1
$\begingroup$

For example $-\tfrac9{16}$.

$$\tfrac{9}{16} = \tfrac{1}{16}+\tfrac12 = 0.1001\,,$$ which when normalised becomes $0.1001\times 2^0$. Can its mantissa be $0.0001001$ in 8 bits?

If so, as $-\tfrac9{16}$ is negative, twos' complement is done to make its mantissa $1.1110111$. How do we normalise that?

$\endgroup$
1
$\begingroup$

Not clear of your process, but here is how it can be done: $\tfrac{9}{16} = \tfrac{1}{16} + \tfrac12$, which gives you $0.1001$ in binary form. To normalize, take the first bit to be 1, which gives you $1.001 \times 2^{-1}$ which makes 3 bits for the mantissa $001$ and some bits for the exponent to represent $-1$ in twos' complement.

IEEE floating point formats don't use twos' complement for negative mantissas: they use an extra bit to indicate the sign (1 means negative).

So in the end, if exponent is 4 bit (and without any excess exponent): $1\,1111\,001$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.