Chapter 2, Problem 7E

Explanation of Solution

Decimal values of all “4” bit 2’s complement number:

To create a table that shows the decimal values of all four-bit 2’s complement numbers, note the below points,

• Four numbers of bits are used for representation.
• The Left most Significant Bit (LSB) is used to denote the sign of the decimal values.
• The sign of the decimal value is positive, if the LSB of four-bit 2’s complement number is zero.
• Convert the binary number into decimal number, if the sign of the decimal value is positive.
• The sign of the decimal value is negative, if the LSB of four-bit 2’s complement number is one.
• If “1” is appeared in the LSB of four-bit 2’s complement number, convert the 2’s complement binary number into one’s complement by transforming the bit from “0” to “1” and from “1” to “0”. Sum “1” to the result and compute the corresponding decimal value.
• Merge the sign and absolute value to compute the corresponding decimal value of the four-bit 2’s complement number.
• The conversion of four-bit 2’s complement numbers into decimal values is depicted in the below table,

Four-bit 2’s complement number	LSB	Computation	Decimal value
0000	0 (positive)	$2^3\times 0+2^2\times 0+2^1\times 0+2^0\times 0=0$	0
0001	0 (positive)	$2^3\times 0+2^2\times 0+2^1\times 0+2^0\times 1=1$	+1
0010	0 (positive)	$2^3\times 0+2^2\times 0+2^1\times 1+2^0\times 0=2$	+2
0011	0 (positive)	$2^3\times 0+2^2\times 0+2^1\times 1+2^0\times 1=3$	+3
0100	0 (positive)	$2^3\times 0+2^2\times 1+2^1\times 0+2^0\times 0=4$	+4
0101	0 (positive)	$2^3\times 0+2^2\times 1+2^1\times 0+2^0\times 1=5$	+5
0110	0 (positive)	$2^3\times 0+2^2\times 1+2^1\times 1+2^0\times 0=6$	+6
0111	0 (positive)	$2^3\times 0+2^2\times 1+{\mathrm{2<}}^{}$