We work in base-10. If $a$ and $b$ are two real normal irrational numbers with decimal expansion written as $a_1a_2\ldots a_n\ldots$ and $b_1b_2\ldots b_n\ldots$ respectively, I write $a\oplus b$ for the number formed by concatenating the n-th digit of $a$ to the left of the n-th digit of $b$. Explicitly $$a\oplus b=a_1b_1a_2b_2\ldots a_nb_n\ldots$$
Question: is $a\oplus b$ normal ?
Intuitively to me the answer is yes. At a minimum I would expect $a\oplus b$ to be simply normal. But then I am not ever certain $a\oplus b$ is irrational. Here is what I know:
$$(a\oplus b)\oplus c=a\oplus (b\oplus c)$$ $$a\oplus b \neq b\oplus a$$ $$a\oplus b=a\oplus c \implies \text{ }b=c$$ $$b\oplus a=c\oplus a \implies \text{ }b=c$$
Edit: Following the suggestion of a comment I have replaced "$+$" with "$\oplus$"
You could e.g. have $b = a$, and then your new number is not normal: the distribution of pairs is wrong.