Chapter 2, Problem 10E

a.

Program Plan Intro

Steps to convert binary number into decimal value:

• If “0” is appeared in the Left most Significant Bit (LSB), it denotes the sign of the decimal value is positive. Binary number can be directly converted into decimal number.
• The sign of the decimal value is negative, if the LSB is one.
  • If “1” is appeared in the LSB, compute 2’s complement of the decimal value.
  • To find 2’s complement, transform the bits from “0” to “1” and from “1” to “0”, which gives 1’s complement and then add “1” to the result.
• Decimal value can be obtained by adding the powers of “2”, which having the coefficients “1”.
• The user includes a minus sign in front of the decimal value, if the original number is negative.

b.

Explanation of Solution

Conversion from the binary number “01011010” to decimal value:

• The LSB of the binary number is “0” and it represents the sign of the decimal value is positive.
• Hence, the user can directly compute the magnitude value as follows,

$\begin{array}{c}{2}^7\times 0+2^6\times 1+2^5\times 0+2^4\times 1\end{array}$

c.

Explanation of Solution

Conversion from the binary number “11111110” to decimal value:

• The LSB of the binary number is “1” and it represents the sign of the decimal value is negative.
• The user should find the 2’s complement representation as follows,
  • Find the 1’s complement by transforming the bits from “0” to “1” and from “1” to “0”.
  • Hence, the 1’s complement of the number is ‘00000001”.
  • Add “1” to the above result. “00000001+1”, which gives “00000010”.
• The 2’s complement representation of the positive number is “00000010”...

d.

Explanation of Solution

Conversion from the binary number “0011100111010011” to decimal value:

• The LSB of the binary number is “0” and it represents the sign of the decimal value is positive.
• Hence, the user can directly compute the magnitude value as follows,

$\begin{array}{l}{2}^{15}\times 0+2^{14}\times 0+2^{13}\times 1+2^{12}\times 1+2^{11}\times 1+2^{10}\times 0+2^9\times 0+2^8\times 1+\\ 2^7\times 1+2^6\times 1+2^{}\end{array}$