Answer:
A) Describe the number 7.0 bit
The exponential value ( E ) = 2
while the significand value ( M ) = 1.112 ≈ 7/4
fractional value ( F ) = 0.112
And, numeric value of the quantity ( V ) = 7
The exponent bits will be represented as : 100----01.
while The fraction bits will be represented as : 1100---0.
<u>B) The largest odd integer that can be represented exactly </u>
The integer will have its exponential value ( E ) = n
hence the significand value ( M )
= 1.11------12 = 2 - 2-n
also the fractional value ( F ) =
0.11------12 = 1 – 2-n
Also, Value, V = 2n+1 – 1
The exponent bits will be represented as follows: n + 2k-1 – 1.
while The bit representation for the fraction will be as follows: 11---11.
<u>C) The reciprocal of the smallest positive normalized value </u>
The numerical value of the equity ( V ) = 22k-1-2
The exponential value ( E ) = 2k-1 – 2
While the significand value ( M ) = 1
also the fractional value ( F ) = 0
Hence The bit representation of the exponent will be represented as : 11---------101.
while The bit representation of the fraction will be represented as : 00-----00.
Explanation:
E = integer value of exponent
M = significand value
F = fractional value
V = numeric value of quantity
A) Describe the number 7.0 bit
The exponential value ( E ) = 2
while the significand value ( M ) = 1.112 ≈ 7/4
fractional value ( F ) = 0.112
And, numeric value of the quantity ( V ) = 7
The exponent bits will be represented as : 100----01.
while The fraction bits will be represented as : 1100---0.
<u>B) The largest odd integer that can be represented exactly </u>
The integer will have its exponential value ( E ) = n
hence the significand value ( M )
= 1.11------12 = 2 - 2-n
also the fractional value ( F ) =
0.11------12 = 1 – 2-n
Also, Value, V = 2n+1 – 1
The exponent bits will be represented as follows: n + 2k-1 – 1.
while The bit representation for the fraction will be as follows: 11---11.
<u>C) The reciprocal of the smallest positive normalized value </u>
The numerical value of the equity ( V ) = 22k-1-2
The exponential value ( E ) = 2k-1 – 2
While the significand value ( M ) = 1
also the fractional value ( F ) = 0
Hence The bit representation of the exponent will be represented as : 11---------101.
while The bit representation of the fraction will be represented as : 00-----00.
