Given a floating-point formal with a k-bit exponent and an n-bit (fraction, write formulas for the exponent E, significant M, the fraction f, and the value V for the quantities that follow. In addition, describe the bit representation. A. The number 7.0
B. The largest odd integer that can be represented exactly
C. The reciprocal of the smallest positive normalized value

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 :  10001.

while The fraction bits will be represented  as : 1100---0.

B) The largest odd integer that can be represented exactly

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.1112 = 2 - 2-n

also the fractional value ( F ) =  

0.1112 = 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.

C) The reciprocal of the smallest positive normalized value

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 : 11101.

while The bit representation of the fraction will be represented as : 0000.

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 :  10001.

while The fraction bits will be represented  as : 1100---0.

B) The largest odd integer that can be represented exactly

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.1112 = 2 - 2-n

also the fractional value ( F ) =  

0.1112 = 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.

C) The reciprocal of the smallest positive normalized value

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 : 11101.

while The bit representation of the fraction will be represented as : 0000.

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