Please show work 1) (-1) s x (1+0.FFFF) x 2 (e-127) In analyzing an IEEE 754 flo
ID: 3725058 • Letter: P
Question
Please show work
1) (-1)s x (1+0.FFFF) x 2(e-127)
In analyzing an IEEE 754 floating point value, if the: sign bit = 0, biased exponent is 131 and fraction is 0000 0000 0000 0000 000
What is the real number value stored(show your work) in memory.
2) In the IEEE 754 Floating Point number representation, what does the bias value do to the True Exponent value?
3) Convert IEEE 754 Floating Point Format memory dump value to Float Point decimalnumber
1 10000111 0011 0000000000000000000 .
4) A number in IEEE floating-point binary notation that has no leading 0s is called a normalized scientific binary floating point number. (T/F)
5) Computer arithmetic that represents numbers in which the binary point is not fixed is called Floating Point. (T/F)
Explanation / Answer
Solution:
This is how IEEE 754 single floating point representation in memory looks like
The format for IEEE 754 single precision is
sign
Exponent
Mantissa
1 bit
8 bits
23 bits
The value of bias = 127,
so actual exponent = 131-127 = 4 = 100
stated problem representation will look like
The format for IEEE 754 single precision is
sign
Exponent
Mantissa
0
10000011
00000000000000000000000
So the number in the normalized format
0 means a number is a positive number
1.00000000000000000000000 * 2^100
= 10000.000000000000000000
The value in decimal is 16
2)
Biased exponent = bias + actual expoenent
3)
1 10000111 0011 0000000000000000000. in decimal
The biased exponent value in decimal is= 135
actual exponent = 135-127 = 7
sign bit 1 means the value is a negative value
in the normalized format the number is
1.0011*2^7
10011000 = 152
4)
False,
The number has to be of 1.bbbbbbb... format
5)
False.
binary point is fixed, that is the whole point of normalized format
I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)
sign
Exponent
Mantissa
1 bit
8 bits
23 bits