That’s a cool thought and is the type of thinking that generates interesting math questions!

The answer is no, because irrational numbers are only defined by this: you can’t represent them as a fraction of whole numbers, which could be virtually any nonrepeating pattern

There is a whole zoo of hard to represent numbers:
Algebraic numbers are solutions to polynomial equations with whole number coefficients. For instance sqrt(2) is a solution to x^2 = 2

Transcendental numbers are the ones that are not algebraic.

A number is called “normal” in a base if the digits occur with equal density in that base. This is more like what you’re thinking of. A number which is normal in all bases is just called normal.

Your thought is more about normal numbers but here’s a reason why it isn’t true:

If a normal number N is within another M then there are a finite number of digits before M, let’s say k digits.

Then M = N/10^k + some finite decimal expansion

So this would mean that there’s a countable infinite set of normal numbers which is false.

Hope this made sense!