Question

Exercise 1.25 Convert the following decimal numbers to unsigned binary numbers Exercise 1.31 Repeat Exercise 1.29, but convert to 8-bit sign/magnitude numbers KExercise 1.32 Repeat Exercise 1.30, but convert to 8-bit sign/magnitude numbers (a) 4210 (b) 6310 Exercise 1.33 Convert the following 4-bit two's complement numbers to 8-bit two's complement numbers. (c) 22910 (d) 84510 (a) 0101 b) 1010 XExercise 1.26 Convert the following decimal numbers to unsigned binary numbers. Exercise 1.34 Convert the following 4-bit two's complement numbers to 8-bit cwo's complement numbers. (a) 140 (a) 0111 (b) 5210 (c) 33910 (b) 1001 (d) 71110 Exercise 1.35 Repeat Excrcise 1.33 if the numbers are unsigned rather than two's complement Exercise 1.27 Repeat Exercise 1.25, but convert to hexadecimal. Exercise 1.36 Repeat Exercise 1.34 if the numbers are unsigned rather than two's complement. Exercise 1.28 Repeat Exercise 1.26, but convert to hexadecimal. Exercise 1.29 Convert the following decimal numbers to 8-bit two's complement numbers or indicate that the decimal number would overflow the range. Exercise 1.37 Base 8 is referred to as octal. Convert cach of the numbers from Exercise 1.25 to octal (a) 4210 Exercise 1.38 Base 8 is referred to as octal. Convert each of the numbers from (b) -6310 Exercise 1.26 to octal (c 12410 Exercise 1.39 Convert each of the following octal numbers to binary, hexadecimal, and decimal (d)-12810 (e) 13310 (a) 42s Exercise 1.30 Convert the following decimal numbers to 8-bit two's complement numbers or indicate that the decimal number would overflow the range (b) 63 (c) 255 (a) 2410 (d) 3047s (b) -5910 Exercise 1.40 Convert each of the following octal numbers to binary, hexadecimal, and decimal. (c) 12810 (d) -15010 (a) 23g (e) 12710 (b) 45

Answer 1

1.25

To convert a decimal to binary number, we follow these steps:

1. Divide the number by 2.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the binary digit.
4. Repeat the steps until the quotient is equal to 0.

Here, we have 42=

2	42	0<---Remainder
2	21	1
2	10	0
2	5	1
2	2	0
2	1	1
	0

a)Therefore, (42)_{10 =} 101010

Similarly, the other conversions would yield:

b)(63)₁₀ = 111111

c) (229)₁₀ = 11100101

d) (845)₁₀ = 1101001101

1.26

a)(14)₁₀ = 1110

b)(52)₁₀ = 110100

c) (339)₁₀ = 101010011

d) (711)₁₀ = 1011000111

1.27 To convert a decimal to hexadecimal, we follow:

1. Divide the number by 16.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the hex digit.
4. Repeat the steps until the quotient is equal to 0.

16	42	10<---Remainder == A
16	2	2
	0

Hence (42)₁₀ = (2A)₁₆

Proceeding in a similar fashion, we have:

b)(63)₁₀ = (3F)₁₆

c) (229)₁₀ = (E5)₁₆

d)(845)₁₀ = (34D)₁₆

1.28

a) (14)₁₀ = (E)₁₆

b) (52)₁₀ = (34)₁₆

c) (339)₁₀ = (153)₁₆

d) (711)₁₀ = (2C7)₁₆

Answer 2

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Answer 3

Venugopal nagaar

Answer 4

Venugopal nagaar