For the function f in 1.6D i.e. $$x=0.b_1b_2....(3)$$ then $$f(x) = y= 0.a_1a_2.....(2)$$ where $a_i=b_i/2$
Here (2),(3) represents the binary and ternary expansions of $x$ respectively.
Show that if $(a,b)$ is any one of the open intervals removed in the construction $K$, show that $f(a)=f(b)$.
Hint: show that a and b can be written $a=0.a_1a_2.....a_n1$, $b=0.a_1a_2.....a_n2$, where each $a_i$ is 0 or 2. Then rewrite the expansion for a using only 0 and 2.
I'm able to prove this for the first interval that is removed $(1/3,2/3)$ but for general case i don't totally get it. How can i able to express $a$ and $b$ in the form given in the hint? I'm lost!! Please help me understand decimal expansion.
Essentially at each step you are removing the interval between the ternary numbers
with each $b_i \in \{0,2\}$. The first could also be written as $0.b_1b_2\ldots b_n10000\ldots_3$ and the second as $0.b_1b_2\ldots b_n12222\ldots_3$ but we need to avoid the digit $1$ if we want to apply the function by dividing by $2$.
Applying the function $f$ to these two end points would give
and these are the same value in binary.
Consider the interval $\left(\frac{7}{27}, \frac{8}{27}\right)$ which you will remove. You can write $\frac{7}{27}$ in ternary as $0.021_3$ i.e. $0.0210000\ldots_3$ but in this case we want to write it as $0.0202222\ldots_3$ so we can divide each digit by $2$ to get $f\left(\frac{7}{27}\right)=0.0101111\ldots_2 = \frac{3}{8}$. Meanwhile $\frac{8}{27}$ is $0.0220000\ldots_3$ so $f\left(\frac{8}{27}\right)=0.0110000\ldots_2 = \frac{3}{8}$ too.