Definition: Using the undefined notions "1" and "successor" (denoted by
′
{\displaystyle '}
N
∗
{\displaystyle \mathbb {N} ^{*}}
Axiom 1.
∃
1
(
1
∈
N
∗
)
.
{\displaystyle \exists 1(1\in \mathbb {N} ^{*}).}
Axiom 2.
∀
a
(
a
∈
N
∗
⟹
∃
b
(
b
=
a
′
)
)
.
{\displaystyle \forall a(a\in \mathbb {N} ^{*}\implies \exists b(b=a')).}
Axiom 3.
¬
∃
a
(
a
′
=
1
)
.
{\displaystyle \lnot \exists a(a'=1).}
Axiom 4.
∀
a
∈
N
∗
(
∀
b
∈
N
∗
(
a
′
=
b
′
⟹
a
=
b
)
)
.
{\displaystyle \forall a\in \mathbb {N} ^{*}(\forall b\in \mathbb {N} ^{*}(a'=b'\implies a=b)).}
Axiom 5.
∀
A
⊆
N
∗
(
(
1
∈
A
)
∧
∀
a
∈
A
(
a
′
∈
A
)
⟹
A
=
N
∗
)
.
{\displaystyle \forall A\subseteq \mathbb {N} ^{*}((1\in A)\land \forall a\in A(a'\in A)\implies A=\mathbb {N} ^{*}).}
We can prove theorems for natural numbers using mathematical induction as a consequence of the fifth Peano Postulate.
Definition: We recursively define addition for the natural numbers as a composition using two more axioms; the other properties of addition can subsequently be derived from these axioms. We denote addition with the infix operator +.
Axiom 6.
∀
a
∈
N
∗
(
a
+
1
=
a
′
)
.
{\displaystyle \forall a\in \mathbb {N} ^{*}(a+1=a').}
Axiom 7.
∀
a
∈
N
∗
(
∀
b
∈
N
∗
(
a
+
b
′
=
(
a
+
b
)
′
)
)
.
{\displaystyle \forall a\in \mathbb {N} ^{*}(\forall b\in \mathbb {N} ^{*}(a+b'=(a+b)')).}
Axiom 6 above relies on the first Peano postulate (for the existence of 1) as well as the second (for the existence of a successor for every number).
Henceforth, we will assume that proven theorems hold for all
a
,
b
,
c
,
…
,
n
{\displaystyle a,b,c,\ldots ,n}
N
∗
{\displaystyle \mathbb {N} ^{*}}
Definition: We similarly define multiplication for the natural numbers recursively, again using two axioms:
Axiom 8.
a
(
1
)
=
a
.
{\displaystyle a(1)=a.}
Axiom 9.
a
b
′
=
a
b
+
a
.
{\displaystyle ab'=ab+a.}
Properties of addition
edit
We start by proving that addition is associative.
Theorem 1: Associativity of addition:
(
a
+
b
)
+
c
=
a
+
(
b
+
c
)
.
{\displaystyle (a+b)+c=a+(b+c).}
Proof: Base case: By axioms 6 and 7,
a
+
(
b
+
1
)
=
(
a
+
b
)
′
{\displaystyle a+(b+1)=(a+b)'}
By axiom 6,
a
+
(
b
+
1
)
=
(
a
+
b
)
+
1
{\displaystyle a+(b+1)=(a+b)+1}
Inductive hypothesis: Suppose that, for
k
′
>
1
{\displaystyle k'>1}
(
a
+
b
)
+
k
=
a
+
(
b
+
k
)
{\displaystyle (a+b)+k=a+(b+k)}
Inductive step: By axiom 7,
(
a
+
b
)
+
k
′
=
[
(
a
+
b
)
+
k
]
′
{\displaystyle (a+b)+k'=\left[(a+b)+k\right]'}
By the inductive hypothesis,
(
a
+
b
)
+
k
′
=
[
a
+
(
b
+
k
)
]
′
{\displaystyle (a+b)+k'=\left[a+(b+k)\right]'}
By axiom 7,
(
a
+
b
)
+
k
′
=
a
+
(
b
+
k
)
′
{\displaystyle (a+b)+k'=a+(b+k)'}
By axiom 7,
(
a
+
b
)
+
k
′
=
a
+
(
b
+
k
′
)
{\displaystyle (a+b)+k'=a+(b+k')}
By induction,
(
a
+
b
)
+
c
=
a
+
(
b
+
c
)
{\displaystyle (a+b)+c=a+(b+c)}
Lemma 1:
a
+
1
=
1
+
a
.
{\displaystyle a+1=1+a.}
Proof: Base case: 1+1=1+1.
Inductive hypothesis: Suppose that, for
k
′
>
1
{\displaystyle k'>1}
k
+
1
=
1
+
k
{\displaystyle k+1=1+k}
Inductive step: By axiom 6,
k
′
+
1
=
(
k
+
1
)
+
1
{\displaystyle k'+1=(k+1)+1}
By the inductive hypothesis,
k
′
+
1
=
(
1
+
k
)
+
1
{\displaystyle k'+1=(1+k)+1}
By theorem 1,
k
′
+
1
=
1
+
(
k
+
1
)
{\displaystyle k'+1=1+(k+1)}
By axiom 6,
k
′
+
1
=
1
+
k
′
{\displaystyle k'+1=1+k'}
By induction,
a
+
1
=
1
+
a
{\displaystyle a+1=1+a}
Theorem 2: Commutativity of addition:
a
+
b
=
b
+
a
.
{\displaystyle a+b=b+a.}
Proof: Base case: By lemma 1,
a
+
1
=
1
+
a
{\displaystyle a+1=1+a}
Inductive hypothesis: Suppose that, for
k
′
>
1
{\displaystyle k'>1}
a
+
k
=
k
+
a
{\displaystyle a+k=k+a}
By axiom 6,
a
+
k
′
=
a
+
(
k
+
1
)
{\displaystyle a+k'=a+(k+1)}
By theorem 1,
a
+
k
′
=
(
a
+
k
)
+
1
{\displaystyle a+k'=(a+k)+1}
By the inductive hypothesis,
a
+
k
′
=
(
k
+
a
)
+
1
{\displaystyle a+k'=(k+a)+1}
By theorem 1,
a
+
k
′
=
k
+
(
a
+
1
)
{\displaystyle a+k'=k+(a+1)}
By lemma 1,
a
+
k
′
=
k
+
(
1
+
a
)
{\displaystyle a+k'=k+(1+a)}
By theorem 1,
a
+
k
′
=
(
k
+
1
)
+
a
{\displaystyle a+k'=(k+1)+a}
By axiom 6,
a
+
k
′
=
k
′
+
a
{\displaystyle a+k'=k'+a}
By induction,
a
+
b
=
b
+
a
{\displaystyle a+b=b+a}
Theorem 3:
a
+
b
=
a
+
c
⟹
b
=
c
{\displaystyle a+b=a+c\implies b=c}
Proof: Base case: Suppose
1
+
b
=
1
+
c
{\displaystyle 1+b=1+c}
By theorem 2,
b
+
1
=
c
+
1
{\displaystyle b+1=c+1}
By axiom 6,
b
′
=
c
′
{\displaystyle b'=c'}
By axiom 4,
b
=
c
{\displaystyle b=c}
Inductive hypothesis: Suppose that, for
k
′
>
1
{\displaystyle k'>1}
k
+
b
=
k
+
c
⟹
b
=
c
{\displaystyle k+b=k+c\implies b=c}
Inductive step: Suppose
k
′
+
b
=
k
′
+
c
{\displaystyle k'+b=k'+c}
By axiom 6,
(
k
+
1
)
+
b
=
(
k
+
1
)
+
c
{\displaystyle (k+1)+b=(k+1)+c}
By theorem 2,
(
1
+
k
)
+
b
=
(
1
+
k
)
+
c
{\displaystyle (1+k)+b=(1+k)+c}
By theorem 1,
1
+
(
k
+
b
)
=
1
+
(
k
+
c
)
{\displaystyle 1+(k+b)=1+(k+c)}
By the base case,
k
+
b
=
k
+
c
{\displaystyle k+b=k+c}
k
′
+
b
=
k
′
+
c
⟹
k
+
b
=
k
+
c
{\displaystyle k'+b=k'+c\implies k+b=k+c}
By the inductive hypothesis,
k
′
+
b
=
k
′
+
c
⟹
b
=
c
{\displaystyle k'+b=k'+c\implies b=c}
By induction,
a
+
b
=
a
+
c
⟹
b
=
c
{\displaystyle a+b=a+c\implies b=c}
Properties of multiplication
edit
Theorem 4: Left-distributivity of multiplication over addition:
a
(
b
+
c
)
=
a
b
+
a
c
{\displaystyle a(b+c)=ab+ac}
Proof: Base case: By axioms 6 and 9,
a
(
b
+
1
)
=
a
b
+
a
{\displaystyle a(b+1)=ab+a}
By axiom 8,
a
(
b
+
1
)
=
a
b
+
a
(
1
)
{\displaystyle a(b+1)=ab+a(1)}
Inductive hypothesis: Suppose that, for
k
′
>
1
{\displaystyle k'>1}
a
(
b
+
k
)
=
a
b
+
a
k
{\displaystyle a(b+k)=ab+ak}
Inductive step: By axiom 7,
a
(
b
+
k
′
)
=
a
(
b
+
k
)
′
{\displaystyle a(b+k')=a(b+k)'}
By axiom 9,
a
(
b
+
k
′
)
=
a
(
b
+
k
)
+
a
{\displaystyle a(b+k')=a(b+k)+a}
By the inductive hypothesis,
a
(
b
+
k
′
)
=
(
a
b
+
a
k
)
+
a
{\displaystyle a(b+k')=(ab+ak)+a}
By theorem 1,
a
(
b
+
k
′
)
=
a
b
+
(
a
k
+
a
)
{\displaystyle a(b+k')=ab+(ak+a)}
By axiom 9,
a
(
b
+
k
′
)
=
a
b
+
a
k
′
{\displaystyle a(b+k')=ab+ak'}
By induction,
a
(
b
+
c
)
=
a
b
+
a
c
{\displaystyle a(b+c)=ab+ac}
Theorem 5:
1
a
=
a
{\displaystyle 1a=a}
Proof: Base case: By axiom 8, 1(1)=1.
Inductive hypothesis: Suppose that, for
k
′
>
1
{\displaystyle k'>1}
1
k
=
k
{\displaystyle 1k=k}
Inductive step: By axiom 6,
1
k
′
=
1
(
k
+
1
)
{\displaystyle 1k'=1(k+1)}
By theorem 4,
1
k
′
=
1
k
+
1
(
1
)
{\displaystyle 1k'=1k+1(1)}
By the base case,
1
k
′
=
1
k
+
1
{\displaystyle 1k'=1k+1}
By the inductive hypothesis,
1
k
′
=
k
+
1
{\displaystyle 1k'=k+1}
By axiom 6,
1
k
′
=
k
′
{\displaystyle 1k'=k'}
By induction,
1
a
=
a
{\displaystyle 1a=a}
Theorem 6:
a
′
b
=
a
b
+
b
{\displaystyle a'b=ab+b}
Proof: Base case: By axiom 8,
a
′
(
1
)
=
a
′
{\displaystyle a'(1)=a'}
By axiom 6,
a
′
(
1
)
=
a
+
1
{\displaystyle a'(1)=a+1}
By axiom 8,
a
′
(
1
)
=
a
(
1
)
+
1
{\displaystyle a'(1)=a(1)+1}
Inductive hypothesis: Suppose that,
k
′
>
1
{\displaystyle k'>1}
a
′
k
=
a
k
+
k
{\displaystyle a'k=ak+k}
Inductive step: By axiom 9,
a
′
k
′
=
a
′
k
+
a
′
{\displaystyle a'k'=a'k+a'}
By the inductive hypothesis,
a
′
k
′
=
a
k
+
k
+
a
′
{\displaystyle a'k'=ak+k+a'}
By axiom 6,
a
′
k
′
=
a
k
+
k
+
(
a
+
1
)
{\displaystyle a'k'=ak+k+(a+1)}
By theorem 1,
a
′
k
′
=
a
k
+
(
k
+
a
)
+
1
{\displaystyle a'k'=ak+(k+a)+1}
By theorem 2,
a
′
k
′
=
a
k
+
(
a
+
k
)
+
1
{\displaystyle a'k'=ak+(a+k)+1}
By theorem 1,
a
′
k
′
=
a
k
+
a
+
k
+
1
{\displaystyle a'k'=ak+a+k+1}
By axiom 9,
a
′
k
′
=
a
k
′
+
k
+
1
{\displaystyle a'k'=ak'+k+1}
By theorem 1,
a
′
k
′
=
a
k
′
+
(
k
+
1
)
{\displaystyle a'k'=ak'+(k+1)}
By axiom 6,
a
′
k
′
=
a
k
′
+
k
′
{\displaystyle a'k'=ak'+k'}
By induction,
a
′
b
=
a
b
+
b
{\displaystyle a'b=ab+b}
Theorem 7: Associativity of multiplication:
(
a
b
)
c
=
a
(
b
c
)
.
{\displaystyle (ab)c=a(bc).}
Proof: Base case: By axiom 8,
a
b
(
1
)
=
a
b
=
a
[
b
(
1
)
]
{\displaystyle ab(1)=ab=a\left[b(1)\right]}
Inductive hypothesis: Suppose that, for
k
′
>
1
{\displaystyle k'>1}
(
a
b
)
k
=
a
(
b
k
)
{\displaystyle (ab)k=a(bk)}
Inductive step: By axiom 9,
(
a
b
)
k
′
=
(
a
b
)
k
+
a
b
{\displaystyle (ab)k'=(ab)k+ab}
By the inductive hypothesis,
(
a
b
)
k
′
=
a
(
b
k
)
+
a
b
{\displaystyle (ab)k'=a(bk)+ab}
By theorem 4,
(
a
b
)
k
′
=
a
(
b
k
+
b
)
{\displaystyle (ab)k'=a(bk+b)}
By axiom 9,
(
a
b
)
k
′
=
a
(
b
k
′
)
{\displaystyle (ab)k'=a(bk')}
By induction,
(
a
b
)
c
=
a
(
b
c
)
{\displaystyle (ab)c=a(bc)}
Theorem 8: Commutativity of multiplication:
a
b
=
b
a
{\displaystyle ab=ba}
Proof: Base case: By axiom 8 and theorem 5,
a
(
1
)
=
a
=
1
a
{\displaystyle a(1)=a=1a}
Inductive hypothesis: Suppose that, for
k
′
>
1
{\displaystyle k'>1}
a
k
=
k
a
{\displaystyle ak=ka}
Inductive step: By axiom 9,
a
k
′
=
a
k
+
a
{\displaystyle ak'=ak+a}
By the inductive hypothesis,
a
k
′
=
k
a
+
a
{\displaystyle ak'=ka+a}
By theorem 6,
a
k
′
=
k
′
a
{\displaystyle ak'=k'a}
By induction,
a
b
=
b
a
{\displaystyle ab=ba}
Theorem 9: Right-distributivity of multiplication over addition:
(
a
+
b
)
c
=
a
c
+
b
c
{\displaystyle (a+b)c=ac+bc}
Proof: By theorems 4 and 7,
(
a
+
b
)
c
=
c
a
+
c
b
{\displaystyle (a+b)c=ca+cb}
By theorem 7,
(
a
+
b
)
c
=
a
c
+
b
c
{\displaystyle (a+b)c=ac+bc}
The set of integers
Z
{\displaystyle \mathbb {Z} }
(
a
,
b
)
≡
(
c
,
d
)
⇔
a
+
d
=
c
+
b
.
{\displaystyle (a,b)\equiv (c,d)\Leftrightarrow a+d=c+b.}
Then the set of rational integers is the set of all equivalence classes of such ordered pairs. We will denote the equivalence class of which some pair (a , b ) is a member with the notation [(a , b )]. Then, for any natural numbers a and b , [(a , b )] represents a rational integer.
Definition: We define addition for the integers as follows:
[
(
a
,
b
)
]
+
[
(
c
,
d
)
]
=
[
(
a
+
c
,
b
+
d
)
]
.
{\displaystyle \left[(a,b)\right]+\left[(c,d)\right]=\left[(a+c,b+d)\right].}
Using this definition and the properties for the natural numbers, one can prove that integer addition is both associative and commutative.
Integer multiplication
edit
Definition: Multiplication for the integers, like addition, can be defined with a single axiom:
[
(
a
,
b
)
]
[
(
c
,
d
)
]
=
[
(
a
c
+
b
d
,
a
d
+
b
c
)
]
.
{\displaystyle \left[(a,b)\right]\left[(c,d)\right]=\left[(ac+bd,ad+bc)\right].}
Again, using this definition and the previously-proven properties of natural numbers, it can be shown that integer multiplication is commutative and associative, and furthermore that it is both left- and right-distributive with respect to integer addition.
![{\displaystyle (a+b)+k'=\left[(a+b)+k\right]'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c5ebdf9af3e658d8f6bb357af39aa69dd6bc20e)
![{\displaystyle (a+b)+k'=\left[a+(b+k)\right]'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa97049dff806bba7a86e854e4720ea460630130)
![{\displaystyle ab(1)=ab=a\left[b(1)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d14bb3bc3a366f46f9946ce2b704e8a95ffbce8)
![{\displaystyle \left[(a,b)\right]+\left[(c,d)\right]=\left[(a+c,b+d)\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/812c01093f785b1a4a8498d59f0a56f519a8aa36)
![{\displaystyle \left[(a,b)\right]\left[(c,d)\right]=\left[(ac+bd,ad+bc)\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f38fdd89ae78edac7e1dca425b610abad4921eb)