axiomatic sets 24 (arithmetics in rational numbers)

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.