[PPT] - Finding the Natural Numbers in the Integers Bernd Schr oder logo1 PowerPoint Presentation

SLIDE 1

logo1 Isomorphism N ⊆ Z

Finding the Natural Numbers in the Integers

Bernd Schr¨

der

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 2

logo1 Isomorphism N ⊆ Z

Definition.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 3

logo1 Isomorphism N ⊆ Z

Definition. Let A and B be sets, let n ∈ N, let ◦1,...,◦n be

binary operations on A and let ∗1,...,∗n be binary operations

n B.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 4

logo1 Isomorphism N ⊆ Z

Definition. Let A and B be sets, let n ∈ N, let ◦1,...,◦n be

binary operations on A and let ∗1,...,∗n be binary operations

n B. Then (A,◦1,...,◦n) is called isomorphic to (B,∗1,...,∗n)

iff

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 5

logo1 Isomorphism N ⊆ Z

Definition. Let A and B be sets, let n ∈ N, let ◦1,...,◦n be

binary operations on A and let ∗1,...,∗n be binary operations

n B. Then (A,◦1,...,◦n) is called isomorphic to (B,∗1,...,∗n)

iff there is a bijective function f : A → B so that for all k ∈ {1,...,n} and all x,y ∈ A we have that f(x◦k y) = f(x)∗k f(y).

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 6

logo1 Isomorphism N ⊆ Z

Definition. Let A and B be sets, let n ∈ N, let ◦1,...,◦n be

binary operations on A and let ∗1,...,∗n be binary operations

n B. Then (A,◦1,...,◦n) is called isomorphic to (B,∗1,...,∗n)

iff there is a bijective function f : A → B so that for all k ∈ {1,...,n} and all x,y ∈ A we have that f(x◦k y) = f(x)∗k f(y). The function f is called an isomorphism.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 7

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 8

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ (A,◦)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 9

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 10

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s x (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 11

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s x y (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 12

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s x y (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 13

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s x y ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 14

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s x y s x◦y ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 15

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s x y s x◦y q f ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 16

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s x y s s x◦y f(x◦y) q f ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 17

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s x y s s x◦y f(x◦y) q q f f ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 18

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s s x f(x) y s s x◦y f(x◦y) q q f f ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 19

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s s s x f(x) y f(y) s s x◦y f(x◦y) q q f f ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 20

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s s s x f(x) y f(y) s s x◦y f(x◦y) q q f f ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 21

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s s s x f(x) y f(y) s s x◦y f(x◦y) = f(x)∗f(y) q q f f ✻ ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 22

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s s s x f(x) y f(y) s s x◦y f(x◦y) = f(x)∗f(y) q q ✐ f f f −1 ✻ ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 23

logo1 Isomorphism N ⊆ Z

Visualizing Isomorphism

✬ ✫ ✩ ✪ ✬ ✫ ✩ ✪ s s s s x f(x) y f(y) s s x◦y f(x◦y) = f(x)∗f(y) q q ✐ ✐ f f f −1 f −1 ✻ ✻ (A,◦) (B,∗)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 24

logo1 Isomorphism N ⊆ Z

Theorem.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 25

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 26

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 27

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 28

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers. Proof.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 29

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Proof. Define f : N → Z by f(n) :=
(n+1,1)
.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 30

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Proof. Define f : N → Z by f(n) :=
(n+1,1)
. Then f is

clearly surjective.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 31

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Proof. Define f : N → Z by f(n) :=
(n+1,1)
. Then f is

clearly surjective. Moreover, f is injective, because

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 32

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Proof. Define f : N → Z by f(n) :=
(n+1,1)
. Then f is

clearly surjective. Moreover, f is injective, because f(n) = f(m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 33

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Proof. Define f : N → Z by f(n) :=
(n+1,1)
. Then f is

clearly surjective. Moreover, f is injective, because f(n) = f(m) ⇒

(n+1,1)
=
(m+1,1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 34

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Proof. Define f : N → Z by f(n) :=
(n+1,1)
. Then f is

clearly surjective. Moreover, f is injective, because f(n) = f(m) ⇒

(n+1,1)
=
(m+1,1)
⇒

(n+1)+1 = (m+1)+1

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 35

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Proof. Define f : N → Z by f(n) :=
(n+1,1)
. Then f is

clearly surjective. Moreover, f is injective, because f(n) = f(m) ⇒

(n+1,1)
=
(m+1,1)
⇒

(n+1)+1 = (m+1)+1 ⇒ n = m

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 36

logo1 Isomorphism N ⊆ Z

Theorem. The set of natural numbers N, equipped with

addition and multiplication, is isomorphic to the subset

(n+1,1)
: n ∈ N
f the integers Z, equipped with addition

and multiplication. The set

(n+1,1)
: n ∈ N
⊆ Z will be

called N, too, and we will use the customary digit and place value notation for these numbers.

Proof. Define f : N → Z by f(n) :=
(n+1,1)
. Then f is

clearly surjective. Moreover, f is injective, because f(n) = f(m) ⇒

(n+1,1)
=
(m+1,1)
⇒

(n+1)+1 = (m+1)+1 ⇒ n = m For the compatibility with the algebraic operations, let m,n ∈ N.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 37

logo1 Isomorphism N ⊆ Z

Proof (cont.).

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 38

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 39

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 40

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 41

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 42

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 43

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
=

f(n)+f(m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 44

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
=

f(n)+f(m) f(nm)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 45

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
=

f(n)+f(m) f(nm) =

(nm+1,1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 46

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
=

f(n)+f(m) f(nm) =

(nm+1,1)
=
nm+n+m+1+1,n+m+1+1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 47

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
=

f(n)+f(m) f(nm) =

(nm+1,1)
=
nm+n+m+1+1,n+m+1+1)
=
(n+1)(m+1)+1,(n+1)+(m+1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 48

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
=

f(n)+f(m) f(nm) =

(nm+1,1)
=
nm+n+m+1+1,n+m+1+1)
=
(n+1)(m+1)+1,(n+1)+(m+1)
=
(n+1,1)
·
(m+1,1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 49

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
=

f(n)+f(m) f(nm) =

(nm+1,1)
=
nm+n+m+1+1,n+m+1+1)
=
(n+1)(m+1)+1,(n+1)+(m+1)
=
(n+1,1)
·
(m+1,1)
=

f(n)f(m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 50

logo1 Isomorphism N ⊆ Z

Proof (cont.). f(n+m) =

(n+m+1,1)
=
(n+m+1+1,1+1)
=
(n+1)+(m+1),1+1)
=
(n+1,1)
+
(m+1,1)
=

f(n)+f(m) f(nm) =

(nm+1,1)
=
nm+n+m+1+1,n+m+1+1)
=
(n+1)(m+1)+1,(n+1)+(m+1)
=
(n+1,1)
·
(m+1,1)
=

f(n)f(m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 51

logo1 Isomorphism N ⊆ Z

Theorem.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 52

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 53

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 54

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds:

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 55

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 56

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 57

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 58

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1).

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 59

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 60

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 61

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 62

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 63

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y. Hence x+y

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 64

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y. Hence x+y = f(n)+f(m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 65

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y. Hence x+y = f(n)+f(m) = f(n+m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 66

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y. Hence x+y = f(n)+f(m) = f(n+m) ∈ N

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 67

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y. Hence x+y = f(n)+f(m) = f(n+m) ∈ N and x·y

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 68

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y. Hence x+y = f(n)+f(m) = f(n+m) ∈ N and x·y = f(n)·f(m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 69

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y. Hence x+y = f(n)+f(m) = f(n+m) ∈ N and x·y = f(n)·f(m) = f(n·m)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 70

logo1 Isomorphism N ⊆ Z

Theorem. The subset N ⊆ Z has the following properties.
1. For all x,y ∈ N, we have x+y ∈ N and xy ∈ N,
2. For all x ∈ Z, exactly one of the following three properties

holds: Either x ∈ N or −x ∈ N or x = 0. Proof (part 1). Let Norig be the “original” natural numbers and let f : Norig → N be the isomorphism from the previous proof. Let x,y ∈ N. There are n,m ∈ Norig with f(n) = x and f(m) = y. Hence x+y = f(n)+f(m) = f(n+m) ∈ N and x·y = f(n)·f(m) = f(n·m) ∈ N.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 71

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition:

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 72

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 73

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 74

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 75

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 76

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 77

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 78

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 79

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 80

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 81

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 82

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 83

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 84

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 85

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 86

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity:

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 87

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 88

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N. Hence −0 is not an element of N either.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 89

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N. Hence −0 is not an element of N either. Finally, suppose for a contradiction that there is an integer x ∈ N so that −x ∈ N.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 90

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N. Hence −0 is not an element of N either. Finally, suppose for a contradiction that there is an integer x ∈ N so that −x ∈ N. Then x =

(a+1,1)
for some a ∈ Norig

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 91

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N. Hence −0 is not an element of N either. Finally, suppose for a contradiction that there is an integer x ∈ N so that −x ∈ N. Then x =

(a+1,1)
for some a ∈ Norig and x =
(1,b+1)
for some

b ∈ Norig.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 92

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N. Hence −0 is not an element of N either. Finally, suppose for a contradiction that there is an integer x ∈ N so that −x ∈ N. Then x =

(a+1,1)
for some a ∈ Norig and x =
(1,b+1)
for some

b ∈ Norig. But then

(a+1,1)
=
(1,b+1)
Bernd Schr¨
der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 93

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N. Hence −0 is not an element of N either. Finally, suppose for a contradiction that there is an integer x ∈ N so that −x ∈ N. Then x =

(a+1,1)
for some a ∈ Norig and x =
(1,b+1)
for some

b ∈ Norig. But then

(a+1,1)
=
(1,b+1)
implies

a+b+1+1 = 1+1

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 94

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N. Hence −0 is not an element of N either. Finally, suppose for a contradiction that there is an integer x ∈ N so that −x ∈ N. Then x =

(a+1,1)
for some a ∈ Norig and x =
(1,b+1)
for some

b ∈ Norig. But then

(a+1,1)
=
(1,b+1)
implies

a+b+1+1 = 1+1, which implies a+b+1 = 1. But 1 is not the successor of any natural number, contradiction.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers

SLIDE 95

logo1 Isomorphism N ⊆ Z

Proof (part 2). Every element satisfies one condition: Let x =

(a,b)
∈ Z. We first show that x is either zero, or x ∈ N or

−x ∈ N. If a = b, then x =

(a,a)
=
(1,1)
= 0. If a > b, in

case b = 1, we have x ∈ N, so we can assume that b > 1. Now N ∋

(a−b+1,1)
=
(a−b+1,1)
+0

=

(a−b+1,1)
+
(b−1,b−1)
=
(a,b)
= x.

Finally, if a < b, then −x =

(b,a)
∈ N.

Exclusivity: First note that 0 is not an element of N. Hence −0 is not an element of N either. Finally, suppose for a contradiction that there is an integer x ∈ N so that −x ∈ N. Then x =

(a+1,1)
for some a ∈ Norig and x =
(1,b+1)
for some

b ∈ Norig. But then

(a+1,1)
=
(1,b+1)
implies

a+b+1+1 = 1+1, which implies a+b+1 = 1. But 1 is not the successor of any natural number, contradiction.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Finding the Natural Numbers in the Integers