Rationalizing An Irrational Number

[trfrm]

Ahlan wa Sahlan.

Excuse me ... .

A set of the rational numbers is denoted as $\mathbb{Q}:=\{m/n~|~m,n\in\mathbb{Z}~\textrm{and}~n\neq0\}$, where $\mathbb{Z}:=\{0,\pm{1},\pm{2},\pm{3},\dots\}$ is the set of integers ... .

A set of the irrational numbers $\mathbb{R}-\mathbb{Q}$ consists of the real numbers which cannot be denoted as $m/n$ such that $m,n\in\mathbb{Z}$ and $n\neq0$ ... .

The problem is in following ... .

For example, in present time, here is $a\in\mathbb{R}-\mathbb{Q}$ which be agreed as an irrational number ... .

Can the irrational number $a$ be rationalized to be a rational number if a person is able to denote the $a$ as $m/n$, where $m,n\in\mathbb{Z}$ and $n\neq0$, in the next time ... ? [Here, the absolute values of $m$ and $n$ may be the very very huge integers ... .]

Thank you for the answer ... . 

Arigatou ikimono gakari.

[trfrm]

Agnus Dei, qui tollis peccata mundi.

Is it all integers or only small ones ? What is small ? What is huge ?

Oh, I'm sorry ... . The term "small" and "huge" are relative ... .

I don't actually undrstand what you are trying to do. You define irrational numbers in a way that mean they are ... irrational. What do you mean by"rationalize"? (I can't follow from your notation)

I just tried to regard an irrational number as a rational number ... .

Here is a simple proof that $ \sqrt{2}$ is irrational: Proof that the square root of 2 is irrational number

You can do this with most (all?) irrational numbers, for example: Proof that $\pi$ is irrational

Thank you very much for the explanation ... . Now, I understand why the square root of 2 and the pi are regarded as the irrational numbers ... .

All rational numbers and all irrational number can form a set of real numbers ... .

There are significant differences between character of $\mathbb{R}$ and character of $\mathbb{Q}$ ... .

If in example we define the set $(-1,1)_\mathbb{Q}:=\{x\in\mathbb{Q}~|~-1<x<1\}\subset\mathbb{Q}$ and the set $(-1,1)_\mathbb{R}:=\{x\in\mathbb{R}~|~-1<x<1\}\subset\mathbb{R}$, then we can see the significant difference ... .

We can find at least one bijective map $\mathbb{R}\to(-1,1)_\mathbb{R}$, in example $f\,:\,x\mapsto\tanh{kx}$ and $g\,:\,x\mapsto\frac{2}{\pi}\arctan{kx}$, where $k$ is a constant ... .

But, however, we cannot find at least one bijective map $\mathbb{Q}\to(-1,1)_\mathbb{Q}$ ... .

How can it ... ? Until now I have not understood ... .

Wal bi Taufiq wal Hidayah.

[trfrm]

Salam damai Kristus.

That is not true. The function $y= \frac{x}{x+1}$ maps the set of all rational numbers one-to-one onto the set of rational numbers between -1 and 1.

Excuse me ... . The plot of f(x) = x/(x + 1) is in the following site ... .

x/(x+1) - Wolfram|Alpha

The domain of the function $f\,:\,\mathbb{Q}\to\mathbb{Q}~:~x\mapsto{x/(x+1)}$ is $\textrm{dom}(f)=(-\infty,-1)_\mathbb{Q}\cup(-1,+\infty)_\mathbb{Q}$ ... , and the range (or image) of the $f$ is $\textrm{im}(f)=(-\infty,-1)_\mathbb{Q}\cup(-1,+\infty)_\mathbb{Q}$ ... . Thus, the function is not bijection from $\mathbb{Q}$ to $(-1,1)_\mathbb{Q}$ ... .

I'm sorry ... .

http://www.thescienceforum.com/mathematics/35211-rationalizing-irrational-number.html#post414277

Nderek langkung.