Chapter 1 – Real Number System
Subject: Real Analysis (Mathematics) Level: M.Sc.
Collected & Composed by: Atiq ur Rehman (), http://www.mathcity.org
The rational number system is inadequate for many purposes, both as a field
and as an order set for many purpose. This leads to introduction of so called irrational
numbers. We can prove in many ways that the rational number system has certain
gaps and hence we fail to use it as an ordered set and as a field. q
Í Theorem
There is no rational p such that p 2 = 2 .
Proof
Let us suppose that there exists a rational p such that p 2 = 2 .
This implies we can write
m
p= where m, n ∈ ¢ & m, n have no common factor.
n
m2
Then p = 2 ⇒ 2 = 2 ⇒ m2 = 2n 2
2
n
⇒ m is even
2
⇒ m is even
⇒ m is divisible by 2 and so m 2 is divisible by 4.
⇒ 2n2 is divisible by 4 and so n 2 is divisible by 2. Q m 2 = 2n 2
i.e. n 2 is even ⇒ n is even
⇒ m and n both have common factor 2.
Which is contradiction. (because m and n have no common factor.)
Hence p 2 = 2 is impossible for rational p. q
Í Theorem
Let A be the set of all positive rationals p such that p 2 < 2 and let B consist of
all positive rationals p such that p 2 > 2 then A contain no largest member and B
contains no smallest member.
Proof
We are to show that for every p in A there exists a rational q ∈ A such that
p < q and for all p ∈ B we can find rational q ∈ B such that q < p .
Associate with each rational p > 0 the number
p2 − 2 2p + 2
q= p− = ……..……. (i)
p+2 p+2
2
2p + 2 2( p 2 − 2)
Then q − 2 =
2
− 2 = ( p + 2) 2 ……………(ii)
p+2
Now if p ∈ A then p 2 < 2 ⇒ p 2 − 2 < 0
p2 − 2
Since from (i) q= p− ⇒ q> p
p+2
2( p 2 − 2) 2 2
, Ch. 01 - Real Number System - 2
p2 − 2
Since form (i) q= p− ⇒ q< p
p+2
2( p 2 − 2)
And >0 ⇒ q2 − 2 > 0 ⇒ q2 > 2 ⇒ q ∈ B
( p + 2) 2
The purpose of above discussion is simply to show that the rational number
system has certain gaps, in spite of the fact that the set of rationals is dense i.e. we
can always find a rational between any two given rational numbers. These gaps are
r+s
filled by the irrational number. (e.g. if r < s then r < < s .) q
2
Í Order on a set
Let S be a non-empty set. An order on a set S is a relation denoted by “ < ” with
the following two properties
(i) If x ∈ S and y ∈ S ,
then one and only one of the statement x < y , x = y , y < x is true.
(ii) If x, y, z ∈ S and if x < y , y < z then x < z .
Í Ordered Set
A set S is said to be ordered set if an order is defined on S.
Í Bound
Let S be an ordered set and E ⊂ S . If there exists a β ∈ S such that
x ≤ β ∀ x ∈ E , then we say that E is bounded above, and β is known as upper
bound of E.
Lower bound can be define in the same manner with ≥ in place of ≤ .
Í Least Upper Bound (Supremum)
Suppose S is an ordered set, E ⊂ S and E is bounded above. Suppose there
exists an α ∈ S such that
(i) α is an upper bound of E.
(ii) If γ < α then γ is not an upper bound of E.
Then α is called the least upper bound of E or supremum of E and is written as
sup E = α .
In other words α is the least member of the set of upper bound of E.
We can define the greatest lower bound or infimum of a set E , which is bounded
below, in the same manner. q
Í Example
Consider the sets
A = { p : p ∈ ¤ ∧ p 2 < 2}
B = { p : p ∈ ¤ ∧ p 2 > 2}
where ¤ is set of rational numbers.
Then the set A is bounded above. The upper bound of A are the exactly the members
of B. Since B contain no smallest member therefore A has no supremum in ¤ .
Similarly B is bounded below. The set of all lower bounds of B consists of A and
r ∈¤ with r ≤ 0 . Since A has no largest member, therefore, B has no infimum in ¤ .
Í Example
If α is supremum of E then α may or may not belong to E.
, Ch. 01 - Real Number System - 3
Í Example
1
Let E be the set of all numbers of the form , where n is the natural numbers.
n
1 1 1
i.e. E = 1, , , ,.........
2 3 4
Then sup E = 1 which is in E, but inf E = 0 which is not in E. q
Í Least Upper Bound Property
A set S is said to have the least upper bound property if the followings is true
(i) S is non-empty and ordered.
(ii) If E ⊂ S and E is non-empty and bounded above then supE exists in S.
Greatest lower bound property can be defined in a similar manner. q
Í Example
Let S be set of rational numbers and
E = { p : p ∈ ¤ ∧ p 2 < 2}
then E ⊂ ¤ , E is non-empty and also bounded above but supremum of E is not in S,
this implies that ¤ the set of rational numbers does not posses the least upper bound
property. q
Í Theorem
Suppose S is an ordered set with least upper bound property. B ⊂ S , B is non-
empty and is bounded below. Let L be set of all lower bounds of B then α = sup L
exists in S and also α = inf B .
In particular infimum of B exists in S.
OR
An ordered set which has the least upper bound property has also the greatest
lower bound property.
Proof
Since B is bounded below; therefore, L is non-empty.
Since L consists of exactly those y ∈ S which satisfy the inequality.
y≤x ∀ x∈B
We see that every x ∈ B is an upper bound of L.
⇒ L is bounded above.
Since S is ordered and non-empty therefore L has a supremum in S. Let us call it α .
If γ < α , then γ is not upper bound of L.
⇒ γ ∉B L B
⇒ α≤x ∀ x∈ B ⇒ α ∈L γ α
Now if α < β then β ∉ L because α = sup L .
We have shown that α ∈ L but β ∉ L if β > α . In other words, α is a lower
bound of B, but β is not if β > α . This means that α = inf B . q
………………………
, Ch. 01 - Real Number System - 4
Í Field
A set F with two operations called addition and multiplication satisfying the
following axioms is known to be field.
Axioms for Addition:
(i) If x, y ∈ F then x + y ∈ F . Closure Law
(ii) x + y = y + x ∀ x, y ∈ F . Commutative Law
(iii) x + ( y + z ) = ( x + y ) + z ∀ x, y, z ∈ F . Associative Law
(iv) For any x ∈ F , ∃ 0 ∈ F such that x + 0 = 0 + x = x Additive Identity
(v) For any x ∈ F , ∃ − x ∈ F such that x + (− x ) = (− x) + x = 0 +tive Inverse
Axioms for Multiplication:
(i) If x, y ∈ F then x y ∈ F . Closure Law
(ii) x y = y x ∀ x, y ∈ F Commutative Law
(iii) x ( y z ) = ( x y ) z ∀ x, y, z ∈ F
(iv) For any x ∈ F , ∃ 1∈ F such that x ⋅1 = 1⋅ x = x Multiplicative Identity
1 1 1
(v) For any x ∈ F , x ≠ 0 , ∃ ∈ F , such that x = x = 1 × tive Inverse.
x x x
Distributive Law
For any x, y , z ∈ F , (i) x ( y + z ) = xy + xz
(ii) ( x + y ) z = xz + yz q
Í Theorem
The axioms for addition imply the following:
(a) If x + y = x + z then y = z
(b) If x + y = x then y = 0
(c) If x + y = 0 then y = − x .
(d) −(− x) = x
Proof
(a) Suppose x + y = x + z .
Since y = 0 + y
= ( − x + x) + y Q −x+x=0
= − x + ( x + y) by Associative law
= − x + ( x + z) by supposition
= (− x + x ) + z by Associative law
= (0) + z Q −x+x=0
=z
(b) Take z = 0 in (a)
x+ y = x+0
⇒ y =0
(c) Take z = − x in (a)
x + y = x + ( − x)
⇒ y = −x
(d) Since (− x) + x = 0
then (c) gives x = −(− x) q
Subject: Real Analysis (Mathematics) Level: M.Sc.
Collected & Composed by: Atiq ur Rehman (), http://www.mathcity.org
The rational number system is inadequate for many purposes, both as a field
and as an order set for many purpose. This leads to introduction of so called irrational
numbers. We can prove in many ways that the rational number system has certain
gaps and hence we fail to use it as an ordered set and as a field. q
Í Theorem
There is no rational p such that p 2 = 2 .
Proof
Let us suppose that there exists a rational p such that p 2 = 2 .
This implies we can write
m
p= where m, n ∈ ¢ & m, n have no common factor.
n
m2
Then p = 2 ⇒ 2 = 2 ⇒ m2 = 2n 2
2
n
⇒ m is even
2
⇒ m is even
⇒ m is divisible by 2 and so m 2 is divisible by 4.
⇒ 2n2 is divisible by 4 and so n 2 is divisible by 2. Q m 2 = 2n 2
i.e. n 2 is even ⇒ n is even
⇒ m and n both have common factor 2.
Which is contradiction. (because m and n have no common factor.)
Hence p 2 = 2 is impossible for rational p. q
Í Theorem
Let A be the set of all positive rationals p such that p 2 < 2 and let B consist of
all positive rationals p such that p 2 > 2 then A contain no largest member and B
contains no smallest member.
Proof
We are to show that for every p in A there exists a rational q ∈ A such that
p < q and for all p ∈ B we can find rational q ∈ B such that q < p .
Associate with each rational p > 0 the number
p2 − 2 2p + 2
q= p− = ……..……. (i)
p+2 p+2
2
2p + 2 2( p 2 − 2)
Then q − 2 =
2
− 2 = ( p + 2) 2 ……………(ii)
p+2
Now if p ∈ A then p 2 < 2 ⇒ p 2 − 2 < 0
p2 − 2
Since from (i) q= p− ⇒ q> p
p+2
2( p 2 − 2) 2 2
, Ch. 01 - Real Number System - 2
p2 − 2
Since form (i) q= p− ⇒ q< p
p+2
2( p 2 − 2)
And >0 ⇒ q2 − 2 > 0 ⇒ q2 > 2 ⇒ q ∈ B
( p + 2) 2
The purpose of above discussion is simply to show that the rational number
system has certain gaps, in spite of the fact that the set of rationals is dense i.e. we
can always find a rational between any two given rational numbers. These gaps are
r+s
filled by the irrational number. (e.g. if r < s then r < < s .) q
2
Í Order on a set
Let S be a non-empty set. An order on a set S is a relation denoted by “ < ” with
the following two properties
(i) If x ∈ S and y ∈ S ,
then one and only one of the statement x < y , x = y , y < x is true.
(ii) If x, y, z ∈ S and if x < y , y < z then x < z .
Í Ordered Set
A set S is said to be ordered set if an order is defined on S.
Í Bound
Let S be an ordered set and E ⊂ S . If there exists a β ∈ S such that
x ≤ β ∀ x ∈ E , then we say that E is bounded above, and β is known as upper
bound of E.
Lower bound can be define in the same manner with ≥ in place of ≤ .
Í Least Upper Bound (Supremum)
Suppose S is an ordered set, E ⊂ S and E is bounded above. Suppose there
exists an α ∈ S such that
(i) α is an upper bound of E.
(ii) If γ < α then γ is not an upper bound of E.
Then α is called the least upper bound of E or supremum of E and is written as
sup E = α .
In other words α is the least member of the set of upper bound of E.
We can define the greatest lower bound or infimum of a set E , which is bounded
below, in the same manner. q
Í Example
Consider the sets
A = { p : p ∈ ¤ ∧ p 2 < 2}
B = { p : p ∈ ¤ ∧ p 2 > 2}
where ¤ is set of rational numbers.
Then the set A is bounded above. The upper bound of A are the exactly the members
of B. Since B contain no smallest member therefore A has no supremum in ¤ .
Similarly B is bounded below. The set of all lower bounds of B consists of A and
r ∈¤ with r ≤ 0 . Since A has no largest member, therefore, B has no infimum in ¤ .
Í Example
If α is supremum of E then α may or may not belong to E.
, Ch. 01 - Real Number System - 3
Í Example
1
Let E be the set of all numbers of the form , where n is the natural numbers.
n
1 1 1
i.e. E = 1, , , ,.........
2 3 4
Then sup E = 1 which is in E, but inf E = 0 which is not in E. q
Í Least Upper Bound Property
A set S is said to have the least upper bound property if the followings is true
(i) S is non-empty and ordered.
(ii) If E ⊂ S and E is non-empty and bounded above then supE exists in S.
Greatest lower bound property can be defined in a similar manner. q
Í Example
Let S be set of rational numbers and
E = { p : p ∈ ¤ ∧ p 2 < 2}
then E ⊂ ¤ , E is non-empty and also bounded above but supremum of E is not in S,
this implies that ¤ the set of rational numbers does not posses the least upper bound
property. q
Í Theorem
Suppose S is an ordered set with least upper bound property. B ⊂ S , B is non-
empty and is bounded below. Let L be set of all lower bounds of B then α = sup L
exists in S and also α = inf B .
In particular infimum of B exists in S.
OR
An ordered set which has the least upper bound property has also the greatest
lower bound property.
Proof
Since B is bounded below; therefore, L is non-empty.
Since L consists of exactly those y ∈ S which satisfy the inequality.
y≤x ∀ x∈B
We see that every x ∈ B is an upper bound of L.
⇒ L is bounded above.
Since S is ordered and non-empty therefore L has a supremum in S. Let us call it α .
If γ < α , then γ is not upper bound of L.
⇒ γ ∉B L B
⇒ α≤x ∀ x∈ B ⇒ α ∈L γ α
Now if α < β then β ∉ L because α = sup L .
We have shown that α ∈ L but β ∉ L if β > α . In other words, α is a lower
bound of B, but β is not if β > α . This means that α = inf B . q
………………………
, Ch. 01 - Real Number System - 4
Í Field
A set F with two operations called addition and multiplication satisfying the
following axioms is known to be field.
Axioms for Addition:
(i) If x, y ∈ F then x + y ∈ F . Closure Law
(ii) x + y = y + x ∀ x, y ∈ F . Commutative Law
(iii) x + ( y + z ) = ( x + y ) + z ∀ x, y, z ∈ F . Associative Law
(iv) For any x ∈ F , ∃ 0 ∈ F such that x + 0 = 0 + x = x Additive Identity
(v) For any x ∈ F , ∃ − x ∈ F such that x + (− x ) = (− x) + x = 0 +tive Inverse
Axioms for Multiplication:
(i) If x, y ∈ F then x y ∈ F . Closure Law
(ii) x y = y x ∀ x, y ∈ F Commutative Law
(iii) x ( y z ) = ( x y ) z ∀ x, y, z ∈ F
(iv) For any x ∈ F , ∃ 1∈ F such that x ⋅1 = 1⋅ x = x Multiplicative Identity
1 1 1
(v) For any x ∈ F , x ≠ 0 , ∃ ∈ F , such that x = x = 1 × tive Inverse.
x x x
Distributive Law
For any x, y , z ∈ F , (i) x ( y + z ) = xy + xz
(ii) ( x + y ) z = xz + yz q
Í Theorem
The axioms for addition imply the following:
(a) If x + y = x + z then y = z
(b) If x + y = x then y = 0
(c) If x + y = 0 then y = − x .
(d) −(− x) = x
Proof
(a) Suppose x + y = x + z .
Since y = 0 + y
= ( − x + x) + y Q −x+x=0
= − x + ( x + y) by Associative law
= − x + ( x + z) by supposition
= (− x + x ) + z by Associative law
= (0) + z Q −x+x=0
=z
(b) Take z = 0 in (a)
x+ y = x+0
⇒ y =0
(c) Take z = − x in (a)
x + y = x + ( − x)
⇒ y = −x
(d) Since (− x) + x = 0
then (c) gives x = −(− x) q
