# Normalising fractional numbers

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?

## 1 Answer

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): $1\,1111\,001$.