The two questions that i am asking are in bold.
To be clear, i am talking about whole number here.
Having seen 3 is everywhere by Numberphile that shows that almost 100% of the whole the numbers have the digit three in them, i have been thinking of approaching the same problem from the different perspective:
We have the following method of generating numbers with 3:
Take any whole number without digit 3, add 3 to the left of it, and then keep adding whatever digits to the left of that 3 for as long as you want.
What this tells me is that i can generate infinite amount of numbers with 3 for each one number without 3. I can then trivially show that each infinite set of numbers with 3 was not double-counted (i.e. any 3 possessing number could have been generated for just 1 number without 3).
For me, this means the the question can then be restated as:
If you have an infinite set and a single number, and you randomly select a single number out of the superset, what are the chances of getting a single number?
However, i've been pointed out that because we are talking about sets both of which are infinite, then we have to take into account that you can divide the set of numbers without 3 into infinitely many subsets, each of which is infinite (which i cannot dispute). And thus there maybe a way to write my generation method in reverse, and if so, then the argument is internally contradictory.
The question then becomes: Is there anything preventing us from generating sets with numbers that have no digit 3, in such a way that each set is infinite, each of these sets can be associated with some number that has the digit 3, and no number without 3 is associated to more than one number that has 3?
Here is how i have attempted to reason about this question, but i maybe missing something.
- The superset of numbers with 3 and the numbers without 3 is the set of all the whole numbers.
- It has already been shown that numbers with 3 have some specific way to be associated with numbers without 3. Thus we can write each number with 3 as TyT₄T₃T₂3Nx…N₄N₃N₂N₁N₀ where N are digits that cannot be 3 and T are digits that can be 3.
- Even with some other form of generating numbers in reverse, it must be possible to represent them this way.
- If we allow infinite sets of numbers without 3 to be generated for each of the numbers with 3, then our x must go to infinity at some point.
- But that would mean that we must allow to position at least one digit 3 past infinity, which is nonsense.
The second question then becomes: What is nonsense? My logic or the possibility of generating infinite number of infinite sets without number 3 for each number with 3 without double counting?