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.