Binary Homework help (Bits)

Problem 1:

a)      7 bits

Total number of values that are to be represented is 128 (0 through 127). Total number of bits used are log₂128= 7

b)     9 bits

Total number of values that are to be represented is 257 (0 through 256). Total number of bits used are log₂257= 9 (rounded off to the next integer)

c)      6 bits

Total number of values that are to be represented is 64 (-32 through 31). Total number of bits used are log₂64= 6

d)     2 bits

Total number of values that are to be represented is 3 (i.e., total number of decks). Total number of bits used are log₂3= 2(rounded off to next integer)

e)      4 bits

Total number of values that are to be represented is 0 (1 through 9). Total number of bits used are log₂9= 4(rounded off to next integer)

f)       5 bits

Total number of values that are to be represented is 27 (deck1 card 1 through deck3 card 9). Total number of bits used are log₂27= 5 (rounded off to next integer)

Problem 2:

a)      The total number of distinct values that can be encoded in a 32 bit word if all the values are to be positive integers is 2^32 (from 0 through 4,294,967,295)

b)     The total number of distinct values that can be encoded will not be affected if negative numbers are allowed. The total number remains to 2^64.

c)      Binary representation of 12942

The binary representation is 11 0010 0010 1010

The total number of bits is 14. To use 16-bit 2’s complement add 2 0’s before this number

Hence the number becomes

0011 0010 0010 1010

Its  1’s complement is 1100 1101 1101 0101

Its 2’s complement is 1100 1101 1101 0110

d)     Binary representation of 1

The binary representation of 1 is 1

The total number of bits is 14. Now to make it 16-bit add 15 zero’s ; it becomes

0000 0000 0000 0001

Its 1’s complement is 1111 1111 1111 1110

Its 2’s complement is 1111 1111 1111 1111

e)      Binary representation of most negative number is

1000 0000 0000 0000

f)       Its decimal equivalent is -32768

g)      The negation of -32768 is its 1’s complement that is

0111 1111 1111 1111. Its decimal equivalent will be 32767

Problem 3:

Hexadecimal representation of

a)      6834ten

Divide 6834 by 16

6834/16 => quotient=427; remainder=2

427/16 => quotient=26; remainder=B

26/16 => quotient=1; remainder=A

1/16 => quotient=0; remainder=1 (quotient=0 stop here)

Hexa-decimal representation= 1AB2 hex

b)     -2048ten

By above method 2048 in hexadecimal form= 800

Now 2048 in binary can be written is 0000 1000 0000 0000

Its 1’s complement is 1111 0111 1111 1111

Its 2’s complement is 1111 1000 0000 0000

The hexadecimal equivalent of this number is F800

c)      1001 0110 1010 1101two

Consider 4 bits at a time and write its equivalent hexadecimal form

96AD

d)     1101 1000 0110 1010 10two

Group all the binary numbers 4 at a time that is

11 0110 0001 1010 1010

Add two 0’s at most significant place

0011 0110 0001 1010 1010                

Its binary equivalent is 361AA

e)      -3ten

Represent 3 in binary form

That is 0000 0011

Now its 1’s complement is 1111 1100

Its 2’s complement is 1111 1101

The hexadecimal equivalent of -3 is FD

Problem 4:

a)      24 ten+31 ten

Operand1= 24ten

Operand2=31ten

Result in decimal form= (24+31)ten= 55ten

Operand1 in binary form = 0001 1000

Operand2 in binary form= 0001 1111

Now adding both the binary numbers we get

0001 1000

0001 1111

0011 0111

So result in binary= 0011 0111

Check it now

0011 0111= (0*2^7)+(0*2^6)+(1*2^5)+(1*2^4)+(0*2^3)+(1*2^2)+(1*2^1)+(1*2^0)

=0+0+32+16+0+4+2+1

=55

b)     75 ten-23 ten

Operand1= 75

Operand2= -23

Result in decimal form=75-23=52

Operand1 in binary form= 0100 1011

Now consider positive part of operand2 i.e., 23

23 in binary form is 0001 0111

To represent -23 we need to consider 2’s complement form

1’s complement= 1110 1000

2’s complement= 1110 1001

Now adding both 75 and -23 we get

0100 1011

1110 1001

(1)   0011 0100

1 is carry. Ignore it.

Result in binary= 0011 0100

c)      13ten-68ten

Operand1=13

Operand2=-68

Result= 13-68=-55

13 in binary= 0000 1101

68 in binary= 0100 0100

-68 in binary can be 2’s complement of 68 =1011 1100

Now add both binary forms;

0000 1101

1011 1100

1100 1001

Result in binary form= 1100 1001

d)     115ten+65ten

115+65= 180

115 in binary= 0111 0011

65 in binary= 0100 0001

Now add both binary forms;

0111 0011

0100 0001

1011 0100

Result in binary form= 1011 0100

e)      In above solution if we observe there is 1 at MSB. But in this case 1 is not a sign bit but it is a carry

Problem 5:

a)    82.75

Consider integral part that is 82

82/2 => quotient=41; remainder=0

41/2 => quotient=20; remainder=1

20/2 => quotient=10; remainder=0

10/2 => quotient=5; remainder=0

5/2 => quotient=2; remainder=1

2/2 => quotient=1; remainder=0  

1/2 => quotient=0; remainder=1 (stop here as quotient=0)

Now 82= 0101 0010

Consider fractional part that is 0.75

0.75*2=1.5 => whole number is 1

0.5*2=1.0 => whole number is 1

0.0*2=0.0 => stop here

Now .75D=1100 0000

Now 82.75D= 0101 0010.1100 0000

b)    -15.01325

Here perform the same steps as above and finally do a 2’s complement

The answer before two’s complement is

Binary representation of integral part 15= 0000 1111

Binary representation of fractional part .01325D = 0000 0011

Now 2’s complement form of 15= 1111 0001

2’s complement form of .01325D=1111 1100

Hence -15.01325 in binary= (1)1111 0001.1111 1101

c)    0110 1101.0111 0100

In this representation separate the integral and decimal parts.

Integral part= 0110 1101

Check for the MSB in the integral part. It is zero (0) here. So the number is positive.

Now the conversion is done by multiplying each binary number with powers of 2. That is,

0110 1101= (0*2^7)+(1*2^6)+(1*2^5)+(0*2^4)+(1*2^3)+(1*2^2)+(0*2^1)+(1*2^0)

=0+64+32+0+8+4+0+1

=109

Now the conversion of fractional part is done by multiplying each binary number with negative powers of 2. That is,

0111 0100=

(0*2^-1)+(1*2^-2)+(1*2^-3)+(1*2^-4)+(0*2^-5)+(1*2^-6)+(0*2^-7)+(0*2^-8)

=0+0.25+0.125+0.0625+0+0.015625+0+0

=0.453125

Hence 0110 1101.0111 0100 in decimal form is 109.453125

d)    1111 1110.0010 0000

Again separate this number into integral part and fractional part

            Integral part is:

                        1111 1110. If we observe, this number had MSB 1. So this number turns to be a negative number. To convert it to positive, we need to follow two’s complement method to convert this number to decimal.

1111 1110 as 1’s complement= 0000 0001

Its 2’s complement= 1’s complement+1 = 0000 0010

This number in decimal form = -2

Now consider the integral part that is

0010 0000= (0*2^-1)+(0*2^-2)+(1*2^-3)+(0*2^-4)+(0*2^-5)+(0*2^-6)+(0*2^-7)+(0*2^-8)

            =0+0+0.125+0+0+0+0+0=0.125

Now adding both integral and decimal parts we get -2+0.125 = -1.875

Hence 1111 1110.0010 0000 in decimal form is -1.875