What is the IEEE standard 32 bit floating point representation of the binary number -31.75?

Answer 1

answer is

0 10000011 00111000000000000000000

to get it first we have to convert 19.5 to binary

19 ->10011 ( 1* 2^4+1*2^3+1*2^2+1*2^1+1*2^0)

0.5-> 0.1 (2^-1=0.5)

10011+0.1=10011.1

The next step is to normalize this number so that only one non zero decimal place is in the number. To do this you must shift the decimal place4 positions to the left. The number4 becomes important so we note it. This process leaves us with the number 1.00111 which is the fraction that is represented in the last 23 bit places in the 32 bit binary. This is then padded with 0's to fill in the full 23 bits - leaving us with 00111000000000000000000.

so now we have first digit as 0 because this z a positive no and the last 23 digits.

We must now derive the middle 8 bits. To do this we take our exponent (4) and add 127 (the maximum number you can express with 8 bits (2^8-1 or the numbers 0 to 127)) which gives us 131. We then express this number as an 8 bit binary. This is 10000011 (or 1*2^7 + 1*2^1+ 1*2^0 or 128+2+1). Now we have the middle bits.

Taken as a whole our bit sequence will be:

0 10000011 00111000000000000000000

Answer 2

Firstly, IEEE is not a standard, it is an organisation (the Institute of Electrical and Electronics Engineers). TheIEEE Standards Organisation is responsible for the standardisation activities of the IEEE. As such, there are many IEEE standards.

There are two official IEEE standards covering 32-bit binary values:

• IEEE 754-1985 (single)
  
• IEEE 754-2008 (binary32)
  

IEEE 754-2008 single-precision binary floating-point format: binary32

• The high-order bit always denotes the sign (0 for positive, 1 for negative).
  
• The next 8 bits denote the exponent. This can either be notated in twos-complement (-128 to +127) or 127-biased (0 to 255). IEEE 754-2008 (binary32) uses the 127-biased form.
  
• The low-order 23 bits denote the normalised mantissa. There's actually 24 bits in the mantissa but the high-order bit is always 1 and can therefore be implied rather than stored.
  

Converting a real decimal number to a real binary number

Converting from decimal to binary32 requires several conversions. First we have to convert the decimal fraction to a binary fraction. Binary fractions are really no different to decimal fractions except we deal in powers of 2 rather than 10. So the decimal value -31.75 simply becomes -11111.11 in binary. Note that the bits to the right of the "binary point" are decreasing powers of 2 (0.5, 0.25, 0.125 and so on) while bits to the left are increasing powers of 2 (1, 2, 4, 8 and so on). That's the easy part dealt with.

Converting a binary number to a binary32 value

Now we have to convert this binary fraction to a binary32 value. We start by normalising the fraction so that it becomes a value within the half-closed range [1.0 : 2.0). Thus -11111.11 becomes -1.111111. The binary point has moved 4 positions to the left, so the exponent is 4. In 127-biased form, an exponent of 0 would be represented by the bias 127, thus an exponent of 4 is represented as 127 + 4 = 131, which is 10000011 in 8-bit binary. That's the exponent dealt with.

Next, we ignore the sign and subtract 1.0 from the normalised binary value (because it is implied), thus we are left with the binary value 111111 (we can also ignore the binary point at this stage).

Finally we combine the two components together with the sign:

Sign-bit: 1
Exponent: 10000011
Mantissa: 11111100000000000000000

Thus -31.75 decimal is 11000001111111100000000000000000 in IEEE 754-2008 (binary32).
Note that the mantissa is padded with zeroes to the right rather than to the left as we normally would. Remember that the high-order bit represents decimal 0.5, followed by 0.25, 0.125 and so on.