# How to calculate a floating point of negative number?

In IEEE double precision n=53. So to represent 16 I can do the following:
The next biggest number from $16=+(.10 \dots 01)_22^5=2^{-1}2^5+2^{-53}2^5=16+2^{-48}$

Now the biggest number from $-16=-16+2^{-49}$, but how to show this formally like i did above for positive $16$ case?

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IEEE double precision numbers store they sign in a separate bit. So, (-16) is the same as (+16) except sign-bit changed. Getting "next" number after (-16) is equivalent to getting "previous" number before (+16) with sign-bit changed.
Next number after (+16) is
$\large+(1.00...01)_2\cdot2^4 = 16\cdot(1+2^{-52}) = 16+2^{-48}$
Previous number before (+16) is
$\large+(1.11...11)_2\cdot2^3 = 16\cdot(1-2^{-53}) = 16-2^{-49}$
Next number after (-16) is
$\large-(1.11...11)_2\cdot2^3 = -16\cdot(1-2^{-53}) = -16+2^{-49}$

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I kind of figured it out already, THANK YOU for the answer. The only thing, everything looks correct but $1$s should be after decimal point, I edited it already :) –  ASROMA Dec 3 '12 at 1:47
@ASROMA - No, my original message was correct. Please read Wikipedia and get the correct version back. :-) –  Egor Skriptunoff Dec 3 '12 at 3:11
We both were correct since i did not see that your power on 2 was one less than mine :-) –  ASROMA Dec 3 '12 at 3:29