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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Johannes Cuno
Welcome toMAT1341
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
1 Ancient Greece
2 Motivation I: Where are we starting?
3 Motivation II: Where are we going to?
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
PlatoAristotleA
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
PlatoAristotleA
AAU ���
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
PlatoAristotleA
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
PlatoAristotleA
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
PlatoAristotleA
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
PlatoAristotleA
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
PlatoAristotleA
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
1 Ancient Greece
2 Motivation I: Where are we starting?
3 Motivation II: Where are we going to?
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
N1
23
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZ
12
3
0−1 −2
−3
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQ
12
3
0−1 −2
−3
1/2 −17/5
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQ
12
3
0−1 −2
−3
1/2 −17/5
We would like to solve the equation
x2 = 2
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQ
12
3
0−1 −2
−3
1/2 −17/5
We would like to solve the equation
x2 = 2
Hippasusof Metapont
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof.
Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q.
Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N.
Therefore,(ab
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2
, ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2
, ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too.
This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c .
Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b
, ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2
, ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even.
In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified.
Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Claim. There is no element x ∈ Q such that x2 = 2.
Proof. Suppose there was such an element x ∈ Q. Being a rationalnumber we may express it as simplified fraction a/b with a ∈ Z andb ∈ N. Therefore,(a
b
)2= 2 , ⇔ a2
b2= 2 , ⇔ a2 = 2b2 .
The right hand side 2b2 of the latter is even, so the left hand side a2
must be even, too. This implies that a itself must be even, whencethere is a c ∈ Z such that a = 2c . Therefore,
(2c)2 = 2b , ⇔ 4c2 = 2b2 , ⇔ 2c2 = b2 .
This implies that b must be even. In fact, both a and b are even,whence the fraction a/b was not simplified. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQ
12
3
0−1 −2
−3
1/2 −17/5
We would like to solve the equation
x2 = 2
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQR
12
3
0−1 −2
−3
1/2 −17/5
e √2 π
e = 2.7182818284...√2 = 1.4142135623... π = 3.1415926535...
We would like to solve the equation
x2 = 2
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQR
12
3
0−1 −2
−3
1/2 −17/5
e √2 π
e = 2.7182818284...√2 = 1.4142135623... π = 3.1415926535...
We would like to solve the equation
x2 = 2 X
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
-
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 2/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@I
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
@@R
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 −2/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
@@R
@@R
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
@@R
@@@@@R
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
@@R
@@@@@R
-
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
@@R
@@@@@R
-@@I
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
@@R
@@@@@R
-@@I
@@I
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 2/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
@@R
@@@@@R
-@@I
@@I
@@I
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
0 1 −1 2 −2 3 −3 4 −4 · · ·
1/2 −1/2 3/2 −3/2 4/2 −4/2 · · ·
1/3 −1/3 2/3 −2/3 3/3 −3/3 4/3 −4/3 · · ·
1/4 −1/4 −2/4 3/4 −3/4 4/4 −4/4 · · ·
1/5 −1/5 2/5 −2/5 3/5 −3/5 4/5 −4/5 · · ·
......
......
......
......
- -@@I
6@@R
@@R-@@
@@@I
@@I
6@@R
@@R
@@@@@R
-@@I
@@I
@@
@@@I
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof.
Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.
Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers.
For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·
a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·
a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·
......
......
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{
1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 1
2 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list.
Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Countability
Claim. The real numbers R are not countable.
Proof. Suppose there was a possibility to count the real numbers.Then create a list comprising all real numbers. For each entry,denote the integer part and every single fractional digit:
a1 . b1,1 b1,2 b1,3 · · ·a2 . b2,1 b2,2 b2,3 · · ·a3 . b3,1 b3,2 b3,3 · · ·...
......
...
For each number i ∈ N, we may now define
ci :=
{1 if bi ,i 6= 12 if bi ,i = 1
The real number 0. c1 c2 c3 · · · is not in our list. Contradiction. �
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQR
12
3
0−1 −2
−3
1/2 −17/5
e √2 π
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQR
12
3
0−1 −2
−3
1/2 −17/5
e √2 π
We would like to solve the equation
x2 = −1
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQRC
12
3
0−1 −2
−3
1/2 −17/5
e √2 π
We would like to solve the equation
x2 = −1
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
NZQRC
12
3
0−1 −2
−3
1/2 −17/5
e √2 π
We would like to solve the equation
x2 = −1 X
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Gauss
NZQRC
12
3
0−1 −2
−3
1/2 −17/5
e √2 π
We would like to solve the equation
x2 = −1 X
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Gauss
NZQRC
12
3
0−1 −2
−3
1/2 −17/5
e √2 π
There is a solution to any equation of the form:
anxn + . . . + a2x
2 + a1x = b(with an 6= 0)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
1 Ancient Greece
2 Motivation I: Where are we starting?
3 Motivation II: Where are we going to?
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
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Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
v
P
vL
ww1
w2
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(x) cos(x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(x) cos(x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(x) cos(x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(2x) cos(x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(3x) cos(x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(4x) cos(x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(4x) cos(2x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(4x) cos(3x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
sin(4x) cos(4x)
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f1
g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f2
g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f3
g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f4
g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f10
g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f20
g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f30
g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
g1
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
g2
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
g3
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
g4
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
g10
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
g20
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
g30
made 2π-periodic made 2π-periodic
Ancient Greece Motivation I: Where are we starting? Motivation II: Where are we going to?
Excursion about Fourier Series
f , g : R→ R . . . . . . . . . . . square integrable 2π-periodic functions
f (x) = x
f40
g(x) = x3
g40
made 2π-periodic made 2π-periodic
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