Real Number System

Quadrant II – Transcript and Related Materials
Programme: Bachelor of Science (First Year)
Subject: Mathematics
Paper Code: MTC 101
Paper Title: DSC 1A: Calculus and Numerical Methods
Unit: Unit 1: Real Number System
Module Name: Algebra of Real Number System
Module No: 01
Name of the Presenter: Ms. Smita Kuncolienkar
Glossary of terms/words: 1
Additional Examples/Illustrations
Set of Natural Numbers denoted by N

N = {1,2 ,3, 4……..}
Set of Integers denoted by Z
Z = {…….-3,-2,-1,0,1,2,3,…….}
Set of Rational Numbers denoted by Q
𝑝
Q = { : 𝑝, 𝑞 ∈ 𝑍 ,𝑞 ≠ 0 }
𝑞
Set of Irrational Numbers denoted by 𝑸𝒄 i.e. the numbers which are not rational Numbers
Real Numbers denoted by R
Note: N ⊂ Z ⊂ Q ⊂ R
Algebraic Properties of R
On the set R, there are two operations defined on it ,denoted by + and . or x called Addition and
Multiplication respectively.
Given 𝑥, 𝑦 ∈ R , their sum is written as 𝑥 + 𝑦 and their product as 𝑥. 𝑦
The operations on R satisfy the following Properties:
• Properties of Addition
• Properties of Multiplication
Properties of Addition (+)
A1: Commutative property i.e. 𝑥 + 𝑦 = 𝑦 + 𝑥 for all 𝑥,y in R
A2: Assosiative property i.e. (𝑥 + 𝑦) + 𝑧 = 𝑥 + (𝑦 + 𝑧) for all 𝑥,𝑦,𝑧 in R
A3: Existence of identity i.e. there exists a unique element 0 in R such that
𝑥 + 0 = 𝑥 = 0 + 𝑥 for all 𝑥 in R
0 is additive identity in R
A4: Existence of inverse i.e.For every 𝑥 in R there exists a unique element
𝒚 in R such that 𝑥 + 𝑦 = 0 = 𝑦 + 𝑥
We denote this 𝑦 by −𝑥
Properties of Multiplication (.)

M1: Commutative property i.e. 𝑥. 𝑦 = 𝑦. 𝑥 for all 𝑥 , 𝑦 in R
M2: Associative property i.e.(𝑥. 𝑦). 𝑧 = 𝑥. (𝑦. 𝑧) for all 𝑥, 𝑦, 𝑧 in R
M3: Existence of identity i.e.there exists a unique element 1 in R such that 𝑥. 1 = 𝑥 =
1. 𝑥 for all x in R
1 is multiplicative identity in R
M4: Existence of inverse i.e. For every 𝑥 in R, 𝑥 ≠ 0 there exists a unique element 𝑦 in R such
that 𝑥. 𝑦 = 1 = 𝑦. 𝑥
1
We denote this 𝑦 by 𝑥 −1 or by
𝑥
Distributive Law
Multiplication (.) is distributive over addition (+)
i.e. For all 𝑥, 𝑦, 𝑧 in R ,we have
D1: 𝑥. (𝑦 + 𝑧) = 𝑥. 𝑦 + 𝑥. 𝑧 (Left Distributive Law)
D2: (𝑥 + 𝑦). 𝑧 = 𝑥. 𝑧 + 𝑦. 𝑧 (Right Distributive Law)
Definition:
(i) Subtraction in 𝑅 is defined as :
𝑥 − 𝑦 = 𝑥 + (−𝑦) for all 𝑥, 𝑦 in 𝑅
(ii) Division in 𝑅 is defined as :
𝑥 ÷ 𝑦 = 𝑥. 𝑦 −1 for all 𝑥, 𝑦 in 𝑅,𝑦 ≠ 0

Some basic techniques of Algebra
Theorem: a) If 𝑧 and 𝑎 are elements in 𝑅 with 𝑧 + 𝑎 = 𝑎 then 𝑧 = 0
b) If 𝑎 and ,𝑏 ≠ 0 are elements in 𝑅 such that 𝑎. 𝑏 = 𝑏 then
𝑎 = 1
c) If 𝑎 ∈ 𝑅 then 𝑎. 0 = 0
Proof: a) Given 𝑧 + 𝑎 = 𝑎 To prove 𝑧 = 0
now 𝑧 = 𝑧 + 0 Using A3
= 𝑧 + (𝑎 + (−𝑎)) Using A4
= (𝑧 + 𝑎) + (−𝑎) Using A2
= 𝑎 + (−𝑎) given
=0 using A4
b) 𝑎 = 𝑎. 1 using M3
1
= 𝑎. (𝑏. 𝑏 ) using M4
1
= (𝑎. 𝑏). (𝑏 ) using M2
1
= 𝑏. (𝑏) given
=1 using M4
c) 𝑎 + 𝑎 . 0 = 𝑎 . 1 + 𝑎 . 0 using M3
= 𝑎 . (1 + 0) using D1
= 𝑎 .1 using A3
= 𝑎 using M3
Hence 𝑎. 0 = 0 using (a)
Paper Code: MTC 101
Module Name: Axioms of Order Structure in R-Part 1
Module No: 02
Glossary of terms/words: 1.
Order Properties of R (Axioms of Order Structure in R)

+
𝑅 Contains a nonempty subset 𝑹 called set of positive real numbers, satisfying the following
properties: + +
O1: If𝑥, 𝑦 ∊ 𝑅 +, then 𝑥 + 𝑦 ∊ +𝑅 (closure property of Addition)
O2: If 𝑥 , 𝑦 ∊ 𝑅 ,then 𝑥. 𝑦 ∊ 𝑅 (closure property of Multiplication)
O3: For any 𝑥 ∊+𝑅 ,one and only one of the following
+ is true:
𝑥 ∊ 𝑅 𝑜𝑟 𝑥 = 0 𝑜𝑟 − 𝑥 ∊ 𝑅 (Trichotomy Law)
+
𝑅 contains positive real numbers +
+
Nonzero numbers which do not belong to 𝑅 are called negative real numbers
I f – 𝑥 ∊ 𝑅 , then 𝑥 is negative real number
𝐼𝑓 𝑥 , 𝑦 ∊ 𝑅 , then 𝒙 is said to be less than 𝒚 (denoted by

Definition: + 𝑥 < 𝑦 ) if
𝑦−𝑥 ∊ 𝑅
If 𝑥 < 𝑦, we write equivalently 𝑦 > 𝑥.
+
Thus 𝑦 > 𝑥 𝑖𝑓 𝑦 − 𝑥 ∊ 𝑅
Remark: 1. In 𝑥 < 𝑦, if we put 𝑥 = 0 we get 0 < 𝑦 which means 𝒚 > 𝟎 which means 𝑦 − 0 = 𝒚 is
positive Real number
+
Similarly put 𝑦 = 0 ,we get 𝒙 < 𝟎 which means 0 − 𝑥 =– 𝑥 ∊ 𝑅 which means 𝒙 is negative real
number
2. With above notation 𝑥 < 𝑦 means 𝑦 − 𝑥 > 0
3 . If 𝑥 < 𝑦 or 𝑥 = 𝑦 ,we write 𝑥 ≤ 𝑦
Using < and > , order properties of 𝑹 can be stated as follows
If 𝑥 , 𝑦 ∊ 𝑅 then
(1) 𝑥 > 0 , 𝑦 > 0 ⇒ 𝑥 + 𝑦 > 0 Closure under Addition
(2) 𝑥 > 0 , 𝑦 > 0 ⇒ 𝑥. 𝑦 > 0 Closure under Multiplication
(3) Either 𝑥 > 0 , 𝑥 = 0 𝑜𝑟 𝑥 < 0 trichotomy Law
Similarly 𝑥 ≥ 𝑦 means 𝑥 > 𝑦 or 𝑥 = 𝑦
Trichotomy Law can also be stated as follows:

If 𝒙, 𝒚 ∊ 𝑹 then either 𝒙 < 𝒚 𝒐𝒓 𝒙 = 𝒚 𝒐𝒓 𝒙 > 𝒚
Proof: Let 𝑥, 𝑦 ∊ 𝑅
If 𝑦 ∊ 𝑅 then −𝑦 ∊ 𝑅
If 𝑥 ∊ 𝑅 , −𝑦 ∊ 𝑅 ⇒ 𝑥 + (−𝑦) ∊ 𝑅 by closure property of Addition
∴ If 𝑥, 𝑦 ∊ 𝑅 then 𝑥 + (−𝑦) = 𝑥 − 𝑦 ∊ 𝑅
Now Consider the real number 𝑥 − 𝑦
By Trichotomy Law ,either 𝑥 − 𝑦 < 0 𝑜𝑟 𝑥 − 𝑦 = 0 𝑜𝑟 𝑥 − 𝑦 > 0
ie. either 𝑥 < 𝑦 𝑜𝑟 𝑥 = 𝑦 𝑜𝑟 𝑥 > 𝑦
EX : Show that if 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥 then 𝑥 = 𝑦 for 𝑥, 𝑦 ∊ 𝑅

Sol/ Consider real number 𝑥 − 𝑦
By Trichotomy Law,either 𝑥 − 𝑦 < 0 or 𝑥 − 𝑦 > 0 or 𝑥 − 𝑦 = 0
Suppose 𝑥 − 𝑦 ≠ 0 then 𝑥 − 𝑦 < 0 or 𝑥 − 𝑦 > 0
If 𝑥 − 𝑦 > 0 then 𝑥 > 𝑦
But this contradicts the given information that 𝑥 ≤ 𝑦
Hence 𝑥 − 𝑦 ⊁ 0
Similarly 𝑥 − 𝑦 ⊀ 0
Hence 𝑥 = 𝑦
EX: Show that if 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥 then 𝑥 = 𝑦 for 𝑥, 𝑦 ∊ 𝑅

Sol/ Consider real number 𝑥 − 𝑦
By Trichotomy Law, either 𝑥 − 𝑦 < 0 or 𝑥 − 𝑦 > 0 or 𝑥 − 𝑦 = 0
Suppose 𝑥 − 𝑦 ≠ 0 then 𝑥 − 𝑦 < 0 or 𝑥 − 𝑦 > 0
If 𝑥 − 𝑦 > 0 then 𝑥 > 𝑦
But this contradicts the given information that 𝑥 ≤ 𝑦
Hence 𝑥 − 𝑦 ⊁ 0
Similarly 𝑥 − 𝑦 ⊀ 0
Hence 𝑥 = 𝑦
Exercise: Use Trichotomy law to prove the following

If 𝑥 ≤ 𝑦 + 𝜖 for every 𝜖 > 0 then 𝑥 ≤ 𝑦 where x , y ∈ R
Paper Code: MTC 101
Module Name: Axioms of Order Structure in R-Part 2
Module No: 03
Glossary of terms/words:
Monotone Law of Addition

Let 𝑥, 𝑦 , 𝑧 ∊ 𝑅 then
(1) 𝑥 < 𝑦 ⇒ 𝑥 + 𝑧 < 𝑦 + 𝑧
(2) 𝑥 > 𝑦 ⇒ 𝑥 + 𝑧 > 𝑦 + 𝑧
+
Proof:(1) If 𝑥 < 𝑦 then 𝑦 − 𝑥 ∊ 𝑅
+
Hence (𝑦 + 𝑧) − (𝑥 + 𝑧) = 𝑦 + 𝑧 − 𝑥 − 𝑧 = 𝑦 − 𝑥 ∊ 𝑅
+
i.e. (𝑦 + 𝑧) − (𝑥 + 𝑧) ∊ 𝑅
i.e.𝑥 + 𝑧 < 𝑦 + 𝑧
+
Proof: (2) If 𝑥 > 𝑦 then 𝑥 − 𝑦 ∊ 𝑅
+
Hence (𝑥 + 𝑧) − (𝑦 + 𝑧) = 𝑥 + 𝑧 − 𝑦 − 𝑧 = 𝑥 − 𝑦 ∊ 𝑅
+
i.e. (𝑥 + 𝑧) − (𝑦 + 𝑧) ∊ 𝑅
i.e. 𝑥 + 𝑧 > 𝑦 + 𝑧
Monotone Law of Multiplication

(i) If 𝑥 < 𝑦 𝑎𝑛𝑑 𝑧 > 0 ⇒ 𝑥. 𝑧 < 𝑦. 𝑧
(ii) If 𝑥 < 𝑦 𝑎𝑛𝑑 𝑧 < 0 ⇒ 𝑥. 𝑧 > 𝑦. 𝑧
Proof: (i) If 𝑥 < 𝑦 𝑎𝑛𝑑 𝑧 > 0 ⇒ 𝑥. 𝑧 < 𝑦. 𝑧

+
Since 𝑥 < 𝑦 ⇒ 𝑦 − 𝑥 ∊ 𝑅
+
+ 𝑧>0⇒𝑧 ∊ 𝑅
(𝑦 − 𝑥). 𝑧 ∊ 𝑅 ………by closure property of Multiplication
+
i.e. 𝑦. 𝑧 − 𝑥. 𝑧 ∊ 𝑅
∴𝑥. 𝑧 < 𝑦. 𝑧
Proof: (ii) If 𝑥 < 𝑦 𝑎𝑛𝑑 𝑧 < 0 ⇒ 𝑥. 𝑧 > 𝑦. 𝑧
+
Since 𝑥 < 𝑦 ⇒ 𝑦 − 𝑥 ∊ 𝑅
+
+ 𝑧 < 0 ⇒ −𝑧 ∊ 𝑅
(𝑦 − 𝑥)(−𝑧) ∊ 𝑅 ………by closure property of Multiplication
+
i.e. 𝑥. 𝑧 − 𝑦. 𝑧 ∊ 𝑅
∴ 𝑥. 𝑧 > 𝑦. 𝑧
Transitive Property of R
(i) 𝑥 < 𝑦 𝑎𝑛𝑑 𝑦 < 𝑧 ⇒ 𝑥 < 𝑧
(ii) 𝑥 > 𝑦 𝑎𝑛𝑑 𝑦 > 𝑧 ⇒ 𝑥 > 𝑧
+
Proof: 𝑥 < 𝑦 ⇒ 𝑦 − 𝑥 ∊ 𝑅
+
𝑦 <𝑧 ⇒ 𝑧−𝑦 ∊ 𝑅
+
∴ (𝑦 − 𝑥) + (𝑧 − 𝑦) ∊ 𝑅 by closure property of addition
+
i.e. (𝑧 − 𝑥) ∊ 𝑅 ⇒ 𝑥 < 𝑧
EXERCISE 1 : Let 𝑥, 𝑦 , 𝑧, 𝑤 ∊ 𝑅 then
𝑥 < 𝑦 𝑎𝑛𝑑 𝑧 < 𝑤 ⇒ 𝑥 + 𝑧 < 𝑦 + 𝑤
Proof : 𝑥 < 𝑦 ∴ 𝑥+𝑧 < 𝑦+𝑧 …… by monotone Law of Addition
𝑧<𝑤 ∴ 𝑧+𝑦 < 𝑤+𝑦 ……. by monotone Law of Addition
i.e. 𝑦 + 𝑧 < 𝑦 + 𝑤 …….. by commutativity of Addiion
∴ 𝑥+𝑧 < 𝑦+𝑤 ……… by transitivity of <
EXERCISE 2 : Let 𝑥, 𝑦 , 𝑧, 𝑤 ∊ 𝑅 then
If 𝑥 > 𝑦 > 0 and 𝑧 > 𝑤 > 0 , then 𝑥. 𝑧 > 𝑦. 𝑤
Proof : Since 𝑥 > 𝑦 and 𝑧 > 0,then 𝑥. 𝑧 > 𝑦. 𝑧 ….. by monotone Law of multiplication
(i)
Also 𝑧 > 𝑤 and 𝑦 > 0,then 𝑧. 𝑦 > 𝑤. 𝑦 … by monotone Law of Multiplication
i.e. 𝑦. 𝑧 > 𝑦. 𝑤 ….. by commutativity of Multiplication (ii)
∴ 𝑥. 𝑧 > 𝑦. 𝑤 ….. by transitivity of > From (i) and (ii)
Programme: Bachelor of Science

Paper Code: MTC-101
Paper Title: Calculus and Numerical Methods
Unit: Real Number System
Module Name: Upper and Lower Bounds of Subsets of ℝ (Part 1)
Name of the Presenter: Mr. Swapnil Sadanand Belekar (Assistant
Professor)
Notes
Upper bound : Let 𝐴 be a non empty subset of ℝ. Let 𝛼 ∈ ℝ. Then 𝛼 is an upper bound of 𝐴
if ∀ 𝑥 ∈ 𝐴, 𝑥 ≤ 𝛼.
Geometrically this means, all the elements of 𝐴 are
to the left of α on the real number line. 𝛼
(∀𝑥 ∈ 𝐴, 𝑥 ≤ 𝛼)
Eg. Upper bound of (2,3) are 3, 3.1, 3.5, 4, 4.5,.etc

Not an upper bound: A real number 𝛼 is not an upper bound of 𝐴 if ∃ 𝑥 ∈ 𝐴 ∋ 𝑥 > 𝛼.
Geometrically this means we can find an
element of the set A to the right of α 𝛼 𝑥
Eg. 2.5 is not an upper bound of the set (2,3) as 2.6 ∈ (2, 3) and 2.6 > 2.5.
NOTE:- Upper bound of a set if exists need not be unique, i.e if 𝛼 is an upper bound of 𝐴
and 𝛼 ′ > 𝛼 then 𝛼 ′ is also an upper bound of 𝐴.
Lower bound: Let 𝐴 be a non empty subset of ℝ. Let 𝛼 ∈ ℝ. Then 𝛼 is a lower bound of 𝐴 if
∀ 𝑥 ∈ 𝐴, 𝛼 ≤ 𝑥.
Geometrically this means all the elements of 𝐴 are
to the right of 𝛼 on the real number line.
𝛼
Eg. Lower bound of (2,3) are 0, 1, 1.5, 2 .etc
Not a lower bound: A real number α is not a lower bound of 𝐴 if ∃𝑥 ∈ 𝐴 ∋ 𝑥 < 𝛼

Geometrically this means we can find an element of ( )
the set 𝐴 to the left of 𝛼. 𝑥 𝛼
Eg. 2.4 is not a lower bound of the set 𝐴 = (2,3), as 2.3 ∈ 𝐴 and 2.3 < 2.4.
NOTE:- Lower bound of a set if exists need not be unique, i.e if 𝛼 is a lower bound of 𝐴 and
𝛼 ′ < α then 𝛼 ′ is also a lower bound of 𝐴.

Paper Code: MTC-101
Professor)
Notes:
BOUNDED ABOVE SET: Let ∅ ≠ 𝐴 be a subset of ℝ . Then 𝐴 is said to be bounded above
in ℝ if ∃ 𝛼 ∈ ℝ such that 𝛼 is an upper bound of 𝐴 (i.e A is bounded above in ℝ if ∃ 𝛼 ∈
ℝ ∋ ∀𝑥 ∈ 𝐴, 𝑥 ≤ 𝛼).
BOUNDED BELOW SET: Let ∅ ≠ 𝐴 be a subset of ℝ . Then 𝐴 is said to be bounded
below in ℝ if ∃ 𝛼 ∈ ℝ such that 𝛼 is a lower bound of 𝐴 (i.e A is bounded below in ℝ if
∃ 𝛼 ∈ ℝ ∋ ∀𝑥 ∈ 𝐴, 𝑥 ≥ 𝛼).
NOT BOUNDED ABOVE SET : Let 𝜙 ≠ 𝐴 be a subset of ℝ. Then 𝐴 is not bounded above
in ℝ if ∀𝛼 ∈ ℝ ∃ 𝑥 ∈ 𝐴 ∋ 𝑥 > 𝛼.
NOT BOUNDED BELOW SET: Let 𝜙 ≠ 𝐴 be a subset ℝ. Then 𝐴 is not bounded below in
ℝ if ∀𝛼 ∈ ℝ ∃ 𝑥 ∈ 𝐴 ∋ 𝑥 < 𝛼.
BOUNDED SET: A set 𝐴 is said to be bounded if it’s both bounded above and bounded
below. Equivalently, a set A is bounded set if ∃ 𝑚, 𝑀 ∈ ℝ ∋ ∀ 𝑥 ∈ 𝐴, 𝑚 ≤ 𝑥 ≤ 𝑀. (i.e, a set
A is bounded set if it has an upper bound and lower bound).
UNBOUNDED SET: A set 𝐴 is said to be unbounded if its either not bounded above or not
bounded below.
Eg. 1) ℕ is bounded below by 1 but not bounded above in ℝ. (Proof will be done later in
Archimedean property module).
2) 𝕎 is bounded below by zero but not bounded above in ℝ.
3) ℤ is neither bounded above nor bounded below in ℝ.
4) ℚ, ℚ𝑐 are neither bounded above nor bounded below in ℝ.
Q. Check whether the following sets are bounded in ℝ or not. Justify.
1) ℝ
Soln: It is neither bounded above nor bounded below and hence an unbounded set.
For, Let 𝛼 ∈ ℝ. Then 𝛼 + 1 ∈ ℝ and 𝛼 + 1 > 𝛼. This implies ℝ is not bounded
above. Also, 𝛼 − 1 ∈ ℝ and 𝛼 − 1 < 𝛼. This shows that ℝ is not bounded below.
2) 𝐴 = {1,2, 3, 4 , 5}
Soln: A is bounded above in ℝ.
For, 1, 5 ∈ ℝ and ∀ x ∈ A, 1 ≤ 𝑥 ≤ 5.
1
3) 𝐴 = { 𝑛 | 𝑛 ∈ ℕ}.
Soln: A is bounded set.

1
For, 0 ≤ 𝑛 ≤ 1 ∀ 𝑛 ∈ ℕ. Thus A is bounded set.
4) 𝜙 .
Soln: Assume ∅ is not bounded. This means it is not bounded above or not bounded
below. Suppose it is not bounded above. Then ∀ 𝛼 ∈ ℝ there exists 𝑥 ∈ ∅ such that
𝑥 > 𝛼. But the existence of 𝑥 𝑖𝑛 ∅ itself is false. Thus our assumption is wrong. Hence
empty set is bounded above set. Similarly, it’s is bounded below and hence, empty set is
bounded set.
5) A = (0, 1)
Soln: For all 𝑥 ∈ 𝐴, 0 ≤ 𝑥 ≤ 1. Hence, A is bounded set.

Paper Code: MTC-101
Professor)
Notes
Result:- A subset of a bounded set is also bounded.
Proof:- Let 𝐴 be bounded set in ℝ and 𝐵 be subset of 𝐴.
Claim: B is bounded.
Let 𝑥 ∈ 𝐵 then 𝑥 ∈ 𝐴 since 𝐵 is a subset of 𝐴.
Since 𝐴 is bounded, ∃ 𝑚, 𝑀 ∈ ℝ ∋ 𝑚 ≤ 𝑥 ≤ 𝑀 ∀ 𝑥 ∈ 𝐴. In particular, 𝑚 ≤ 𝑥 ≤ 𝑀 ∀ 𝑥 ∈ 𝐵.
∴ 𝐵 is bounded.
Result:- Let 𝑆 be a non-empty subset of ℝ which is bounded above and 𝐴 be the set of all
upperbounds of 𝑆 then 𝐴 is bounded below in ℝ.
Proof:- Let 𝑥 ∈ 𝐴
Then 𝑥 is an upper bound of 𝑆
∴ 𝑥 ≥ 𝑦 ∀𝑦 ∈ 𝑆
In particular, 𝑥 ≥ 𝑦0 for some fixed 𝑦0 ∈ 𝑆 (this is possible since 𝑆 ≠ ∅)
Since 𝑥 is arbitrary,
𝑦0 is a lower bound of 𝐴
∴ 𝐴 is bounded below in ℝ.
Result:- Let 𝑆 be a non empty subset of ℝ which is bounded below and 𝐴 be the set of all
lowerbounds of 𝑆 then 𝐴 is bounded above in ℝ.
Proof:- Let 𝑥 ∈ 𝐴.
Then 𝑥 is a lower bound of 𝑆
∴ 𝑥 ≤ 𝑦 ∀𝑦 ∈ 𝑆
In particular, 𝑥 ≤ 𝑦0 for some fixed 𝑦0 ∈ 𝑆 (this is possible since 𝑆 ≠ ∅)
Since 𝑥 is arbitrary,
𝑦0 is a upper bound of 𝐴
∴ 𝐴 is bounded above in ℝ.

Paper Code: MTC-101
Professor)
Notes
Theorem: A finite subset of ℝ is bounded.
Proof : Let 𝐴 be a finite subset of ℝ .
Let 𝑚 = min 𝐴 & 𝑀 = 𝑚𝑎𝑥 𝐴.
Let 𝑥 ∈ 𝐴 then 𝑚 ≤ 𝑥 ≤ 𝑀.
∴ 𝐴 is bounded.
WAYS TO GENERATE MORE BOUNDED SETS

Theorem: Le t 𝐴 & 𝐵 be bounded subsets of ℝ then,
I. 𝐴 ∪ 𝐵 is bounded (i.e union of two bounded sets is bounded)
II. 𝐴 ∩ 𝐵 is bounded (i.e intersection of two bounded sets is bounded)
Proof:
since 𝐴 is bounded, ∃ 𝑚1, 𝑀1 ∈ ℝ ∋ 𝑚1 ≤ 𝑥 ≤ 𝑀1 , ∀ 𝑥 ∈ 𝐴} ----- 1
since 𝐵 is bounded, ∃ 𝑚2, 𝑀2 ∈ ℝ ∋ 𝑚2 ≤ 𝑥 ≤ 𝑀2 , ∀ 𝑥 ∈ 𝐵} ----- 2

let 𝑀 = max{𝑀1 , 𝑀2 } & m = min {𝑚1 , 𝑚2 }
I) let 𝑥 ∈ 𝐴 ∪ 𝐵 then 𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵.
If 𝑥 ∈ 𝐴 then 𝑚 ≤ 𝑚1 ≤ 𝑥 ≤ 𝑀1 ≤ 𝑀 i.e 𝑚 ≤ 𝑥 ≤ 𝑀
If 𝑥 ∈ 𝐵 then 𝑚 ≤ 𝑚2 ≤ 𝑥 ≤ 𝑀2 ≤ 𝑀 i.e 𝑚 ≤ 𝑥 ≤ 𝑀
This implies ∀𝑥 ∈ 𝐴 ∪ 𝐵 , ∃ 𝑚 & 𝑀 ∈ ℝ ∋ 𝑚 ≤ 𝑥 ≤ 𝑀.
∴ 𝐴 ∪ 𝐵 is a bounded set.
II) If 𝐴 ∩ 𝐵 = 𝜙 , then nothing to prove as ∅ is always bounded.
Let 𝐴 ∩ 𝐵 ≠ ∅.
Let 𝑥 ∈ 𝐴 ∩ 𝐵
since 𝐴 ∩ 𝐵 ⊆ 𝐴
∴ 𝑥 ∈ 𝐴.
since 𝐴 is bounded by 1 & 2, ∃ 𝑚, 𝑀 ∈ ℝ ∋ 𝑚1 ≤ 𝑥 ≤ 𝑀1
⟹ 𝐴 ∩ 𝐵 is bounded.

Paper Code: MTC-101
Module Name: Lub of Subsets of ℝ (Part 1)
Professor)
Notes
SUPREMUM :- Let 𝜙 ≠ 𝐴 is a subset of ℝ which is bounded above and 𝛼 ∈ ℝ. Then 𝛼 is
said to be supremum of 𝐴 (denoted by sup 𝐴) if
a) 𝛼 is an upper bound of 𝐴.
b) If 𝛽 is any other upper bound of 𝐴 then 𝛼 ≤ 𝛽.
Supremum is also called least upper bound (𝑙𝑢𝑏).
INFIMUM :- Let 𝜙 ≠ 𝐴 is a subset of ℝ which is bounded below and 𝛼 ∈ ℝ . Then 𝛼 is
said to be infimum of 𝐴 (denoted by inf 𝐴) if
a) 𝛼 is an lower bound of 𝐴.
b) If 𝛽 is any other lower bound of 𝐴 then 𝛽 ≤ 𝛼.
Infimum is also called greatest lower bound (𝑔𝑙𝑏).
Eg. 𝐴 = (0,1)
sup 𝐴 = 1 and inf 𝐴 = 0
NOTE:-
1) sup 𝐴 need not be an element of 𝐴
2) inf 𝐴 need not be an element of 𝐴 (eg. refer above)
3) If sup 𝐴 ∈ 𝐴 then sup 𝐴 is the maximum element of 𝐴.
4) If inf 𝐴 ∈ 𝐴 then inf 𝐴 is the minimum element of A.
Eg. 𝐴 = [0,1],
sup A = 1 = max A & inf 𝐴 = 0 = min 𝐴
Question:- Find 𝑠𝑢𝑝 & 𝑖𝑛𝑓 of the following sets.

𝜋
1) 𝐴 = {cos 𝑥 ∶ 0 < 𝑥 < 2 }
2) 𝐵 = {𝑥 ∈ ℝ ∶ 𝑥 2 − 𝑥 − 6 = 0}
Solution:-
𝜋
1) 0 < 𝑥 < 2
𝜋 𝜋
Then, 𝑐𝑜𝑠 0 > 𝑐𝑜𝑠𝑥 > 𝑐𝑜𝑠 2 (since 𝑐𝑜𝑠 is decreasing on (0, 2 ))
1 cos 𝑥
𝜋
i.e 1 > 𝑐𝑜𝑠𝑥 > 0 ∀ 𝑥 ∈ (0, 2 )
∴ sup 𝐴 = 1 , inf 𝐴 = 0 0 𝜋
2
2) 𝐵 = {𝑥 ∈ ℝ /𝑥 2 − 𝑥 − 6 = 0}
= {𝑥 ∈ ℝ /(𝑥 + 2)(𝑥 − 3) = 0}
= {𝑥 ∈ ℝ /𝑥 = −2 𝑜𝑟 𝑥 = +3}
= {−2, 3}
sup 𝐵 = 3, inf 𝐵 = −2.

Paper Code: MTC-101
Module Name: Lub of Subsets of ℝ (Part 2)
Name of the Presenter: Mr. Swapnil Sadanand Belekar (Assistant Professor)
Notes
Question: Show that 𝑙𝑢𝑏 of the set (0,1) = 1.
Proof:- Let A = (0, 1).
Notice, ∀ 𝑥 ∈ 𝐴 , 0 < 𝑥 < 1
⟹ 1 is an upper bound of 𝐴 ----------------------(I)
Let 𝛽 be any other upper bound of 𝐴.
Claim:- 𝛽 ≥ 1
Assume 𝛽 < 1.
1
Since 𝛽 is an upper bound of 𝐴 and ∈ 𝐴, we have
2
1
0 < 2 ≤ 𝛽.
⟹ 𝛽 > 0.
Thus we have, 0 < 𝛽 < 1.
1+𝛽 1+𝛽
Now 1 > >𝛽>0 (Notice is the midpoint of 1 and 𝛽)
2 2
1+𝛽
(Reason for > 𝛽:
2
1−𝛽
𝛽 < 1 ⟹ −𝛽 > −1 ⟹ 1 − 𝛽 > 0 ⟹ > 0.
2
1+𝛽 1−𝛽 1+𝛽
Now, −𝛽 = >0 ⟹ > 𝛽 ).
2 2 2
1+𝛽 1+𝛽
⟹ ∈ 𝐴 and >𝛽
2 2
This is a contradiction to the fact that 𝛽 is an upper bound of 𝐴

∴𝛽≥1 ------------------------------------ (II)
From (I) and (II), it follows that,
1 is the 𝑙𝑢𝑏 of 𝐴.
Quadrant II - Transcript and Related Materials

Paper Code: MTC 101
Paper Title: DSC1A Calculus and Numerical methods
Module Name: Archimedean Property of R - Part 1
Module No: 22
Name of the Presenter: Ms.Namrata D. Khorje
Notes:
Archimedean Property 1 :
The set N of natural numbers is not bounded above in R.
i.e given any x ∈ R, there exists n ∈ N such that n > x.
Proof : Let if possible N is bounded above.
∴ By Least Upper Bound property of R, ∃ α ∈ R such that α = sup(N).
⇒ α − 1 is not an upper bound of N as α − 1 < α
⇒ ∃ n0 ∈ N such that n0 > α − 1
⇒ n0 + 1 > α
This is a contradiction as α is an upper bound of N and n0 + 1 ∈ N as n0 ∈ N
∴ N is not bounded above in R.
Archimedean Property 2 :
Given x, y ∈ R with x > 0 ∃ n ∈ N such that nx > y.
Proof : We will prove the result by contradiction.
For each n ∈ N, ∃ x, y ∈ R with x > 0 such that nx ≤ y

y
⇒ n≤ ∀n∈N
x
y
⇒ is an upper bound for N.
x
⇒ N is bounded above.
This is contradiction as N is not bounded above.
∴ Given x, y ∈ R, x > 0 ∃ n ∈ N such that nx > y.


Paper Code: MTC 101
Module Name: Archimedean Property of R - Part 2
Module No: 23
Notes:
Applications of Archimedean Property :
1
1. Given > 0, there exists n ∈ N such that < .
n
Proof : Use Archimedean Property 2 with x = and y = 1.
Then ∃ n ∈ N such that n > 1.

1
i.e ∃ n ∈ N such that < .
n
1
2. Let x ≥ 0. Then x = 0 iff for each n ∈ N we have x ≤ .
n
Proof : Let x = 0
Given x ≥ 0.
1
0< ∀n∈N
n
1
⇒ x< ∀n∈N
n
1
Conversely assume that x ≤ ∀n∈N
n
Claim : x = 0
Let if possible x > 0.

1
Then ∃ n0 ∈ N such that < x.
n0
1
This is a contradiction as x ≤ ∀n∈N
n
Therefore our assumption that x > 0 is wrong.
Hence x = 0.

1
3. 0 is infimum of A = n∈N
n
(i) ∀ n ∈ N, n > 0
1
⇒ ∀ n ∈ N, > 0.
n
∴ 0 is lower bound of A.
(ii) Let β be any other lower bound of A.
Claim : β ≤ 0.
Let if possible β > 0.

1
∴ ∃ n0 ∈ N such that < β.
n0

1 1
This is a contradiction as β is lower bound of A and as n0 ∈ N , ∈ n∈N
n0 n
Therefore our assumption that β > 0 is wrong.
Hence β ≤ 0.

1
From (i) and (ii), 0 is infimum of A = n∈N .
n

Paper Code: MTC 101
Paper Title: DSC 1A Calculus and Numerical Methods
Unit: 1
Module Name: Archimedian property of ℝ, Part 3
Module No: 24
Name of the Presenter: Mr. Brandon Fernandes
Notes
Proposition: The set ℤ of integers is neither bounded above nor bounded
below.
Proof:
Claim 1: ℤ is not bounded above in ℝ .
If 𝛼 ∈ ℝ is an upper bound of ℤ then ∀ 𝑛 ∈ ℕ ⊂ ℤ we have 𝑛 ≤ 𝛼
⇒ ℕ is bounded above in ℝ .
⇒⇐ The Archimedian property which states that ℕ is not bounded above in ℝ.

Hence ℤ is not bounded above in ℝ . - - - - - - - - - - - - - - (1)
Claim 2: ℤ is not bounded below in ℝ .
If 𝛽 ∈ ℝ is an lower bound of ℤ then ∀ 𝑛 ∈ ℕ ⊂ ℤ we have −𝑛 ∈ ℤ and hence

𝛽 ≤ −𝑛
i.e. ∀ 𝑛 ∈ ℕ we have 𝑛 ≤ −𝛽
⇒ ℕ is bounded above in ℝ .
⇒⇐ The Archimedian property which states that ℕ is not bounded above in ℝ.
Hence ℤ is not bounded below in ℝ . - - - - - - - - - - - - - - (2)
Hence the set ℤ of integers is neither bounded above nor bounded below.
By (1) & (2)
Result: If 𝑥 ∈ ℝ is fixed then ∃ 𝑚, 𝑛 ∈ ℤ ∋ 𝑚 < 𝑥 < 𝑛 .

Proof:
Claim 1: x < n
Suppose ∄ n ∈ ℤ ∋ x < n
⇒ ∀n ∈ ℤ x≥n
⇒⇐ ℤ is not bounded above in ℝ.
This proves our claim that x < n - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (1)
Claim 2: m < x
Suppose ∄ m ∈ ℤ ∋ m < x
⇒ ∀m ∈ℤ m≥x
⇒⇐ ℤ is not bounded below in ℝ.
This proves our claim that m < x - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (2)
Hence 𝑚 < 𝑥 < 𝑛 by (1) & (2)

1
Result: Let 𝑎, 𝑏 ∈ ℝ . Show that if 𝑎 < 𝑏 + ∀ 𝑛 ∈ ℕ then 𝑎 ≤ 𝑏 .
𝑛
Proof:
1
Assume 𝑎 < 𝑏 + ∀ 𝑛 ∈ ℕ and 𝑎 > 𝑏
𝑛
⇒ 𝑎−𝑏 > 0
Hence by Archimedian property, given (𝑎 − 𝑏) > 0
⇒ ∃ 𝑛 ∈ ℕ ∋ 𝑛(𝑎 − 𝑏) > 1
1
⇒ ∃ 𝑛 ∈ ℕ ∋ (𝑎 − 𝑏) >
𝑛
1
⇒∃𝑛∈ℕ ∋ 𝑎 >𝑏+
𝑛
This contradicts our assumption.
Hence our assumption must be false.
∴𝑎≤𝑏

Paper Code: MTC 101
Unit: 1
Module Name: Intervals and their types
Module No: 25
Name of the Presenter: Poonam Govind Dangui
Notes:
Definition of an Interval:
A subset 𝐽 ⊂ ℝ is said to be an interval if 𝑎, 𝑏 ∈ 𝐽 and 𝑎 < 𝑥 < 𝑏 implies 𝑥 ∈ 𝐽.
Hence ∀𝑎, 𝑏 ∈ 𝐽 and 𝑎 < 𝑥 < 𝑏 we have 𝑥 ∈ 𝐽.
Types of intervals:
Let 𝑎 ≤ 𝑏 be real numbers. Define,
1. [𝑎, 𝑏]: = {𝑥 ∈ ℝ ∕ 𝑎 ≤ 𝑥 ≤ 𝑏}
2. (𝑎, 𝑏): = {𝑥 ∈ ℝ ∕ 𝑎 < 𝑥 < 𝑏}
3. [𝑎, 𝑏): = {𝑥 ∈ ℝ ∕ 𝑎 ≤ 𝑥 < 𝑏}
4. (𝑎, 𝑏]: = {𝑥 ∈ ℝ / 𝑎 < 𝑥 ≤ 𝑏}
5. [𝑎, ∞): = {𝑥 ∈ ℝ ∕ 𝑥 ≥ 𝑎}
6. (𝑎, ∞): = {𝑥 ∈ ℝ ∕ 𝑥 > 𝑎}
7. (−∞, 𝑏]: = {𝑥 ∈ ℝ ∕ 𝑥 ≤ 𝑏}
8. (−∞, 𝑏): = {𝑥 ∈ ℝ ∕ 𝑥 < 𝑏}
9. (−∞, ∞): = ℝ
Note:
• The first 4 types of intervals [𝑎, 𝑏], (𝑎, 𝑏), [𝑎, 𝑏), (𝑎, 𝑏] are called finite or
bounded intervals.
• The last 5 types of intervals [𝑎, ∞), (𝑎, ∞), (−∞, 𝑏], (−∞, 𝑏), (−∞, ∞)
are called infinite or unbounded intervals.
• The intervals [𝑎, 𝑏], [𝑎, ∞)𝑎𝑛𝑑 (−∞, 𝑏] are called closed intervals.
• The intervals [𝑎, 𝑏), (𝑎, 𝑏] are called semi-open (or semi-closed) intervals.
• The intervals (𝑎, 𝑏), (𝑎, ∞)𝑎𝑛𝑑 (−∞, 𝑏) are called open intervals.
• If a = 𝑏, then
[𝑎, 𝑏] = [𝑎, 𝑎] = {𝑎} and (𝑎, 𝑏) = (𝑎, 𝑎) = ∅ are both intervals.

Paper Code: MTC 101
Unit: 1
Module Name: Nested Interval Theorem
Module No: 26
Name of the Presenter: Poonam Govind Dangui
Notes:
Nested interval theorem states that,
Let 𝑱𝒏 ≔ [𝒂𝒏 , 𝒃𝒏 ] be intervals in ℝ such that 𝑱𝒏+𝟏 ⊆ 𝑱𝒏 for all 𝒏 ∈ ℕ. Then
⋂ 𝑱𝒏 ≠ ∅.
𝑱𝟏 = [𝒂𝟏 , 𝒃𝟏 ], 𝑱𝟐 = [𝒂𝟐 , 𝒃𝟐 ], … . . , 𝑱𝒏 = [𝒂𝒏 , 𝒃𝒏 ], 𝑱𝒏+𝟏 = [𝒂𝒏+𝟏 , 𝒃𝒏+𝟏 ], … ..
𝑱𝒏+𝟏 ⊆ 𝑱𝒏 for all 𝒏 ∈ ℕ means … . . ⊆ 𝑱𝒏+𝟏 ⊆ 𝑱𝒏 ⊆ ⋯ ⊆ 𝑱𝟑 ⊆ 𝑱𝟐 ⊆ 𝑱𝟏 (Nested
sequence)
Proof:
We will make use of LUB (least upper bound) property of ℝ which states that,
Given any non-empty subset of ℝ which is bounded above in ℝ has a least upper
bound that is, Supremum.
Consider a set, 𝐴 = {𝑎𝑛 /𝑛 ∈ ℕ}
Then 𝐴 is a non-empty subset of ℝ.
Claim: 𝑏𝑘 is an upper bound of 𝐴 ∀𝑘 ∈ ℕ.
That is, 𝑎𝑛 ≤ 𝑏𝑘 ∀𝑛, 𝑘
If 𝑘 ≤ 𝑛, then [𝑎𝑛 , 𝑏𝑛 ] ⊆ [𝑎𝑘 , 𝑏𝑘 ] (by definition of Nested interval)
Then, 𝑎𝑛 ≤ 𝑏𝑛 ≤ 𝑏𝑘 …………………… (1)
If 𝑘 > 𝑛, then [𝑎𝑘 , 𝑏𝑘 ] ⊆ [𝑎𝑛 , 𝑏𝑛 ] (by definition of Nested interval)
Then, 𝑎𝑛 < 𝑎𝑘 ≤ 𝑏𝑘 …………………… (2)
Hence from (1) and (2), 𝑎𝑛 ≤ 𝑏𝑘 ∀𝑛, 𝑘.
Therefore, by the LUB property, ∃ 𝑐 ∈ ℝ ∋ 𝑐 = 𝑆𝑢𝑝 𝐴.
We will show that 𝑐 ∈ 𝐽𝑛 ∀𝑛
Since 𝑐 is an upper bound of 𝐴, 𝑎𝑛 ≤ 𝑐 ∀𝑛
Since each 𝑏𝑛 is an upper bound of 𝐴 and 𝑐 is a least upper bound of 𝐴, 𝑐 ≤ 𝑏𝑛 ∀𝑛
Therefore, 𝑎𝑛 ≤ 𝑐 ≤ 𝑏𝑛 ∀𝑛
This implies, 𝑐 ∈ [𝑎𝑛 , 𝑏𝑛 ] = 𝐽𝑛 ∀𝑛
Hence,⋂ 𝐽𝑛 ≠ ∅.

Paper Code: MTC 101
Module Name: Absolute Value of a real number and their properties - Part 3
Module No: 29
Notes:
Triangle Inequality :
|x + y| ≤ |x| + |y| ∀ x, y ∈ R.
Proof : We will prove that max{(x + y), −(x + y)} ≤ |x| + |y|
We know, x ≤ |x| and y ≤ |y|
∴ x + y ≤ |x| + |y| (1)

Also −x ≤ |x| and −y ≤ |y|
∴ −(x + y) ≤ |x| + |y| (2)

From (1) and (2), max{(x + y), −(x + y)} ≤ |x| + |y|
⇒ |x + y| ≤ |x| + |y| ∀ x, y ∈ R
Applications of Triangle Inequality :

1. |x| − |y| ≤ |x − y| ∀ x, y ∈ R.

Proof : We will prove that −|x − y| ≤ |x| − |y| ≤ |x − y|
Consider,
|y| = |y − x + x|
≤ |y − x| + |x| (By Triangle Inequality)
= |x − y| + |x| ( |x| = | − x| )
∴ −|x − y| ≤ |x| − |y| (i)
Also,
|x| = |x − y + y|
≤ |x − y| + |y| (By Triangle Inequality)
|x| − |y| ≤ |x − y| (ii)
From (i) and (ii),

−|x − y| ≤ |x| − |y| ≤ |x − y|

⇒ |x| − |y| ≤ |x − y| ∀ x, y ∈ R.
2. |x − y| ≤ |x| + |y| ∀ x, y ∈ R.
Proof : |x − y| = |x + (−y)| ≤ |x| + | − y|
∴ |x − y| ≤ |x| + |y| (∵ | − y| = |y|)


Paper Code: MTC 101
Module Name: Absolute Value of a real number and their properties - Part 4
Module No: 30
Notes:
Generalization of Triangle Inequality :
If x1 , x2 , . . . . . . xn are any real numbers then
|x1 + x2 + . . . . . . + xn | ≤ |x1 | + |x2 | + . . . . . . + |xn |
Proof : We will prove the result by Induction.
Let P (n) : |x1 + x2 + . . . . . . + xn | ≤ |x1 | + |x2 | + . . . . . . + |xn |
For n = 2, we have
P (2) : |x1 + x2 | ≤ |x1 | + |x2 | (By Triangle Inequality)
∴ P (2) is true.
Assume P (k) is true for some k ∈ N.
i.e P (k) : |x1 + x2 + . . . . . . + xk | ≤ |x1 | + |x2 | + . . . . . . + |xk |
To prove that P (k + 1) is true.
i.e to prove that P (k + 1) : |x1 + x2 + . . . . . . + xk + xk+1 | ≤ |x1 | + |x2 | + . . . . . . + |xk | + |xk+1 |
|x1 + x2 + . . . . . . + xk + xk+1 | ≤ |x1 + x2 + . . . . . . + xk | + |xk+1 | (By Triangle Inequality)
|x1 + x2 + . . . . . . + xk + xk+1 | ≤ |x1 | + |x2 | + . . . . . . + |xk | + |xk+1 |
∴ P (k + 1) is true.
Hence by induction the result is true for all n ∈ N.

Necessary and Sufficient condition for Boundedness of a set in
terms of Absolute value of a real number :
A non-empty set A is bounded iff there exists a real number M > 0 such that |x| ≤ M for all x ∈ A.
Proof : Let A be bounded in R.
Therefore there exists α, β ∈ R such that α ≤ x ≤ β ∀x∈R
Let M = max{|α|, |β|}
So x ≤ β ≤ |β| ≤ max{|α|, |β|} = M .
⇒ x≤M (i)
Now −α ≤ |α| ≤ max{|α|, |β|} = M
∵ α≤x
⇒ −α ≥ x
⇒ M ≥ −α ≥ −x
⇒ M ≥ −x
⇒ −M ≤ x (ii)
From (i) and (ii), we have
−M ≤ x ≤ M
⇒ |x| ≤ M ∀ x ∈ A.
Conversely, if there exists M > 0 such that |x| ≤ M ∀x∈A
i.e there exists M > 0 such that −M ≤ x ≤ M ∀x∈A
⇒ A is bounded.

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Quadrant II – Transcript and Related Materials

Programme: Bachelor of Science (First Year)

Paper Code: MTC 101

Paper Title: DSC 1A: Calculus and Numerical Methods

Unit: Unit 1: Real Number System

Module Name: Algebra of Real Number System

Name of the Presenter: Ms. Smita Kuncolienkar

Set of Natural Numbers denoted by N

Properties of Multiplication (.)

Multiplication (.) is distributive over addition (+)

i.e. For all 𝑥, 𝑦, 𝑧 in R ,we have

D1: 𝑥. (𝑦 + 𝑧) = 𝑥. 𝑦 + 𝑥. 𝑧 (Left Distributive Law)

D2: (𝑥 + 𝑦). 𝑧 = 𝑥. 𝑧 + 𝑦. 𝑧 (Right Distributive Law)

𝑥 − 𝑦 = 𝑥 + (−𝑦) for all 𝑥, 𝑦 in 𝑅

(ii) Division in 𝑅 is defined as :

𝑥 ÷ 𝑦 = 𝑥. 𝑦 −1 for all 𝑥, 𝑦 in 𝑅,𝑦 ≠ 0

Programme: Bachelor of Science (First Year)

Paper Code: MTC 101

Paper Title: DSC 1A: Calculus and Numerical Methods

Unit: Unit 1: Real Number System

Module Name: Axioms of Order Structure in R-Part 1

Name of the Presenter: Ms. Smita Kuncolienkar

Order Properties of R (Axioms of Order Structure in R)

𝐼𝑓 𝑥 , 𝑦 ∊ 𝑅 , then 𝒙 is said to be less than 𝒚 (denoted by

Using < and > , order properties of 𝑹 can be stated as follows

Trichotomy Law can also be stated as follows:

EX : Show that if 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥 then 𝑥 = 𝑦 for 𝑥, 𝑦 ∊ 𝑅

EX: Show that if 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥 then 𝑥 = 𝑦 for 𝑥, 𝑦 ∊ 𝑅

Exercise: Use Trichotomy law to prove the following

Programme: Bachelor of Science (First Year)

Paper Code: MTC 101

Paper Title: DSC 1A: Calculus and Numerical Methods

Unit: Unit 1: Real Number System

Module Name: Axioms of Order Structure in R-Part 2

Name of the Presenter: Ms. Smita Kuncolienkar

Monotone Law of Addition

Monotone Law of Multiplication

Proof: (i) If 𝑥 < 𝑦 𝑎𝑛𝑑 𝑧 > 0 ⇒ 𝑥. 𝑧 < 𝑦. 𝑧

Programme: Bachelor of Science

Eg. Upper bound of (2,3) are 3, 3.1, 3.5, 4, 4.5,.etc

Not a lower bound: A real number α is not a lower bound of 𝐴 if ∃𝑥 ∈ 𝐴 ∋ 𝑥 < 𝛼

Programme: Bachelor of Science

Soln: A is bounded set.

Programme: Bachelor of Science

Programme: Bachelor of Science

WAYS TO GENERATE MORE BOUNDED SETS

since 𝐴 is bounded, ∃ 𝑚1, 𝑀1 ∈ ℝ ∋ 𝑚1 ≤ 𝑥 ≤ 𝑀1 , ∀ 𝑥 ∈ 𝐴} ----- 1

since 𝐵 is bounded, ∃ 𝑚2, 𝑀2 ∈ ℝ ∋ 𝑚2 ≤ 𝑥 ≤ 𝑀2 , ∀ 𝑥 ∈ 𝐵} ----- 2

Programme: Bachelor of Science

Question:- Find 𝑠𝑢𝑝 & 𝑖𝑛𝑓 of the following sets.

Programme: Bachelor of Science

This is a contradiction to the fact that 𝛽 is an upper bound of 𝐴

Programme: Bachelor of Science (First Year)

Proof : Let if possible N is bounded above.

∴ By Least Upper Bound property of R, ∃ α ∈ R such that α = sup(N).

⇒ α − 1 is not an upper bound of N as α − 1 < α

⇒ ∃ n0 ∈ N such that n0 > α − 1

This is a contradiction as α is an upper bound of N and n0 + 1 ∈ N as n0 ∈ N

Then ∃ n ∈ N such that n > 1.