Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

THE REAL NUMBER SYSTEM
Subham Karmakar
June 9, 2021

Properties of R, the set of real numbers
Let us consider the binary operation 0 +0 (usual addition) and ’ · ’ (usual multiplication) on R
...
’+’ and ’·’ are commutative on R
...
e
∀x, y ∈ R, x + y = y + x and x · y = y · x
2
...
i
...
∃ 0, 1 ∈ R such that 0 + x = x and 1 · x = x ∀x ∈ R
...

4
...

c is called the additive inverse of b
...

For every t ∈ R − {0}, ∃ u ∈ R such that u · t = 1
...
’·’ is distributive over ’+’ on R i
...
To show any mathematical
system is a field, we must verify whether the above axioms holds in the system, if any one
of the above axioms doesn’t holds, we conclude it is not a field
...
For any a, b ∈ R, exactly one of the following holds:(a) a < b
(b) a = b
(c) a > b
This is known as Law of Trichotomy
...
a, b, c ∈ R and a < b ⇒ a + c < b + c
8
...
x, y, z ∈ R with x < y and y < z ⇒ x < z
...

Definition : Let S ⊆ R
...

A set which is not bounded above is called unbounded above
...
is bounded above, 1 being an upper bound
...



(c) Z, Q, R are unbounded above
...

Definition : Let S ⊂ R
...
Since ∀n ∈ N,
≤ 1
...
For
n
n
arbitarily choosen ε > 0, ∃ 1 ∈ S such that 1 > 1 − ε
...



1
(b) Let T = 3 − : n ∈ N
...

n
n
n
Let ε > 0 be arbitrarily choosen,
 
1
1
1
1
1
3− >3−ε
if − > −ε ⇒ < ε ⇒ n > ⇒ n ≥
+ 1 = p (say)
n
n
n
ε
ε
1
1
1
So 3 − , 3 −
,3−
,
...

p
p+1
p+2
Thus sup T = 3
2

Note : LUB of a bounded above set may or may not be a member of the set
...

10
...

Example: Let A = {x ∈ Q : x > 0 ∧ x2 < 2}
√
∴ x ∈ A ⇒ x > 0 ∧ x2 < 2 ⇒ 0 < x < 2
...

By Completeness Axiom, sup A ∈ R
...
We show that u ∈
/ Q
...
Then u > 0 and u ≤ 2 and u ∈ Q ⇒ 0 < u < 2
...
Then v > 0 and v ∈ Q
Let v =
3 + 2u
4 + 3u
4 − 2u2
2(2 − u2 )
v−u=
−u=
=
>0⇒v>u
3 + 2u
3 + 2u
3 + 2u
u2 − 2
16 + 24u + 9u2
−
2
=
< 0 ⇒ v2 < 2
Also, v 2 − 2 =
9 + 12u + 4u2
(3 + 2u)2
Thus v ∈ A which contradicts the fact that u = sup A
...

Therefore R is a complete ordered field and Q is an incomplete ordered field
...
Archimedean Property of Real Numbers
There are two ways to state the Archimedean property
(a) The set N of natural numbers is unbounded above
...

If possible, let N be bounded above
...
Let sup N = µ
Then
i
...
corresponding to ε = 1, ∃ p ∈ N suc that p > µ − 1 ⇒ p + 1 > µ and p + 1 ∈ N
This contradicts the fact that µ = sup N
...

(b) (equivalent statement): If x, y ∈ R and x > 0, then ∃n ∈ N such that nx > y
...

Let us assume that y > 0
...

Then the set S = {nx : n ∈ N} is a non-empty bounded above subset of R, y being an
upper bound
...

Let sup S = u
...
∀ n ∈ N, nx ≤ u
ii
...

Also (m + 1)x ∈ S, which contradicts the fact that u = sup S
...

Remark: From the above property it follows that for any ε > 0, ∃ p ∈ N such that

1
<ε
p

12
...

Proof: Let x, y ∈ R and x < y, then y − x > 0
...


√
√
√
√
Also x, y√
⇒ x + 2 < y + 2 ⇒ ∃r ∈ √
Q such that x + 2 < r < y + 2
⇒ ∃ r − 2 ∈ R\Q such that x < r − 2 < y
...

Corollary: Between any two distinct real number, there lies infinitly many rational and infiniely many urrational numbers
...
Linear Continuum
There is one-to-one correspondence between the set R and the set of points on a straight
line

---