + k2 where n(m) = m − k1 −k
2 . On the other hand we have that B ⊆ A
because given an arbitrary element 2n + k2 ∈ B 0 this element lies in A0 as 2n + k2 = 2m(n) + k1 where
1
m(n) = n + k2 −k
2 .
§2.1 The empty set.
empty set.
We write ∅ to denote the set containing no elements and refer to it as the
∅ := {y : y 6= y}.
Any set containing elements is said to be non-empty.
Theorem 2.11. Let A be a set. Then
(a) ∅ ⊆ A;
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Number Theory ArielU 2016
Elad Aigner-Horev
(b) if in addition A 6= ∅ then ∅ ⊂ A.
Example 2.12.
∅ ∈ ∅ - false
∅ ∈ {∅} - true
∅ ⊂ ∅- false
∅ ⊆ {∅} - true (see Theorem 2.11)
∅ ⊂ {∅} - true (see Theorem 2.11)
§2.2 Operations on sets.
Definition 2.13. Let A and B be two sets.
1. The intersection of A and B is the set denoted A ∩ B given by
A ∩ B := {x : x ∈ A and x ∈ B}.
2. The union of A and B is the set denoted A ∪ B given by
A ∪ B := {x : x ∈ A or x ∈ B}.
Definition 2.14. Two sets A and B satisfying A ∩ B = ∅ are called disjoint.
Definition 2.15. Let A and B be two sets. We write A \ B to denote the set
A \ B := {x : x ∈ A and x ∈
/ B}
Definition 2.16. The symmetric difference of two sets A and B is given by
A∆B := (A \ B) ∪ (B \ A).
Definition 2.17. Let U be some universe in which a set A is taken. The complement of A is the set
A := U \ A.
Example 2.18. Let us have U = N and let A = {2k : k ∈ N} so that A is the even natural numbers.
Then A = {2k + 1 : k ∈ N} is the odd natural numbers.
§2.3 Cartesian product. We have said that {a, b} = {b, a}. That is we ignore the ordering of the
elements for sets. To express that ordering matters we use (·). For instance (a, b) is an ordered pair of
elements. It is ordered in the sense that (a, b) and (b, a) are considered to be different elements. Also,
unlike sets, here we allow repetitions. For instance, while {a, a} = {a} the ordered pair (a, a) is not
equivalent to (a).
Definition 2.19. A k-tuple is an ordered list of k elements.
Example 2.20.
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Number Theory ArielU 2016
Elad Aigner-Horev
(a) - 1-tuple
(a, b), (a, a), (b, a) - 2-tuples (pairs)
(a, a, a, a), (a, b, a, a), (b, a, a, a) - 4-tuples
Definition 2.21. Let A and B be two sets. Then the Cartesian product of A and B is defined to be
the set of 2-tuples
A × B := {(a, b) : a ∈ A and b ∈ B}.
In particular,
A × A = {(a, a0 ) : a, a0 ∈ A}.
Definition 2.22. The k-fold Cartesian product of a set A with itself is given by
Ak := {(a1 , . . . , ak ) : ai ∈ A ∀i ∈ [k]}.
More generally we have
AN = {(an )n∈N : ai ∈ A ∀i ∈ N}.
and
AZ = {(an )n∈Z : ai ∈ A ∀i ∈ Z}.
Example 2.23. The plane is the set R2 = R × R.
{(x, y) ∈ R2 : x2 + y 2 < 1}.
Example 2.24. The set {0, 1}3 is also known as the cube.
§3. The natural numbers
The definition we provided so far for the natural numbers is that of dots along a line separated by a
single unit of length. Almost all of the words just used are undefined. What is a dot really? To expel all
these uncertainties we now provide a proper definition for the natural numbers using sets.
The idea behind this definition is exceedingly simple.
• The empty set ∅ has no elements. This we can identify with 0 in N.
• The set {∅} (or more generally {a}) has a single element which may identify with 1 in N.
• The set {∅, {∅}} (or more generally {a, b}) has two elements which we may identify with 2 in N.
A closer look at the above sets reveals the following pattern.
• Define 0 to be the set ∅, i.e., 0 := ∅.
• Next, define 1 := 0 ∪ {∅} = ∅ ∪ {∅} = {∅}.
• 2 := 1 ∪ {1} = {∅} ∪ {{∅}} = {∅, {∅}}.
• 3 := 2 ∪ {2} = {∅, {∅}, {∅, {∅}}}, and so on.
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Number Theory ArielU 2016
Elad Aigner-Horev
At this stage note that
0 ∈ 1 ∈ 2 ∈ 3.
That is we are replacing the natural order of the naturals which we are accustomed to write
0<1<2<3
using ∈. Let us make all this precise then.
Definition 3.1. The successor of a set X is given by S(X) := X ∪ {X}.
Definition 3.2.
(N.1) 0 is natural.
(N.2) If n is natural then its successor S(n) is natural as well.
(N.3) All naturals can be generated through (N.1) and (N.2).
N := "the least set" generated by (N.1) (N.2).
Successors of finite sets are finite. However, we think of N as an infinite set. A fair question then is
when does the process captured by (N.1) and (N.2) turns infinite? In this course we shall not provide an
answer to this. Suffice to say that this is the part where we resort to the following.
Axiom of infinity. Infinite sets exist.
We are unable to prove that infinite sets exist and so at this point the axiom of infinity comes into play.
The naturals then is considered to be the "smallest" infinite set. In this course we will not clarify the
notion behind "smallest". Instead we will pose the following question: which set is "larger" the naturals
or the natural even numbers?
§4. Well-ordering of the natural numbers
Unlike the sets Z, Q, R the natural numbers are special. Indeed, the natural numbers have the following
important property.
Well-ordering principle (WOP). every non-empty set of N has a least element.
At this stage of the course we shall accept this as an axiom. Nevertheless, the following is a simple
demonstration of its power. This being our first proof in the course we shall write this in an informal
manner with an emphasis on the ideas instead on brevity.
√
Theorem 4.1.
2 is irrational.
Proof. We claim that there exists no x ∈ Q such that x2 = 2. We prove this using the WOP. Assume
towards a contradiction that there exists an x ∈ Q such that x2 = 2. That is x = m
n for some m, n ∈ Z and
m 2
2
2
n 6= 0 and such that m = 2n . As n = 2 we may assume without loss of generality that m, n ∈ Z+ .
It follows that
S = {{m, n} : m2 = 2n2 , m, n ∈ Z+ } =
6 ∅
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Number Theory ArielU 2016
Elad Aigner-Horev
We cannot appeal to the WOP with S as S is a set of unordered pairs and not a subset of N. To be able
to appeal to the WOP we consider the projection of S namely
S 0 = {n ∈ Z+ : ∃m ∈ Z+ s.t. {m, n} ∈ S}.
The set S 0 is non-empty (since S is non-empty) and S 0 ⊆ Z+ ⊆ N. By the WOP S 0 has a least element.
Let n∗ denote this element and let m∗ ∈ Z+ be a positive integer such that {m∗ , n∗ } ∈ S. Put another
way {m∗ , n∗ } is a pair in S whose n-part is least amongst all pairs in S.
To get a contradiction we show that there exists a pair {p, q} ∈ S such that q < n∗ . How can we come
up with the numbers p and q? As the only restriction we have is on q it is better to start with q. There
are two immediate candidates for q that we can think of. The first is n∗ − m∗ . Surely n∗ − m∗ < n∗ .
However, as (m∗ )2 = 2(n∗ )2 we have n∗ < m∗ so that n∗ − m∗ < 0 and in particular n∗ − m∗ ∈
/ Z+ so
with such a q we fall out of S.
The second candidate is m∗ −n∗ . This number is surely positive. But do we have m∗ −n∗ < n∗ ? Suppose
not; then m∗ ≥ 2n∗ . Consequently, (m∗ )2 ≥ (2n∗ )2 contradicting the assumption that (m∗ )2 = 2(n∗ )2 . In
particular, we have just established that
m∗ < 2n∗ .
(4.2)
With q chosen to be m∗ − n∗ the number p is determined and we calculate it as follows.
2q 2 = 2(m∗ − n∗ )2
= 2(m∗ )2 − 4m∗ n∗ + 2(n∗ )2
recall that 2(n∗ )2 = (m∗ )2 so that
= 2 · 2(n∗ )2 − 4m∗ n∗ + (m∗ )2
= 4(n∗ )2 − 2(2n∗ )(m∗ ) + (m∗ )2
= (2n∗ − m∗ )2 .
| {z }
this is p
This suggests that we should take p = 2n∗ − m∗ . We can do so providing that p > 0. To have the latter
we require that 2n∗ > m∗ . This we have already seen in (4.2).
We have just seen that the pair {p, q} = {m∗ − n∗ , 2n∗ − m∗ } ∈ S and q < n∗ a contradiction to the
choice of {m∗ , n∗ }.
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