KAMP
Kategori Umum => Matematika => Topik dimulai oleh: trfrm pada Juli 20, 2018, 01:05:12 PM
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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 ... . :)
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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 ... .
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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