The multiplication of natural numbers is defined same as that of the addition.
The operation $g$ of the multiplication is a function satisfying following conditions.
(You must remember the function $g$ is also a set.)
(1)$g(a,0)=0$
(2)$g(a,b^+)=g(a,b)+a$
,where $a,b\in\mathbb{N}$ and $b^+$ is the successor set of $b$ . ($b^+=b\cup\left\{b\right\}$)
These mean simply ;
$a\times 0=0$ ,
$a\times (b+1)=(a\times b)+a$ .
For example,
$g(2,3)=2\times(2+1)=(2\times 2)+2=((2\times 1)+2)+2$
$=(((2\times 0)+2)+2)+2=0+2+2+2=6$ .
In addition, the exponent of natural numbers is extended.
By using the multiplication, the function $e$ of the exponent satisfies following conditions.
(1)$e(a,0)=1$ ,
(2)$e(a,b^+)=g(e(a,b),a)$ .
That is,
$a^0=1$ ,
$a^{b+1}=(a^b)\times a$ .
If $a=2,b^+=3$ ,then
$e(2,3)=g(e(2,2),2)=g(g(e(2,1),2),2)=g(g(g(e(2,0),2),2),2)=1\times 2\times 2\times 2$ .
These are called a finite recursion formula.
2016/06/06
axiomatic sets 17 (arithmetic in natural numbers)
The set of all natural numbers is closed under addition (and multiplication).
We have already defined one equation of addition.
\[ n+1=n\cup\left\{ n \right\} \qquad (n\in\mathbb{N}, 0\in\mathbb{N},1=\left\{ 0 \right\}) \]
Since, for any natural number $n$ , there must be an $n\cup\left\{ n \right\}$ by axiom of infinity,
and for any natural number $m\ne 0$ , there must be an $n$ such that $n\cup\left\{ n \right\}=m$ ,
this equation is true.
We want to extend this to the definition of "$a+b$".
For this purpose the function "$f$" is needed.
A collection of a pair of two natural numbers is a set.
(i.e. Cartesian Product of two $\mathbb{N}$s is a set. )
\[ \mathbb{N}\times\mathbb{N}=\left\{\left\{ a,b \right\}: a\in\mathbb{N},b\in\mathbb{N} \right\} \]
In axiomatic set theory, as the function $f$ is also a set,
\[ f:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{N} , \]
and
\[ \forall a\forall b\exists f[f(\left\{a,b \right\})\in\mathbb{N}] , \]
and $f$ has to be satisfied the following two conditions;
(1)$f(\left\{a,0\right\})=a$
(2)$f(\left\{a,b^+\right\})=(f(\left\{a,b\right\}))^+$
,where $b^+$ is the successor set of $b$ . Namely, $b^+=b\cup\left\{ b \right\} $ .
You may see it is not easy to understand. However, its mean, simply,
(1)$a+0=a$ ,
(2)$a+(b+1)=(a+b)+1$ .
If $b$ is not $0$, there exists a $c$ such that $c^+=c+1=b$ .
Thus,
\[ a+(b+1)=(a+b)+1=(a+c^+)+1=(a+(c+1))+1=((a+c)+1)+1 . \]
Same operations take us to
\[ a+b=(\cdots(a+0)+1)+1)\cdots )+1 . \]
,where $(\cdots (0+1)+1)\cdots )+1=b$ .
This is the definition of an arithmetic of addition in axiomatic set theory.
We have already defined one equation of addition.
\[ n+1=n\cup\left\{ n \right\} \qquad (n\in\mathbb{N}, 0\in\mathbb{N},1=\left\{ 0 \right\}) \]
Since, for any natural number $n$ , there must be an $n\cup\left\{ n \right\}$ by axiom of infinity,
and for any natural number $m\ne 0$ , there must be an $n$ such that $n\cup\left\{ n \right\}=m$ ,
this equation is true.
We want to extend this to the definition of "$a+b$".
For this purpose the function "$f$" is needed.
A collection of a pair of two natural numbers is a set.
(i.e. Cartesian Product of two $\mathbb{N}$s is a set. )
\[ \mathbb{N}\times\mathbb{N}=\left\{\left\{ a,b \right\}: a\in\mathbb{N},b\in\mathbb{N} \right\} \]
In axiomatic set theory, as the function $f$ is also a set,
\[ f:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{N} , \]
and
\[ \forall a\forall b\exists f[f(\left\{a,b \right\})\in\mathbb{N}] , \]
and $f$ has to be satisfied the following two conditions;
(1)$f(\left\{a,0\right\})=a$
(2)$f(\left\{a,b^+\right\})=(f(\left\{a,b\right\}))^+$
,where $b^+$ is the successor set of $b$ . Namely, $b^+=b\cup\left\{ b \right\} $ .
You may see it is not easy to understand. However, its mean, simply,
(1)$a+0=a$ ,
(2)$a+(b+1)=(a+b)+1$ .
If $b$ is not $0$, there exists a $c$ such that $c^+=c+1=b$ .
Thus,
\[ a+(b+1)=(a+b)+1=(a+c^+)+1=(a+(c+1))+1=((a+c)+1)+1 . \]
Same operations take us to
\[ a+b=(\cdots(a+0)+1)+1)\cdots )+1 . \]
,where $(\cdots (0+1)+1)\cdots )+1=b$ .
This is the definition of an arithmetic of addition in axiomatic set theory.