Binary Addition: (1+(2^-51+2^-53)-1) by hand in IEEE double

[Guest]

(1+(2^-51+2^-53)-1) Do this by hand in IEEE double precision.

How do you go about adding this? I want to know the process not the answer.
Thanks

 

[arthur ohlsten]

1=1.00000000000000000000000000
2=10
2^-1=.1
2^-2=.01
2^-51=.0000000000000000000000000000000000000000000000000010000000000
2^-53=.0000000000000000000000000000000000000000000000000000100000000

answer=.00000000000000000000000000000000000000000000000000101

I think this is what you want
Arthur