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2016/09/02

axiomatic sets 23 (types of numbers)

There are some kinds of numbers. Let us list these samples up.

The empty set means the number "0", that is, \phi=\left\{ \right\}=0 .
We will see that "0" is 0_N  in natural numbers.

The natural number "1" is the set whose member is only the empty set, 1_N=\left\{\phi\right\} .
The natural number "2" is the set whose members are natural numbers "0" and "1",
2_N=\left\{0_N,1_N \right\}=\left\{\phi,\left\{\phi\right\} \right\} .

The integer "0" is the ordered pair class 0_Z=[(0_N,0_N)] ,
but it's class has only one element (0_N,0_N) .

The non negative integer "1" is [(1,0)] ,where "1" is the natural number 1_N, and "[ ]" means a equivalent class. Therefore,
 1_Z=[(1_N,0_N)]=[(\left\{\phi\right\},\phi)]=[(\left\{\left\{ \right\}\right\},\left\{ \right\})]
As the equivalent relation is for two integers (a_N,b_N),(c_N,d_N) ,
a_N+d_N=b_N+c_N  must be satisfied,
you have to note that (1_N,0_N)\sim (2_N,1_N)\sim (3_N,2_N)\sim\cdots .
Those are in 1_Z ,and we use every elements as 1_Z .

The non negative integer "2" is also the ordered pair class [(2_N,0_N)]=2_Z ,
The negative integer "-2" is [(0_N,2_N)]=-2_Z .

The rational number "0" is also the ordered pair class [(0,a)] ,where a  is an arbitrary integer, but not zero, and "[ ]" means a equivalent class.
 0_Q=[(0_Z,a_Z)]=[([(0_N,0_N)],[(a_N,0_N)])] .
The equivalent relation on rational numbers is for two rational nummbers (p_Z,q_Z),(r_Z,s_Z)
(and q_Z,s_Z\ne 0), p_Zs_Z=q_Zr_Z must be satisfied.

The non negative rational number "1" is [(1,1)] ,where "1" is the integer number "1_Z".
Therefore,
1_Q=[(1_Z,1_Z)]=[([(\left\{\phi\right\},\phi)],[(\left\{\phi\right\},\phi)])] ,
and for x\in 1_Q ,
x\sim (1_Z,1_Z)\sim (-1_Z,-1_Z)\sim (2_Z,2_Z)\sim (-2_Z,-2_Z)\sim (3_Z,3_Z)\sim\cdots  
By same way, 2_Q=[(2_Z,1_Z)]=\left\{(2_Z,1_Z),(-2_Z,-1_Z),(4_Z,2_Z),(-4_Z,-2_Z),(6_Z,3_Z),\cdots \right\} .

You will find these kinds of numbers are constructed by \phi  and axioms of sets.
Please try to make some types of numbers.









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