I'm taking a numerical analysis course. This is one of the examples in my professor's notes:
"Suppose we have 10 bits of precision and
\begin{gather}
x = (1.01011100 ∗ ∗)_2 × 2^{E},
\end{gather}
\begin{gather}
y = (1.01011000 ∗ ∗)_2 × 2^{E}
\end{gather}
where the $*$ stands for inaccurate bits (i.e. garbage) that say were generated in
previous floating point computations. Then, in this 10 bit precision arithmetic
\begin{gather}
z = x − y = (1.00 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗)_2 × 2^{E−6}
\end{gather}
We end up with only 2 bits of accuracy in z. Any further computations using
z will result in an accuracy of 2 bits or lower!"
I'm confused about why the exponent ends up being $E-6$ as well as how you determine how many garbage bits end up in the final answer? For example, why couldn't the answer have been $(1.0000 ∗ ∗ ∗ ∗ ∗ ∗)_2 × 2^{E−6}$?
When you do the subtraction you have $$\ \ \ \ 1.01011100**\\ -\underline{1.01011000**}\\ \ \ \ \ 0.00000100**$$ You now normalize the number by multiplying the mantissa by $2^6$ to make the leading $1$ be in the ones place. That means you need to reduce the exponent by $6$ to keep the value the same. When you do that you bring in six more bits of $*$.