I'm trying to establish whether restricting the set of numbers in $(0,1)$ to those whose even bits in the binary representation (where "binary representation" means the unique binary representation not end in an infinite sequence of $1s$) all equal to $0$ says anything about the rationality of the number. Just looking for a sanity check here.
My thought is that you can still have irrational numbers with all even bits equal to $0$. If you take, e.g. $\frac{\pi}{5} \in (0,1)$, which is irrational, let its binary representation be denoted by $0.x_1x_2x_3x_4x_5\ldots$.
If we have a function $\phi$ that "turns off" every even bit, then $\phi(\frac{\pi}{5}) = 0.x_10x_30x_5\ldots$ would still be irrational, right?
We could also define a function $\psi$ that inserts a $0$ after every bit, so that $\psi(\frac{\pi}{5}) = 0.x_10x_20x_30x_40x_5\ldots$ then the binary representation of $\psi(\frac{\pi}{5})$ is $0$ in all even bits and $\psi(\frac{\pi}{5})$ is irrational.
The main reason I'm not sure about whether this argument follows, is that maybe $\phi$ or $\psi$ are not well-defined. If they're not, I'm not sure how to rigorously argue that there are irrationals whose binary representation has all even bits equal to $0$.
"My thought is that you can still have irrational numbers with all even bits equal to 0." : TRUE
"If you take, e.g. $\pi/5 \in (0,1)$ , which is irrational , let its binary representation be denoted by $0.x_1x_2x_3x_4x_5\cdots$"
"If we have a function $\phi$ that "turns off" every even bit, then $X=\phi(\pi/5)=0.x_10x_30x_5\cdots$ would still be irrational, right?" : not necessarily. The remaining Digits might be all $0$ , hence $X=0$ or the remaining Digits might have a pattern , hence $X$ might be rational or the remaining Digits might not have a Pattern , hence $X$ might be irrational or even transcendental.
To see that , consider the number $X=\phi(\pi/5)$ : If it is $0$ or rational , we have the Example. In Case $X$ is irrational , then consider the number $X/2$ , where the Digits will be shifted right , hence all odd bits are $0$. We can then see that $Y=\phi(X/2)$ will have all even bits $0$ too , hence over all , $Y$ will be Zero.
We can tweak this to generate rational numbers like this :
$Y=\phi(Z+X/2)$ , where $Z$ is some rational number with $2N$ bits , where only odd $N$ bits are $1$.
Then $Y$ will get all $0$ bits from $X/2$ & then it will get few $1$ bits from $Z$ , overall , we will get rational number with few $1$ bits.
"We could also define a function $\psi$ that inserts a $0$ after every bit , so that $\psi(\pi/5)=0.x_10x_20x_30x_40x_5\cdots$ , then the binary representation of $\psi(\pi/5)$ is $0$ in all even bits and $\psi(\pi/5)$ is irrational." : true
The way to see that is like this :
Consider Base 4 number representation. Use only the Digits $0$ & $2$ [ & never use $1$ & $3$ ] to make irrational numbers. We can then convert the Base 4 number to Base 2 , where the even bits are always zero. Yet we will have irrational number.
Eg $0.2020020002000200002000002\cdots$ in Base 4 , when converted to Base 2 , will change $0$ to $00$ & $2$ to $10$ , hence all even bits will be $0$ in that irrational number.
Similar Idea will work with Base 8 or Base 16 (Hexa-Decimal)
"The main reason I'm not sure about whether this argument follows, is that maybe $\phi$ or $\psi$ are not well-defined." : the 2 functions are well-defined , no worries there. In terms of guaranteeing irrationality , $\phi$ will not work , while $\psi$ will work.
You can see that when you check $\phi(\phi(\pi-3)/2)$ & $\phi(\psi(\pi-3)/2)$ & $\psi(\phi(\pi-3)/2)$ & $\psi(\psi(\pi-3)/2)$ & ETC