My Math Forum Floating Point Numbers - Single Precision Format (help!)

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 December 11th, 2007, 01:26 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) E-B = 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, 09:30 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,868 Thanks: 1833 Are you sure it was supposed to be 104, not 10^4?
December 12th, 2007, 01:21 PM   #3
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Quote:
 Originally Posted by skipjack Are you sure it was supposed to be 104, not 10^4?
Yes, sorry, my mistake.

It's supposed to be 10^4.

 December 12th, 2007, 10:36 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,868 Thanks: 1833 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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### 214.75 into single precision format

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