Defining an infinite list of irrational numbers

Question

We can form a list of binary real numbers

$ 0.d_{1,1} d_{2,1} d_{3,1} \dots$

$ 0.d_{1,2} d_{2,2} d_{3,2} \dots$

$ 0.d_{1,3} d_{2,3} d_{3,3} \dots$

$\dots$

where each $d_{x,y} $ is a single digit $\{0,1\}$ and

$$ d_{x,y} = rand() : x \in \mathbb N \land y \in \mathbb N$$

or$$ d_{x,y} = \lfloor \left(x + 1\right)^{\sqrt 2} {\left(y + 1\right)^\pi} \rfloor\pmod 2 : x \in \mathbb N \land y \in \mathbb N$$

or$$ d_{x,y} = \lfloor{\frac{(y-1)\pmod {2^x}}{2^{x-1}}} \rfloor : x \in \mathbb N \land y \in \mathbb N$$

Would we then have an infinite list of irrational numbers?

I just want to confirm that each row will have an infinite number of digits as opposed to a finite (though large) number of digits.

Answer 1

Partial answer

First of all note that a sequence of digits (regardless of base)

$$0.a_1a_2a_3\cdots$$

defines a rational number if and only if the sequence $(a_i)$ is eventually cyclic.

$d_{x,y} = rand() : x \in \mathbb N \land y \in \mathbb N$

Obviously this depends on what $rand()$ is. If it is a probabilistic function then you need a probablistic context. Generally the answer is "no" if $rand()$ can return a sequence of $0$'s for example. However if $rand()$ is truely random then the probability of getting a rational is $0$.

$ d_{x,y} = \lfloor \left(x + 1\right)^{\sqrt 2} {\left(y + 1\right)^\pi} \rfloor\pmod 2 : x \in \mathbb N \land y \in \mathbb N$

This one is tricky. I don't have an answer for it yet.

$ d_{x,y} = \lfloor{\frac{(y-1)\pmod {2^x}}{2^{x-1}}} \rfloor : x \in \mathbb N \land y \in \mathbb N$

No, there are no irrationals here. For a fixed $y$ the sequence $d_{x,y}$ is eventually constant $0$. That happens whenever $2^{x-1}>y-1$.