Do irational numbers have "uncountably" many digits?
The answer is NO they do not.
Imagine we take ONE natrual number and create new unique natrual numbers through some process. We take two "1" and continuously add a 0 between them, so it goes 101,1001,10001, this creates an "infinite" set of natrual numbers. We map one element in this set to "pie" and then map the rest to all its " (1)derivatives" which is "pie" with one digit changed. We then take one "(1)derivative" pie and use "101" as a new "1" so 1010101 and map it to this random (1)derivative pie and all its (1) derivatives. Then we use 1001 as "one" but map it to pie derivatives with TWO numbers permanently changed and all its (1)derivatives. Then take pie and its (2)derivatives and map it to 202, 2002, thwn 303 etc. I haven't fully fleshed this out but it's clear to see that through some process like this you can map EVERY irational number between 3 and 4 using only natural numbers.
The answer is NO they do not.
Imagine we take ONE natrual number and create new unique natrual numbers through some process. We take two "1" and continuously add a 0 between them, so it goes 101,1001,10001, this creates an "infinite" set of natrual numbers. We map one element in this set to "pie" and then map the rest to all its " (1)derivatives" which is "pie" with one digit changed. We then take one "(1)derivative" pie and use "101" as a new "1" so 1010101 and map it to this random (1)derivative pie and all its (1) derivatives. Then we use 1001 as "one" but map it to pie derivatives with TWO numbers permanently changed and all its (1)derivatives. Then take pie and its (2)derivatives and map it to 202, 2002, thwn 303 etc. I haven't fully fleshed this out but it's clear to see that through some process like this you can map EVERY irational number between 3 and 4 using only natural numbers.
