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__________________________________________________________________________________________________________________________
How to Define a Real Number Through the Concept
of Dedekind Cut?
Kallol Paula and Dipanwita Paul Ghoshb
a
b
Department of Mathematics, Jadavpur University, Kolkata 700032, India.
email: [email protected]
Department of Mathematics, Victoria Institution (College), Kolkata 700009, India.
e-mail:[email protected]
Abstract - The aim of this article is to provide a logical building of the real number system starting from the
rational number system. The real number system plays most vital role in the study of calculus and real
analysis in undergraduate classes, in fact, real numbers play an important role in school mathematics also.
This article is mainly intended for undergraduate students. Assuming that the rational number system is the
smallest ordered field we talk of the existence of gaps in them and try to fill them up by the notion of Dedekind
cut and thus land up in the system of real numbers. The system of real numbers is shown to be a complete
ordered field. We then define a real number as an element of a complete ordered field after proving that any
two complete ordered fields are order isomorphic.
1
Introduction
The number system of mathematics begins with the counting numbers 1, 2, 3, . . . which are called natural
numbers and is denoted by N. If m, n ∈ N then we cannot always solve the equation x + m = n in N and so
from N we move to the set of integers Z in which we can solve the above type of equations. Again if a, b ∈ Z
and a 6= 0 then we cannot always solve the equation ax = b in Z. So from Z we move to the set of rational
numbers Q by taking the solutions of the above equations. Consider the equation x2 − p = 0, then it is not
always solvable in Q. The question is can we move from Q to R by taking care of solutions of above equations ?
The answer is no. Even if we take care of solutions of all polynomial equations over Q then also we donot land
up in R! The existence of real numbers like π, e justifies that. So what is actually lacking in Q ? The answer
is “Every non-empty bounded above subset of Q does not have a least upper bound in Q”. We move from Q to
R to ensure that every non-empty bounded above subset of R has a least upper bound in R. Moving from Q
to the set of real numbers R is not as simple as moving from N to Z or from Z to Q. It requires the concept of
Dedekind cut or Cauchy sequence of rationals or sequence of nested intervals. One can say that N is incomplete
w.r.t. ‘+’, Z is incomplete w.r.t. ‘.’, Q is incomplete w.r.t. ‘ <’ and R is complete w.r.t all of them. So the
extension from the set of natural numbers to the set of real numbers is three fold.
The first extension may be described as the completion of N, the set of natural numbers with respect to
addition - i.e., the minimum enlargement of N to ensure the solvability of all equations of the type x + n = m
with m, n ∈ N. The extended set is the set of integers Z. The second extension may be described as the
completion of Z with with respect to multiplication - i.e., the minimum enlargement of Z to attain the solvability
of all equations of the form bx = a with a, b ∈ Z and b 6= 0. The extended set is the set of rational numbers
Q. The third extension may be described as the completion of Q with respect to order - i.e., minimum
enlargement of Q to insure that every nonempty bounded above subset has a least upper bound. The resulting
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set is the set of real numbers R. The first two extensions are algebraic extensions whereas the third extension
is analytic extension. Our aim is to talk briefly of construction of real numbers from rational numbers done by
Richard Dedekind [1](published in 1872) which involves the concept of Dedekind cut. The construction of real
numbers can also be done
• through Cauchy sequence of rational numbers (George Cantor (1872))
• through nested sequence of closed rational intervals
• by using decimal expansions (Weierstrass-Stolz (1886)).
2
Rational Number System
The system of rational numbers can be characterised as the smallest ordered field. Then the question naturally
arises what is a field? what is an ordered field? and finally what is meant by smallest ordered field. We anwer
them one by one.
Definition 2.1 By a field we mean a non-empty set F along with two binary opeartions ‘+’ and ‘.’ satisfying
the following properties:
1.
(i)
(ii)
(iii)
(iv)
2. (i)
(ii)
(iii)
(iv)
3.
x + (y + z) = (x + y) + z
x+y =y+x
∃ 0 ∈ F 3 x + 0 = x ∀x ∈ F.
for each x ∈ F ∃ z ∈ F 3 x + z = 0
x.(y.z) = (x.y).z
x.y = y.x
∃ 1 ∈ F 3 1.x = x ∀x ∈ F.
for each x 6= 0 ∃ z ∈ F 3 z.x = 1
x.(y + z) = x.y + x.z
A non-empty subset S of F is said to be a subfield of F if S itself is a field w.r.t. the operations induced on S by
the operations on F.
Definition 2.2 A field F is said to be ordered if there exists a subset P of F such that
• If a, b ∈ P then a + b ∈ P.
• If a, b ∈ P then a.b ∈ P.
• For each element x of F exactly one of the following three holds
x = 0, x ∈ P, − x ∈ P
The third condition is known as the law of trichotomy.
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The set P is known as the set of positive elements of the field F.
In an ordered field F we can define a relation ‘ <’ by
a < b if f b − a ∈ P.
The above conditions can be restated as
• If a > 0 and b > 0 then a + b > 0.
• If a > 0 and b > 0 then a.b > 0.
• For each element x of F exactly one of the following three holds
x = 0, x > 0, − x > 0
Clearly Q is an ordered field with P being the set of positive rational numbers. Now to answer the question on
“smallest ordered field” we mention what is meant by order-isomorphisms between two ordered fields F and F 0 .
Definition 2.3 Let F and F 0 be two ordered fields. A mapping f : F −→ F 0 is said to be an order-isomorphism
from F onto F 0 iff
• f is bijective i.e., f is both one-one and onto
• f (x + y) = f (x) + f (y) and f (x.y) = f (x).f (y) for all x, y ∈ F
• f (P ) = P 0 i.e., f maps positive elements of F onto positive elements of F 0 .
If f satisfies the first two conditions then it is known as field isomorphism from F onto F 0 .
One can easily check that if f is an order-isomorphism from F onto F 0 then
x < y in F if f f (x) < f (y) in F 0
and f −1 is an order-isomorphism from F 0 onto F.
We say that F and F 0 are same from the point of view of an ordered field.
So by the characterisation “ Q is the smallest ordered field ” we mean that Q is an ordered field and if F is any
ordered field then F contains a subfield which is order-isomorphic to Q i.e., F contains a copy of Q.
2.1
Archimedean Property
One of the most important property of rational numbers is the Archimedean Property which is given in the
following theorem
Theorem 2.4 Archimedean Property: Let a and b be two rational numbers with a > 0, b > 0. Then there
exists a natural number n such that na > b.
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Proof. Let a = p/q and b = r/s where p, q, r, s ∈ N. Our aim is to find some n ∈ N such na > b i.e., nps > rq.
Choose n = 2rq so that clearly na > b.
The Archimedean property can be interpreted as “ given two unequal curves, surfaces or bodies the larger of the
two can become smaller than the quantity obtained by a suitable number of repetitions of the smaller one.” If
we look at two unequal straight line segments then the interpretation becomes more clear.
We now state the theorem which characterizes Archimedean property.
Theorem 2.5 Let F be an ordered field. Then the following are equivalent:
• F is Archimedean.
• for each x ∈ F there exists nx ∈ N such that x ≤ nx i.e., the set of natural numbers is unbounded above.
• for each x ∈ F there exists nx , mx ∈ Z such that x ≤ nx and mx ≤ x i.e., the set of integers is both
unbounded above and unbounded below.
• inf{1/n : n ∈ N } = 0.
• inf{1/2n : n ∈ N } = 0.
Proof. Easy.
We would like to mention here that not every ordered field is Archimedean, a brief outline of an ordered field
which is non-Archimedean will be given at the end.
2.2
Existence of numbers which are not rational
One may question about the need of numbers which are not rational. Consider a square of unit length. Then
the square of length of the diagonal is 2 units, but we know there exists no rational number x for which x2 = 2.
So you need such numbers even for elementary school geometry. If you study trigonometry then numbers such
as π and cos r (r is a non-zero rational) are required which fails to be rational, in fact all trigonometric functions
at non-zero rational values fail to be rational. As calculus involves trigonometry so we need those numbers in
calculus also. Thus one has to move on from the system of rational numbers. We here show that there are
simple polynomial equations over the rationals whose roots are not rational numbers.
Theorem 2.6 If the polynomial equation dn xn + dn−1 xn−1 + . . . + d1 x + d0 = 0 with integral coefficients has
a rational root ab then a|d0 and b|dn .
Proof. Suppose the polynomial equation dn xn + dn−1 xn−1 + . . . + d1 x + d0 = 0 with integral coefficients has
a rational root ab . Then we get
an
an−1
a
+ dn−1 n−1 + . . . . . . + d1 + d0
n
b
b
b
⇒ dn an + bdn−1 an−1 + . . . + bn−1 d1 a + bn d0
⇒ dn an + bdn−1 an−1 + . . . . . . + bn−1 d1 a
⇒ bn d0 + bn−1 d1 a + . . . . . . + bdn−1 an−1
dn
=
0
= 0
= −bn d0
= −dn an
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From the last two equations we get for some integers k1 and k2
ak1 = −bn d0
and
bk2 = −dn an .
As a and b has no common factor this shows that a|d0 and b|dn .
Corollary 2.7 If the polynomial equation xn + dn−1 xn−1 + . . . + d1 x + d0 = 0 with integral coefficients has a
rational root then it must be an integer which divides d0 .
Proof. Let the rational root be
divides d0 .
a
b.
Then a|d0 and b|1 so that b = 1. Thus the rational root is an integer which
As an application of this corollary we see that the equation xn − p = 0 has no rational root for n > 1 and p a
1
1
argument wil ensure that for a ∈ N, a n is a rational number
prime. So p n is not a rational number. Further
√
√
2
if a is of the form bn for some b ∈ N. Thus 2, 5, 7 3 are all irrational numbers.
A more detailed discussion on nature of different types of real numbers can be found in our earlier article [2].
2.3
Gaps between the rational numbers and the notion of Dedekind cut
Consider A = {x ∈ Q : x ≤ 0 or (x > 0 and x2 < 2)} and B = {x ∈ Q : x > 0 and x2 > 2}.
We show that A has no least upper bound in Q and B has no greatest lower bound in Q.
If possible let c be the lub of A in Q. Then by the law of trichotomy either c2 = 2 or c2 < 2 or c2 > 2. Cleary
first option is not possible. Suppose c2 < 2. We claim that there exists some n ∈ N such that (c + 1/n)2 < 2.
This is possible if c2 + 2c/n + 1/n2
i.e., if c2 + 1/n(2c + 1)
i.e., if 1/n(2c + 1)
i.e., if n.(2 − c2 )
<
<
<
>
2
2
2 − c2
2c + 1 − this is possible by Archimedean property.
This contradicts the fact that c is the lub A and so the option c2 < 2 is ruled out.
Next we consider the case c2 > 2. We claim that there exists some n ∈ N such that (c − 1/n)2 > 2.
This is possible if c2 − 2c/n + 1/n2
i.e., if c2 − 2c/n
i.e., if 2c/n
i.e., if n.(c2 − 2)
>
>
<
>
2
2
c2 − 2
2c − this is possible by Archimedean property.
Clearly c − 1/n is an upper bound of A which contradicts the fact that c is the lub A.
Thus A has no lub in Q. Similarly one can show that B has no glb in Q.
Now it is easy to check that
• A 6= φ, B 6= φ.
• A ∪ B = Q, A ∩ B = φ.
• a < b, ∀ a ∈ A, b ∈ B.
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but there exists no rational number c such that a ≤ c ≤ b for all a ∈ A, b ∈ B.
Let us think that the rational numbers are represented by dots on a straight line with positive rational numbers
representing dots on the right hand side of the dot ( representing 0 ) and the negative rational numbers
represnting dots on the left hand side of the dot ( representing 0 ). Then A consists of dots coming from left
hand side and we cannot find a dot to the right of which there is no element of A. B consists of dots coming
from right hand side and we can not find a dot to the left of which there is no element of B. So we can say that
there is a gap between elements of A and B. If we consider A0 = {x ∈ Q : x < 2} and B 0 = {x ∈ Q : x ≥ 2} then
A0 and B 0 satisfies all the above conditions but there is no gap between A0 and B 0 i.e., there exists c(= 2) ∈ Q
such that a ≤ c ≤ b for all a ∈ A0 , b ∈ B 0 .
Dedekind√constructed the real numbers by filling those gaps. He actually thought of the pair (A, B) as the real
number 2 and the pair (A0 , B 0 ) as the real number 2. If the pair (A, B) has a gap then it does not represent
a rational number and if the pair has no gap then it represents a rational
number. Note that B = Ac = Q − A
√
0
0c
0
and B = A = Q − A so that one can think that the real number ‘ 2’ is somehow represented by the subset
A of rational numbers and the real number ‘2’ is represented by the subset A0 . Dedekind introduced the notion
of cut now known as “Dedekind cut” which is defined as
Definition 2.8 Dedekind cut.
A set α of rational numbers is said to be a cut if
• α 6= φ and α 6= Q ( this means α is neither empty nor the whole set Q ).
• if p, q ∈ Q be such that q < p and p ∈ α then q ∈ α. ( this means if α contains a rational number then it
contains all rational numbers less than that )
• if p ∈ α then there exists some rational number q > p such that q ∈ α ( this means α contains no largest
rational number).
Take α = A, A0 , then both A, A0 are Dedekind cuts but if we take α = {x ∈ Q : x ≤ 3} then α is not a cut. A0
is known as rational cut as it gives nothing new.
Let R denotes the collection of all Dedekind cuts. We define addition, ordering and multiplication of cuts as
follows.
Definition 2.9 Addition of cuts. Let α, β be two cuts. Then α + β is defined as
α + β = {r ∈ Q : r = p + q
for some p ∈ α q ∈ β}.
Ordering of cuts. Define α < β iff there exists a rational number p such that p ∈ β but p ∈
/ α.
Rational cut. A cut α is said to be a rational cut iff there exists a rational number p such that
α = {r ∈ Q : r < p}.
In this case α is written as p∗ .
Multiplication of cuts. If α ≥ 0∗ , β ≥ 0∗ then define α.β as
α.β = {r ∈ Q : (r < 0) or (r = pq for some p ∈ α, q ∈ β p ≥ 0, q ≥ 0)}
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For any two cuts α and β we define
α.β
= (−α).β if α < 0∗ , β ≥ 0∗
= α.(−β) if α ≥ 0∗ , β < 0∗
= (−α).(−β) if α < 0∗ , β < 0∗
Verifying that the above notions are indeed welldefined one can prove that R is an ordered field (after some
rigorous calculations) with the set of positive elements P as
P = {α ∈ R : α > 0∗ }.
3
Completeness Property of R
Now we can show that in R there is no such gap ( as described above ). This is what is known as
Theorem 3.1 Dedekind’s completeness theorem.
Let A and B be two sets of real numbers such that
• A 6= φ, B =
6 φ.
• A ∪ B = R, A ∩ B = φ.
• α < β, ∀ α ∈ A, β ∈ B.
then there exists one and only one real number γ such that α ≤ γ ≤ β for all α ∈ A, β ∈ B.
Proof. Let γ be the set of all rational p such that p ∈ α for some α ∈ A. Then we note that
• γ 6= φ as A 6= φ.
• If β ∈ B and q ∈
/ β then q ∈
/ α for any α ∈ A( as α < β ). So q ∈
/ γ and γ 6= Q.
• If p ∈ γ and q < p then p ∈ α for some α ∈ A. So q ∈ α and hence q ∈ γ.
• If p ∈ γ then p ∈ α for some α ∈ A, hence there exists q > p such that q ∈ α and hence q ∈ γ.
Thus γ is a real number. Clearly α ≤ γ, ∀α ∈ A. If there were some β ∈ B for which β < γ then there would
be a rational number p ∈ γ such that p ∈
/ β. But if p ∈ γ then p ∈ α for some α ∈ A and so β < α which
contradicts the fact that every element of A is less than every element of B. So γ ≤ β, ∀β ∈ B.
We finally show that γ is unique. If possible there are two numbers γ1 and γ2 such that γ1 < γ2 . Then there
exists γ3 such that γ1 < γ3 < γ2 . Then γ3 ∈ A ( as γ3 < γ2 ) and γ3 ∈ B ( as γ1 < γ3 ). This is a contradiction
as A ∩ B = φ. Thus γ is unique.
Remark. The existence of γ shows that gaps which we found in the rational number system are now filled. It
follows that either A has the largest element or B has the smallest element. Further if we tried to repeat the
process which led us to the reals from the rational, by constructing cuts then every cut will correspond to some
real cut so that nothing new will be obtained. For this reason the theorem is called the completeness theorem for
real numbers.
Thus the real number system (R, +, ., <) can be defined as a set R alongwith two binary operations ‘+’ and ‘.’
and a relation ‘ <’ satisfying the following conditions:
A. ( Field )
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• x + (y + z) = (x + y) + z
• x+y =y+x
• ∃0∈F
3 x + 0 = x ∀x ∈ F
• for each x ∈ F ∃ z ∈ F
3 x+z =0
• x.(y.z) = (x.y).z
• x.y = y.x
• ∃1∈F
3 1.x = x ∀x ∈ F
• for each x 6= 0 ∃ z ∈ F
3 z.x = 1
• x.(y + z) = x.y + x.z
B. ( Ordering on field )
• If x > 0 and y > 0 then x + y > 0.
• If x > 0 and y > 0 then x.y > 0.
• For each element x of F exactly one of the following three holds
x = 0, x > 0, − x > 0
C. ( Dedekind’s Completeness property ) If A and B be two sets of real numbers such that
• A 6= φ, B 6= φ.
• A ∪ B = R, A ∩ B = φ.
• x < y, ∀ x ∈ A, y ∈ B.
then there exists one and only one real number z such that x ≤ z ≤ y for all x ∈ A, y ∈ B.
If R satisfies A then R is called a field, if R satisfies both A and B then R is called an ordered field and if R
also satisfies C then R is called a complete ordered field.
The completeness property (mentioned in C) can be replaced by the following :
Least upper bound Property : Every nonempty subset of R which is bounded above has a least upper
bound in R. (this property is equivalent to greatest lower bound property which says that every nonempty
subset of R which is bounded below has a greatest lower bound in R.)
Theorem 3.2 Let R be an ordered field. Then R satisfies Dedekind’s completeness property iff R satisfies least
upper bound property.
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Proof. We first show that Dedekind’s completeness property implies Least upper bound property.
Let E be a non-empty set of real numbers which is bounded above. We show that lub of E exists in R. Consider
A = {α : ∃ x ∈ E
3 α < x} and B = R − A.
Clearly no member of A is an upper bound of E and every member of B is an upper bound of E. We note that
• A ∪ B = R and A ∩ B = φ.
• As E 6= φ so there exists x ∈ E. For every α < x we have α ∈ A. Thus A 6= φ.
• As E is bounded above so there exists β ∈ R such that x ≤ β ∀x ∈ E. So β ∈ B and B 6= φ.
• If α ∈ A and β ∈ B then there exists x ∈ E such that α < x. As β is an upper bound of E so x ≤ β. Thus
α < β for all α ∈ A and β ∈ B.
Thus by Dedekind’s completeness property there exists γ ∈ R such that α ≤ γ ≤ β for all α ∈ A, β ∈ B. We
show that γ is lub of E.
If γ ∈ A then there exists x ∈ E such that γ < x. Choose some γ 0 such that γ < γ 0 < x, then γ 0 ∈ A which is a
contradiction.
So γ ∈ B. As γ ≤ β for all β ∈ B so γ is least of all upper bounds.
Thus Dedekind’s completeness property implies Least upper bound property.
We next show that Least upper bound property implies Dedekind’s completeness property.
Let A and B be two sets of real numbers such that
• A 6= φ, B 6= φ.
• A ∪ B = R, A ∩ B = φ.
• α < β, ∀ α ∈ A, β ∈ B.
Consider the set A which is non-empty bounded above and so by least upper bound property lub of A exists
and equals γ (say). Clearly α ≤ γ for all α ∈ A. We claim γ ≤ β for all β ∈ B, if not, then β < γ contradicts
the fact that γ is lub of A as β is an upper bound of A. Thus α ≤ γ ≤ β for all α ∈ A, β ∈ B. The uniqueness
of γ follows from uniqueness of lub.
Thus Least upper bound property implies Dedekind’s completeness property. This completes the
proof.
One important property of a complete ordered field is that it is Archimedean.
Theorem 3.3 Every complete ordered field is Archimedean.
Proof. If possible let R be a complete ordered field which is not Archimedean. Then there exists a, b ∈ R with
a > 0, b > 0 such that na ≤ b for all n ∈ N. So the set A = {na : n ∈ N } is bounded above by b and hence has
a lub, say, c in R. Thus (n + 1)a ≤ c for all n ∈ N i.e., na ≤ c − a for all n ∈ N. This shows that c − a is an
upper bound of A contradicting the fact c is lub of A. Thus R is Archimedean.
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Our next aim is to show that any two complete ordered fields are isomorphic so that we can define real number
as an element of a complete ordered field. As the rational number system Q is characterised as the smallest
ordered field so an Archimedean ordered field F contains Q, to be precise F contains an isomorphic image of Q
but we prefer to use the notion F contains Q. We next show that Q is dense in F i.e., given any two elements
a, b ∈ F with a < b then there exists an element c of the rational subfield Q such that a < c < b.
Theorem 3.4 If F is an Archimedean ordered field and a, b ∈ F with a < b then there exists an element c of
the rational subfield Q such that a < c < b.
Proof. Let a > 0. Then b − a > 0 and so by Archimedean property there exists s ∈ N such that s(b − a) > 1.
So
sb > sa + 1 . . . . . . (1)
Again by Archimedean property there exists k ∈ N such that k.1 = k > sa. Consider S = {m ∈ N : m > sa}.
Then S is a nonempty subset of N and so S has a least element, say, r so that r > sa ≥ r − 1. Then sa + 1 ≥ r
and so from (1) we get
sa < r ≤ sa + 1 < sb ⇒ a < r/s < b.
If a ≤ 0 then there exists a positive integer k such that a + k > 0 and so as above there exists r/s ∈ Q such
that a + k < r/s < b + k and so a < r/s − k < b.
This completes the proof.
Remark. As every complete ordered field is Archimedean so the rational subfield is dense in a complete ordered
field.
We now sketch a proof of the fact that
Theorem 3.5 Any two complete ordered fields are order isomorphic.
Proof. Let F and F 0 be two complete ordered fields containing the rational subfields Q and Q0 resp. Let Z
and Z 0 denote the domain of integers contained in Q and Q0 resp. Define a mapping φ : F −→ F 0 as follows :
φ(n) = n0 ∀ n ∈ Z
φ(m/n) = m0 /n0 ∀ m/n ∈ Q
So far we have defined φ on Q and one can easily check that it is an isomorphism from Q onto Q0 .
If an element c is in F - Q then considering the set Ic = {m/n ∈ Q : m/n < c} one can show by using lub
property that c = lub Ic . Now consider the set φ(Ic ) = {m0 /n0 ∈ Q0 : m/n ∈ Ic } in F 0 . We show that φ(Ic ) is
bounded above. There exists m0 /n0 ∈ Q such that c < m0 /n0 and so m0 /n0 < m00 /n00 for all m0 /n0 ∈ φ(Ic ).
This shows that φ(Ic ) is bounded above and so lub φ(Ic ) exists in F 0 . Let c0 = lub φ(Ic ).
We define φ(c) = c0 . Thus φ is defined on F and with some rigorous calculation one can show that φ is an order
isomorphism from F onto F 0 .
We are now in a position to define a real number
Definition 3.6 (Real number) A real number can be defined as an element of a complete ordered field and
R is a complete ordered field if
• (I) R is an ordered field and
R satisfies Dedekind’s completeness property.
OR
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• (II) R is an ordered field and
R satisfies least upper bound property.
OR
• (III) R is an ordered field and
R satisfies greatest lower bound property.
Remark 3.7 There are many other equivalent conditions for Dedekind’s completeness property. We state them
in the form of theorem
Theorem 3.8 Let R be an ordered field. Then the following conditions are equivalent:
1. R is Dedekind complete i.e. if A, B ⊂ R with A ∩ B = φ, A ∪ B = R, A 6= φ, B 6= φ and a < b ∀a ∈
A and b ∈ B then there exists a unique element ξ ∈ R such that a ≤ ξ ≤ b ∀a ∈ A and b ∈ B.
2. R satisfies greatest lower bound ( glb ) property i.e., every non-empty subset of R which is bounded below
has a greatest lower bound in R.
3. R satisfies least upper bound ( lub ) property i.e., every non-empty subset of R which is bounded above has
a least upper bound in R.
4. Every open cover of a closed bounded subset of R admits of a finite subcover.
5. Every bounded monotone sequence in R is convergent.
6. Every bounded sequence in R has a cluster point in R or every bounded sequence in R has a convergent
subsequence..
7. Every sequence in a closed bounded subset A of R has a cluster point in A.
8. Every infinite bounded subset of R has a limit point in R.
If further we assume that the field R is Archimedean then we can add two more equivalent
conditions
9. R satisfies Cantor’s intersection property i.e. if {[an , bn ]}∞
n=1 is a nested sequence of closed bounded
intervals with inf{bn −an : n ∈ N } = 0 then there exists a unique element ξ ∈ R such that ξ ∈ ∩∞
n=1 [an , bn ].
10. Every Cauchy sequence in R is convergent.
We end this discussion with example of a non-Archimedean ordered field taken from [3].
Example 3.9 (A non-Archimedean ordered field) Let D be the set of all sequences {an } of rational numbers having only a finite number of non-zero members. We define {an } = {bn } iff an = bn for all n ∈ N. We
introduce two binary operaions on D by the following definitions:
{an } + {bn } = {an + bn } ∀ n ∈ N
X
{an }.{bn } = {cn } where cn =
ar bs .
r+s=n+1
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Then it is easy to check that (D, +, .) is an integral domain with additive and multiplicative identity elements as
0 = {0, 0, . . .} and 1 = {1, 0, 0, . . .} respectively. Defining P to be the set of all elements {an } of D such that the
last non-zero member of {an } is a positive rational we can show that D is an ordered integral domain. Consider
the quotient field Q(D). Defining P̃ to be the set of all elements {an }/{bn } of Q(D) such that {an }.{bn } ∈ P
we can verify that Q(D) is an ordered field.
Take 1 = {1, 0, 0, . . .}, x = {0, 1, 0, 0 . . .}, then there exist no natural number n such that n.1 > x, thus Q(D) is
not an Archimedean field.
Acknowledgement.
We would like to thank Professor Abhijit Das Gupta, University of Detroit Mercy, Michigan for his invaluable
suggestion while preparing this article.
References
[1] Dedekind R., Essays on the theory of numbers, Dover Publications, 1963.
[2] Paul K. and paul Ghosh D., Numbers: Rational, Irrational or Transcendental? Sutra: International Journal
of Mathematical Science Education, Vol. 3 No. 1 (2010) pp. 24 - 30, (ISSN 0974-3340).
[3] Stoll R.R., Set theory and logic, Eurasia Publishing House(Pvt.) Ltd., 1986.