# Convert decimal number in binary double precision, how to avoid the loss of the last digits after normalization?

I have the decimal number: $$0.023$$, and I want to convert in a binary number with $$52$$ bit of mantissa in Double Precision:

• if I go to convert, using this utility here, in non-normalized form, with $$52$$ bits I have:

$$0.0000010111100011010100111111011111001110110110010001$$

so when I go to normalize I obtain, the integer part not stored, and then 52 digits, I have done the zero-filled after the shifting to the left:

$$1.0111100011010100111111011111001110110110010001\underbrace{000000}_{\text{zero fill}} \times 2^{-6}$$

and that's the result expected to me! This is my proceeding!

• Instead, if I go to convert, using this utility here, directly in a normalized form, I have:

$$1.0111100011010100111111011111001110110110010001\underbrace{01101}_{\star}$$

• How is it possible to have that, not all zeros 5 digits in $$\star$$?
• Is there exists an additional register to memorize that five digits and then retrieve them after the shifting of the normalization, and therefore append them?

Please, can you help me? Many thanks!