```How to represent \$0.148 * 2^{14}\$ in normalized floating point arithmetic with the format

1 - Sign bit
7 - Exponent in Excess-64 form
8 - Mantissa

\$(0.148)_{10} = (0.00100101\;111...)_2\$

We shift it 3 bits to left to make it normalized \$(1.00101\;111)_2 * 2^{11}\$.

Exponent = \$11+64 = (75)_{10} = (1001011)_2\$ and Mantissa = \$(01001\;111)_2\$.

So floating point representation is \$(0\;1001011\;00101111)_2 = (4B2F)_{16}\$ **Representation A**

But if we store the denormalized mantissa into 8 bit register, then it won't have stored the last three \$1\$s and then the mantissa would have normalized from \$(0.00100101)_2\$ to \$(1.00101\;000)_2\$ by inserting 3 \$0\$s instead of \$1\$s.

The representation would have been \$(0\;1001011\;00101000)_2 = (4B28)_{16}\$ **Representation B**

So while normalizing, does the processor takes into account the denormalized mantissa bits beyond 8 bits too? Or just rounds it off? Which one is correct: **A** or **B**?

Does it store the mantissa in fixed point representation? How does it all work?```