2
Numbers
2.1 Prime Numbers
Theorem: There are an infinite number of primes
Proof: Suppose, for contradiction the number of primes if finite.
Hence, we can make a list p1 < p2 < p3 < … < pn
Any larger number is divisible by p1 … pn
Consider A = (p1 . p2 . … . pn) + 1
A > pn
But it is not divisible by p1 … pn
Contradiction
Original statement is false, QED
Fundamental Theorem Of Arithmetic
Any number N has a unique fractionalisation into primes i.e. N = p1a x p2b x …
Where a,b,c are natural numbers and form a unique combination.
Uniqueness if N = p1a’ x p2b’ x …
Then a = a’
b = b’
e.g. 2a x 3b = 2p x 3q
Cancelling factor of 2 gives a = p
Cancelling factor of 3 gives b = q
2.2 Proof By Induction
We want to prove P(n) n
n
1
S n   k  n(n  1)
2
k 1
We need to prove 2 things
1)
P(1) is true
2)
P(r)  P(r+1)
P(1)
S1 
1
.1 . 2  1
2
r 1
r
k
k
Which is true
S r 1   k  k  r  1
S r 1  S r  (r  1)
1
S r 1  .r (r  1)  (r  1)
2
1
S r 1  .(r  1)( r  2)
2
Which is of the correct form, QED
2.2 Rationals and Irrationals
m
Q   m, n  Z , n  0
n
2 ,  , e These are not rational numbers but appear in geometry
Also the appear in limits of sequences.
2.3 Rational And Irrationals Numbers
Proof that 2 is irrational
Lemma (A theorem used to prove another theorem) Even2 = Even, Odd2 = Odd,
conversely n 2 Even  nEven and n 2 Odd  nOdd
p
, p, q  N
q
2. Assume all common factors in q and p have cancelled, therefore p and q
cannot both be even.
3.
p2
2 2
q
4.
1. Suppose for contradiction
4.
2 is rational, i.e.
2
p 2  2q 2  Even
p is even and can be written as p =2r
4r 2  2q 2
q 2  2r 2  Even
q is even
5. p,q both even, Contradiction.
3 is proved to be irrational on class work.
It can also be proved using the fundamental theorem of arithmetic.
If p 2  2q 2
p  2 a x3 b
q  2 n x3 m
2 2 a x3 2b  2 2 n 1 x3 2 m
2a  2n  1
2b  2m
As 2a = 2n+1 is insoluble, there is a contradiction. This principle can be generalised
to all non square roots.
Proof e is irrational.

1
1 1
e    1  1    ...
2! 3!
n  0 n!
Suppose, for contraction
e
p
q
p  eq
Multiplying each side by (q-1)!
pq  1! q (q  1)!e
Gives
p (q  1)!  q!e
p(q – 1)! Is an integer.
 1

1 1
1


q!e! q!1  1    ...  q!

 ...
2! 3!


 q  1! q  2!

Integer
+ R
1
1

 ...
q  1 q  1q  2
1 1
1
R   2  3  ...
q q
q
1
1
q
R

1
1 q 1
1
q
R
Integer = Integer + R
2 Satisfies x2 -2 =0
It is an algebraic number, satisfying a polynomial equation.
anxn + an-1xn-1 + … + a1x + a0 = 0
For n<4 x has a solution in terms of a proof.
n>5 No root solution is possible (Abel proved this)
 , e are not algebraic but transcendental.
Irrationals are either algebraic or transcendental.
Contradiction
2.4 Decimal Representation
Every real no x has a decimal representation.
x = a0. a1a2a3…
This representation is not unique as 0.999… = 1
1
1
1

9  2  ...  9 10  1
1
 10 10

1
10
This is the only essential example of non uniqueness.
The decimal representation for a rational number repeats periodically.
Hence periodic indicates a rational number.
e.g.
x  0.121212...
100 x  12.121212...
100 x  12  x
99 x  12
12
4
x

99 33
Generally, the repeating decimal can be represented
x  0.b1b2 ...bn b1b2 ...bn b1 ...
10 n x  b1b2 ...bn  x
x
b1b2 ...bn
10 n 1
For the decimal notation we use base 10
e.g.
127.2  1x10 2  2 x101  7 x10 0  2 x10 1
Binary representation we use base 2:
i.e.
x  a1 a 2 a3 ...
ai  0,1
e.g.
x  101.011
x  1x 2 2  0 x 21  1x 2 0  0 x 2 1  1x 2 2  1x 2 3
1 1
3
x  4  0 1 0    5
4 8
8
e.g.
x  0.111...
1 1
1
 2  3  ...
2 2
2

1
x   n 1
n 1 2
x
You can have any base as long as

a
x   nn
an = 0,1 … (b-1)
n 1 b
0<x<1
Ternary Representation (Base 3)

a
x   nn
n 1 3
a n  0,1,2
e.g.
x  0.201
2 0
1
 2  3
1
3 3
3
2
1 19
x  0

3
27 27
x
2.5. Rational Approximations
x   and irrational
p
x
q
p
x    but how small is  .
q
X
p
/q
_
p+1
Expect  
1
2q
/q
2.6 Countability
Reals
Rationals
Integers
Proper Fractions
Irrationals
Algebraic
Transcendental
How numerate are rationals, algebraic and transcendental numbers?
Definition:
Set A is countable is its elements can be put in an exhaustive ordered
list a1 , a2 ,...an 
Equivalently if there exists a bijection f : N  A
e.g.
As the even numbers can be placed in an ordered list they are countable. 2,4,6,...
As the bijection f(n) = 2n exists, even numbers are countable.
e.g.
All integers Z 0,1,1,2,2,... are countable as:
f(2n) = n
N Z
f(2n-1) = -(n-1)
n>1
i.e.
Even N to count positive
Odd N to count negative
e.g.
All rationals Q
m

Q   m, n  N , n  0 
n

1
1
2
1
3
1
4
1
1
2
2
2
3
2
4
2
1
3
2
3
3
3
4
3
1
4
2
4
3
4
4
4
There exists a zigzag path through the entire array, so countable.
There exists a bijection N  NxN
Pairs (m,n)
n  2 a 3b 5 c...
f m, n NxN  N
f (m, n)  2 m 2 n
Defines a bijection N  NxN
2 m'3n'  2 m 3n
 m  m'
 n  m'
Actually f  S  N but any subset of N is also countable, i.e. bijection S  N
NxN  N
Bijection
N x N x N is also countable
N x N x N … x N (n times) is also countable, using theorem of arithmetic.
Similarly Z x Z x Z … x Z (n times) is also countable.
Algebraic Numbers:
x satisfies the equation:
a n x n  a n 1 x n 1  ...  a1 x  a 0  0
x  Z,
Where i
n N
Algebraic numbers are sets of all x satisfying all such equations for all n. This set is
countable.
Define the height of each equation by
hn  a n  an1  ... a0
hn  N
For fixed hn there is a fixed number of values an … a0 and each has a finite number of
solutions. Hence algebraic numbers are labelled by:
N
x
Height
Z
an – a0
x
N
N
Solution
Hence Countable
Alternate Proof
Assume for contradiction points in [0,1] can be listed
r1, r2, r3, r4, …
1
to r1
10
1
to r2
Assign length 10 2
...
1
to rn
10 n

Total Length
1
 10
n 1
n

1
 1 Contradiction
9
R x R < Points (x,y) in plane
[0,1] x = 0.a1a2a3…
y = 0.b1b2b3…
Points (x,y)in plane may be labelled by 0.a1b1a2b2…
RxR  R
What about irrationals?
R  Q  R \ Q
Real = Rations and (Reals that are not rational)
Theorem
If A and B are countable, then A  B is also countable.
Proof
A  a1 , a 2 ,...
B  b1 , b2 ,...
A  B  a1 , b1 , a 2 , b2 ,...Countable
2.7 Real Numbers
What distingushs R from Q?
S  x x 2  2 If x is real there is an upper limit, if x is rational there is not.


A  1,2,...10
1 is smallest element (min)
10 is largest element (max)
Closed Interval [0,1] = x x  ,0  x  1
Min 0, Max 1
Open Interval (0,1) = x x  ,0  x  1
No min, no max
Definition:
For A  R if there exists b   such that a < b for all a there is an upper bound for a.
Can have b  a
e.g. x = 1 is upper bound for [0,1] and (0,1)
If b  a then it must also be a max. Similarly for lower bound.
Definition:
If A is bounded above set in R then the least upper bound of A is
bounded such that b < c for any upper bound c.
Least Upper Bound [0,1] = 1 [0,1] max
Least Upper Bound (0,1) = 1 (0,1) no max.
Equivalently: If b us the least upper bound then there exists a  A such that
b   a  b
For any   0
If not a  b  
so b   is least upper bound. Contradiction.
e.g.
n


A  x x 
,n N
n 1


a 1
a  A
Least Upper Bound = 1  A because:
For   0 find n such that 1   
 n  11     n
n
1
n 1
n1     1    n
1    n
1
n
 1
Proves 1 is the least upper bound.
e.g.
S  x x  Q, x 2  2
Theorem:
S has no rational least upper band (proof later)


The Completeness Axiom – Defining Property Of Real Numbers
If A   is bounded above then A has a least upper bound which is real (this would
not be satisfied by rationals).
Now we to show that numbers like

2 exist in the real line.

Proof Consider S  x x  Q, x 2  2
Completeness Axiom  S has a least upper bound l and we can show l2 = 2
We show l2 < 2 and l2 > 2 are impossible hence l2=2
i)
Let l2 < 2
2
We define  so 0 <  < 1 such that l     2
If we assume l 2  4 so l  2
So l is not the upper bound
l   2  l 2  2l   2
l   2  l 2  4   2
l   2  l 2  5a
l   2  2, if ,  2  l
2
5
ii)
Let l2 > 2
2
We define  so 0 <  < 1 such that l     2
So l is not the least upper bound
l   2  l 2  2l   2
l   2  l 2  2l
l   2  l 2  4l
l   
2
l2  2
 2, if ,  
4
Leaving l2 = 2 as on point solution in the real line
This proof also shows S  x x  Q, x 2  2 has no least upper bound.


Nested Interval Property Of Reals
a1
I1
a2
I2
b1
b2
etc
I n  an , bn   x an  x  bn , x  
This nested representation is show the decimal line up works.
I 1  I 2  I 3 ...
If I 1  I 2  I 3 ... there exists    , such that   I n n
If
I n  bn  a n  0
as, n  ,
 , unique
This is the defining property of the reals.
Another proof if the unaccountability of the reals.
Assume for contradiction x  [0,1] maybe arranged in a list x1, x2, x3, …
0,1  I1  I 2  I 3 ...
x1  I 1
Where x 2  I 2
x3  I 3
By the nested interval property there exists    such that   I n n
So cannot be in the original list.
Contradiction.
Archimedean Property Of Reals
Proved using the completeness axiom
a, b  
ab
a, b  0
na  b
Where, n  N
2.8
Cantor Set
 0  [0,1]
 1  2 3
1  0,    , 
 3  3 3
 1   2 3 6 7  8 9
 3  0,    ,    ,    , 
 9  9 9  9 9  9 9
The Cantor Set is formed by continuously deleting the middle third to get
 0  1   2   3 ...
1 2 4
8
 

3 9 27 81
2
3

1  2  2   2 
1         ... 
The length of each section removed is equal to

3  3  3   3 





1 1 
1
3 2
 1   
 3
So the set as no length, yet it contains:
1 2 1 2 7 8
0,1, , , , , , ...
3 3 9 9 9 9

   k
k 1
As the elements in the Cantor Set can be expressed as i.e
x  k , k
The Cantor Set can be thought of in ternary form by removing all elements that have a
1 in there representation.
So 0.a1a2a3… where ai = 0,2 is a member
Hence 0.02020202… is a member
Which is ¼
As the Cantor Set has a bijection with the binaries which has a bijection with the reals
it is uncountable.