Given two numbers $x$ and $y$, which $x,y\leq1$, I can express them in binary number $x=0.a_{1}a_{2}...$, $y=0.b_{1}b_{2}...$. Now I proceed in this way:
The new number $xy=c_{0}+0.c_{1}c_{2}...$, where:
$c_{0}=0+h_{0}$
$c_{1}=0+h_{1}$
$c_{2}=a_{1}b_{1}+h_{2}$
$c_{3}=a_{1}b_{2}+a_{2}b_{1}+h_{3}$
$\vdots$
Where $h_{i}$ is the carrying obtained from $i+1$ equation, with the expression $\lfloor\frac{1}{2}(\sum_{j+k=i+1}a_{j}b_{k}+h_{i+1})\rfloor$
Now the problem arise, if I let all $a_{i}=b_{i}=1$ and try to calculate the $h_{1}$ starting from $h_{i}$ ,say $i=10000$, the $h_{1}$ is huge. In fact I can calculate the limit when $i$ tends to infinity, which is infinite. That means $xy$ is unbounded but $xy=1$ in this case($0.1111...=1$). So what is the problem here? I clearly make some simple mistake but I cannot spotted it.
Edit: Answer to the comment. If we truncated the $x,y$ to n-order, that is, $x=0.a_{1}a_{2}...a_{n}$, $y=0.b_{1}b_{2}...b_{n}$. Then we have the following result:
$c_{n}=a_{1}b_{n-1}+...a_{n-1}b_{n}+h_{n}$
$c_{n+1}=a_{1}b_{n}+...a_{n}b_{1}+h_{n+1}$
$\vdots$
$c_{2n-1}=a_{n-1}b_{n}+a_{n}b_{n-1}+h_{2n-1}$
$c_{2n}=a_{n}b_{n}$
While we can see that the formula is correct. The sequence $h^{(m)}_{n}$ itself converge, but the limit when $m$ goes to infinity does not for $n=m+1,m+2...$. It may be possible to take the limit of $h^{(m)}_{n}$ to the sequence $h_{n}$. But intuitively it is weird that there are infinite carrying inside but disappear at $c_{0}$. Also why we can take $h_{2n-1}=1$(and $h_{2n}=0$) in this case?