1 of 2

# Normalizing the mantissa in floating point representation

I went through a question which asked to represent $$0.145 * 2^{14}$$ in normalized floating point arithmetic with the format

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


$$(0.145)_{10} = (0.00100101001...)_2$$ (say A). We shift it 3 bits to left to make it normalized $$(1.00101001)_2 * 2^{11}$$.

Exponent = $$11+64 = (75)_{10} = (1001011)_2$$ and Mantissa = $$(01001001)_2$$.

So floating point representation should be $$(0\;1001011\;00101001)_2 = (4B29)_{16}$$.

But the solution considered $$(0.145)_{10} = (0.00100101)_2$$(upto 8 bits only). Shifting it to left by inserting zeroes to the right makes it $$(01001000)_2$$ instead of $$(01001001)_2$$. And the representation becomes $$(4B28)_{16}$$.

So while normalizing, does the processor takes into account the mantissa bits beyond 8 bits too? Or just rounds it off?

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