Arithmetics in rational numbers are defined as follow.
Given two rational numbers $x=(a,b)$ and $y=(m,n)$ ,where $a,b,m,n$ are integers and $b,n$ are not zero.
The addition is
\[ x+y=(a,b)+(m,n)=(an+bm,bn) . \]
The multiplication is
\[ xy=(a,b)(m,n)=(am,bn) \]
,where "+" is the addition and "$an$" means the multiplication of integers "$a$" and "$n$".
The magnitude relations "$\gt$", "$=$", and "$\lt$" are
\[ x\gt y \quad \mbox{if and only if }\quad an\gt bm , \]
\[ x= y \quad \mbox{if and only if }\quad an= bm \]
,and
\[ x\lt y \quad \mbox{if and only if }\quad an\lt bm . \]
We say that $x$ is from above "greater than", "equivalent to" ,and "less than" $y$ .
The exponents are;
\[ x^0=1 \]
,where "1" is the rational number "1", (i.e. $1=[(1,1)]$ .)
\[ x^1=x=(a,b) , \]
\[ x^2=xx=(aa,bb) , \]
\[ x^3=(xx)x=((aa)a,(bb)b) \]
,and so on.
Immediately we will get $x^rx^s=x^{r+s}$ .
If $a$ and $b$ are not zero ,then
\[ x^{-1}=(a,b)^{-1}=(b,a) . \]
This is an inverse element of multiplication.
That is,
\[ xx^{-1}=(a,b)(b,a)=(ab,ab)=1 \]
,where "1" is the rational number "1".
We will ordinarily denote $x^{-1}=\frac{1}{x}$ .
More over, $x^{-2}=\frac{1}{x^2}$ , $x^{-3}=\frac{1}{x^3}$ and so on.
More generally, we will write $x^{-1}y=yx^{-1}=\frac{y}{x}$ .
(It is called the quotient. )
The definition of $x^{c}\quad (0\lt c\lt 1)$ is complicated and not easy.
Given a rational number $w(\gt 0)$ and $w^2=ww=z$ ,
then we will define and denote $z^{\frac{1}{2}}=w$ .
In same way, if $w^3=z$, then $z^{\frac{1}{3}}=w$ ,and so on.
However, for any rational number $z$ , we know that there does not exist a rational number $z^{\frac{1}{2}}$ .
(in the preceding post, we have proven $2^{\frac{1}{2}}=\sqrt{2}$ is not a rational number. )
Unfortunately, we can not always assert $x^{c}\quad (0\lt c\lt 1)$ is in rational numbers.
(Although it is necessary to prove, almost numbers $x^{c}$ are not rational numbers. )
In addition, we have known many numbers such $\pi$ and $e$ are not rational numbers.
Thus irrational numbers and real numbers will be needed.
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.
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.