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December 11th, 2007, 12:26 PM  #1 
Newbie Joined: Dec 2007 Posts: 5 Thanks: 0  Floating Point Numbers  Single Precision Format (help!)
How would a computer store the following floating point number in Single Precision format: 3.08 x 10^4 I just don't know where to start, help appreciated. Im stuck on this as i dont know how to start it off as there are two numbers. Do i just multiply those two numbers together, which will be "320.32" and then just work it out in the same process as below? I can work out the following though, as there is only one thing to work out: a) 214.75 = 11010110.11 x = 1.101011011 x 27 Mantissa = 1101 0110 1100 0000 0000 0000 (24 bits) EB = 7 E=B+7 =127+7 = 134 E= 134 = 1000 0110 S = 1 E = 1000 0110 M= 1101 0110 1100 0000 0000 0000 1100 0011 0101 0110 1100 0000 0000 0000 C 3 5 6 C 0 0 0 00 C0 56 C3 
December 11th, 2007, 08:30 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,819 Thanks: 2158 
Are you sure it was supposed to be 104, not 10^4?

December 12th, 2007, 12:21 PM  #3  
Newbie Joined: Dec 2007 Posts: 5 Thanks: 0  Quote:
It's supposed to be 10^4.  
December 12th, 2007, 09:36 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,819 Thanks: 2158 
3.08 × 10^4 = 30800, which is 111100001010000 as a binary integer. Can you finish from there? (Note: Single Precision probably means "single precision floating point", for which the mantissa's magnitude typically lies between 1/2 and 1 by suitable choice of exponent.) 

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