# conversion of floating point to 8- bit binary word

What is the representation of $-(52.625)_{10}$ in 8 bit word? In 8 bit word , first bit represents the sign of the no. The next three bits represents the exponent and the last 4 bits the mantissa..Now since there is no bits to represent the sign of the exponent, how do we go about it?Also there is this concept of biasing .. when do we use biasing ? Here in 8 bit representation ,since the exponent has three bits the bias is 3... And in 32 bit representation the bias is 127 as it has 8 bits for exponent. In the above example if we add biasing the exponent of 2 becomes 6+3=9. And the binary form of 9 is 1001 which occupies 4 bits, but we have only 3 bits for exponent ... Now how do I represent the above no. ? Please correct me if I have gone wrong anywhere..

• Note( errata) : the exponent of 2 is 5+3=8 whose binary form is 1000... Aug 28, 2017 at 20:10

The exponent bias is to make it so you don't need to store the sign of the exponent. With three bits you can have numbers from $000_2=0$ to $111_2=7$. If we decide on a bias of $3$ we would represent an exponent of $-1$ as $-1+3=2_{10}=010_2$ and an exponent of $2$ as $101_2$ You can't store $52.625$ in an $8$ bit word using the format you describe. The exponent should be $5$ because $2^5 \lt 52 \lt 2^6$ but if we bias up by $3$ we get $8$, which doesn't fit into three bits. The largest number you can represent is $1.1111_2 \cdot 2^4=11111_2=31_{10}$ assuming your mantissas are normalized and you don't store the leading $1$. The Wikipedia article is a good one.