Cosider that $0 < x < 1$ and in base $10$, $x = 0.abcde....$ and in base $2$ $x = 0.uvwz......$ and let $2x = h.ijklmn.....$
Now $x \ge \frac 12 \iff a \ge 5\iff 2x \ge 1 \iff u = 1 \iff h = 1$.
And $x < \frac 12 \iff a < 5\iff 2x < 1 \iff u= 0 \iff h=0$.
So $u = h$ and that is why you can do this for the first decimal.
Now we get rid of the first digit $h$ (you weren't do doing that) and we let $y = 0.ijklm...$
Now if $x \ge \frac 12$ then $y = 0.ijklmn....2*(x -\frac 12)$. And if $x < \frac 12$ then $y = 0.ijklmn = 2*x$ and $0 \le y < 1$. Let $x' = x - \frac 12$ if $x \ge \frac 12$ or let $x' = x$ if $x < \frac 12$. Either way, $y = 2x'$.
And now let $2y = \alpha.\beta\gamma....$
Now $x' \ge \frac 14 \iff y \ge \frac 12 \iff 2y \ge 1 \iff i \ge 5 \iff v = 1\iff \alpha = 1$. And this is only if $\frac 14 \le x < \frac 12$ or $\frac 34 \le x < 1$
And $x < \frac 14 \iff y < \frac 12 \iff 2y < 1 \iff i < 5 \iff v=0 \iff \alpha =0$. And this is only if $0 < x < \frac 14$ or $\frac 12 \le x < \frac 34$.
Either way $\alpha = v$ and that is why we can get the second digit.
Inductively we can carry that on forever (or until we hit $0$).
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There are two things you are supposed to keep track of the binary decimal, which starts at $0.$ and the base $10$ decimal.
Watch:
A: $0. = 0$ B: $0.25$.
Multiply the B by 2:
A: $0. = 0$ B: $0.5 $
Take the digit from B and put it on A: The digit is $0$.
A: $0.0 = 0$ B: $0.5$ (that doesn't look like much but the A has gotten one digit longer)
Take the fraction left in A as your new starting point of A. The fraction part is $0.5$.
A: $0.0 = 0$ B: $0.5$.
Multiple by 2:
A: $0.0 = 0$ B: $1.0$
Take the digit from B and put it on A: The digit is $1$.
A: $0.01 = \frac 14$ B: $0.0$ (But this time the $1$ goes from the $B$ to the $A$. And notice the B is reduced).
Take the fraction left in A as your new starting point of A. The fraction part is $.0$.
A: $0.01 = \frac 14$ B:$0.0$.
3)repeat until you either get to 0 or a periodic number.
We've reached $0$. We are done.
Let's do a more illuminating example:
A: $0.$ B: $0.375$
A: $0.$ B: $0.75$ (Double the B)
A: $0.0$ B: $0.75$ (Move the $0$ to the A)
A: $0.0$ B: $1.50$(Double the B).
A: $0.01$ B: $.50$(Move the $0$ to the A)
A: $0.01$ B: $1.0$(Double the B)
A: $0.011$ B: $.0$.(Move the $1$ to the A)
We've reached zero so we stop.
Basically with have $0.375 = 0.375$
$0.75 = 0.375*2 = 0.375*2 - 0$
$1.5 = 0.375*4 - 0*2$
$.5 = 0.375*4 - 0*2 - 1$
$1.0 = 0.375*8 - 0*4 - 1*2$
$.0 = 0.375*8-0*4 - 1*2 - 1$
So $0.375*8 = 0*4 + 1*2 + 1$
$0.375 = \frac {0*4 + 1*2+1}{8} = \frac 02 + \frac 14 + \frac 18$.
This works because way place a $1$ on the decimal if and only if what we have left is greater than $\frac 12$. But the important part is if it is we then remove that half and have double what is left.
"Trying to understand why all this with the multiplying by two actually works, I do this: "
$0.25=0.01=\frac{1}{2}(0+\frac{1}{2}(1+\frac{1}{2}*0))$
Nothing wrong with that.
$0.375 = 0.011 = \frac 12(0 +\frac 12(1 + \frac 12(1)))$