The existence of certain numbers i will try to make my question clear.
Whe know the existence of $\phi=1.61803398874989484_\ldots$
or $\pi=3.14159265359_\ldots$ and we know that the decimals numbers goes to infinity.
But, let's say i write a randon number like : 
$$0.0031881208316410990203632024503368214675523306840544$$
Does it exists?
Wolfram Alpha engine says it is : 
$$\frac{20751887915564}{6509128421234234}$$
But this engine cant handle well some others numbers like : 
$$0.2009999978498162165563223555516789561561$$
The question is :
if i write any number with , lets say a billion randon decimals, this number will 'exist'?
 A: My understanding of your question is: does any number that can be written with a finite amount of digits, can be written in the form $\frac{a}{b}$ with $a \in \mathbb{Z}, b \in \mathbb{N}, b \neq 0$. The answer is yes!
Suppose $x$ can be written with a finite amount of digits (say $n$ digits). Then
$$x = \frac{x\cdot 10^n}{10^n} = \frac{a}{b},$$ and $a\in \mathbb{Z},b \in \mathbb{N}$. Of course this fraction may not be irreducible but it is still a "rational representation" of your number.
A: If you write a number with exactly 1 billion digits after the decimal, it exists because it can be expressed as the rational number (number without the decimal)/(10^1-billion).
Similarly any terminating decimal can be expressed as a rational by removing the decimal point and dividing by 10^(# of decimal places).
Even for an infinite decimal we know it exists. This is because the (standard) definition of a real number is an convergent sequence of rational numbers. The sequence we get by continuously adding another decimal place is convergent and so the number exists.
