Number system

Data Representation

and

Number System

0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 1 0 1

Data Representation

• Data:

Data are numbers and other binary-coded information that are

operated on to achieve required computational results.

• Control Information

Control information is a bit or a group of bits used to specify the

sequence of command signals.

Data Representation

• Bit:

Binary Digit. 0/1• A group of bits in a computer are used to represent many

different things.

It can represent a number.

It can represent a character.

It can represent an instruction.

• Byte:

A group of 8 bits is a byte.

• Nibble

A group of 4 bits is a nibble.

Number System: Radix / Base

• Radix / Base:

A number system of base, or radix, r is a system that uses distinct

symbols for r digits.

In this system the number of states each digit has is determined by the

base or radix.

• Based on the radix there are four number systems

o Decimal

o Binary

o Octal

o Hexadecimal

• Number System is a code representing quantity.

Decimal system

Radix / base = 10

• For example :

253 means

2*102 + 5*101 + 3*100

0

1

2

3

4

5

6

7

8

9

• 1,10,100 (from R to L) are the “weights”

• 10 digits, values 0 through 9

• After 9 comes 10 (double digits)

Binary system

• For example :

1011 means 1*23 + 0*22 + 1*21 + 1*200

1

Radix / base = 2

• 2 digits, values 0 and 1

• 1, 2, 4, 8 are the weights

• After 1 comes 10

• Count 0 1 10 11 100 101 110 111 1000

Binary system

• Used to do calculations in all computers

• Used to store values in memory and on disk

• Not practical for people

• Input-Output done in decimal for user

• Software translates in both directions

Octal system

• For example :

253 means : 2*82 + 5*81 + 3*80

Radix / base = 8

0

1

2

3

4

5

6

7

• 1, 8, 64 are the weights

• 8 digits, values 0 through 7

• After 7 comes 10

• Used to display memory addresses in some

older computers

Hexadecimal system

• For example :

3B6 means 3*162 + B*161 + 6*1600

1

2

3

4

5

6

7

8

9

A

B

C

D

E

F

Radix / base = 16

• 1, 16, 256 are the weights

• 16 digits, values 0 - 9 and A-F

• After F comes 10

• Used to display memory addresses in most

modern computers e.g., 3C0F 95EA

Categorizing the Conversion Rule

• Converting from one number system to the other system

can be categorized as

• Any radix to Decimal system

• Decimal system to any radix

• Octal to binary and hexadecimal to binary

• Any radix to Any radix (other than binary)

For example: (00101010)2

Binary to Decimal Conversion

Approach is from LSB to MSB

For Integers

0 0 1 0 1 0 1 0

27 26 25 24 23 22 21 20

Equals:

27 * 0 + 26 * 0 + 25 * 1 + 24 * 0 + 23 * 1 + 22 * 0 + 21 * 1 + 20 * 0

= 32 + 8 + 2 = 42

(00101010)2 = 4210

Exercise

• 1010002 = ( ??? )10

• 10010112 =

• 1000112 =

• 0110112 =

01001011

Least Significant BitMost Significant Bit

4010

7510

From Decimal to Binary

• For example: 5610 = ( ???? )2

2 ) 56 0

2 ) 28 0

2 ) 14 0

2 ) 7 1

2 ) 3 1

2 ) 1 1

0

For Integers

• Divide by 2 until you reach zero, and then collect the

remainders in reverse.

Least significant bit

(0111000)2

Exercise

• (48)10 = ( ????? )2

(00110000)2

Binary to Decimal

(10110.10101)2

= (1 * 16) + (0 * 8) + (1*4) + (1*2) + (0*1) + ( 1 / 2) + (0 / 4)

+ ( 1 / 8) + ( 0 / 16) + (1 / 32)

= (22.65675) 10

Binary point

With Fraction

1 0 1 1 0 . 1 0 1 0 1

24 23 22 21 20 . 2-1 2-2 2-3 2-4 2-5

From Decimal to Binary

For example:

(56.6875)10 = ( ???? )2

• Convert the integer and fraction part separately

(56) 10 = (111000)2

With Fraction

• For fraction part, multiply the fraction part by 2, and each time

discard the integer so obtained.

• Collect this discarded integer part as the binary equivalent.

• Repeat this process until zero or until the required accuracy.

0.6875 0.5000

x 2 x 2

1.3750 1.0000

0.3750

x 2

0.7500

0.7500

x 2

1.5000

(56.6875)10 = (111000.1011)2

.1011

Exercise

• (48.3125)10 = ( ????? )2

(00110000.0101)2

From Decimal to Radix r

• Separate the integer part and fraction part

• Convert the integer part and then fraction part separately.

• Rule For converting the integer part:

Conversion of a decimal to a base r is done by successive

divisions by r and accumulating the remainders.

This is repeated until the quotient becomes zero.

Collect remainders in the reverse order.

• Rule For converting the decimal part:

Conversion of a decimal to a base r is done by successive

multiplications by r and accumulating the remainders.

• This process is repeated until the fraction parts becomes

zero or number of digits gives the required accuracy

• Take the integer outputs in the forward direction

From Radix r to Decimal

• Beginning with the rightmost digit multiply each nth digit

by r(n-1), and add all of the results together (considering the

position just before the decimal point as the first position.

N = AnRn + An-1Rn-1 + …….A2R2+ A1R1 +A0R0. A1R-1+

A2R-2 +…….

• N - Number

• An - Digit in that position (nth Position)

• R - Radix or base of the system

• - Radix Point

Decimal to Octal Conversion

• Convert the fraction and integer part separately.

• For Integer part:

o The Division Method: Divide by 8 until you reach

zero, and then collect the remainders in reverse.

8 ) 478 6

8 ) 59 3 8 ) 7 7

0(736)8

• For example: (478.5)10 = ( ?? )8

• For Fraction part:

o The Multiplication Method: Multiply the fraction part

successively by 8 and accumulate the remainders until

you reach zero.

0.5

x 8 (736.4)8

4.0

Octal to Decimal Conversion

For example: (736.4)8

8-1.808182

4.637

Equals: 7* 82 + 3 * 81 + 6 * 80 + 4 * 8-1

= 448+24+6+0.5

= (478.5)10

• To convert to base 10, beginning with the rightmost digit

multiply each nth digit by 8(n-1), and add all of the results

together.

Exercise

• (0.40)8 = ______10 0.50

HexaDecimal to Decimal Conversion

• To convert to base 10, beginning with the rightmost digit

multiply each nth digit by 16(n-1), and add all of the results

together.

For example: 1F416

1 F 4

162 161 160

Equals: 1 * 162 + F * 161 + 4 * 160

= 256 + 15*16 + 4

=(500)10

Decimal to Hexa Conversion

• The Division Method. Divide by 16 until you reach zero,

and then collect the remainders in reverse.

For example: 12610 = 7E16

16) 126 14 = E

16) 7 7

0

A 10

B 11

C 12

D 13

E 14

F 15

Exercise

• (AF)16 = ______10

175

Binary to Octal

For example: (1 010 101 111)2

Grouping : 1 010 101 111

1 2 5 7

(1010101111)2 = (1257)8

• Group the binary number into groups of 3 bits starting from

the least significant bit, and convert it into its decimal equivalent.

Octal to Binary

• Take each digit one by one from the string of digits and

convert each digit into its respective binary number, as a

group of three bits.

(257)8 = ____2

7 is converted as 111



(257)8 = (010101 111)2 Binary Triplet Method

Binary -Coded Octal Numbers

Three-bit Group Decimal Digit Octal Digit

000 0 0

001 1 1

010 2 2

011 3 3

100 4 4

101 5 5

110 6 6

111 7 7

001 000 10

010 100 24

Binary to Hexadecimal

For example: (1010 1111 0110 0011)2

Grouping : 1010 1111 0110 0011

A F 6 3

(1010111101100011)2 = (AF63)16

• Group the binary number into groups of 4 bits starting

from the least significant bit, and convert it into its decimal

equivalent.

A 10

B 11

C 12

D 13

E 14

F 15

Hexadecimal to Binary

• Take each digit one by one from the string of digits and

convert each digit into its respective binary number, as a

group of four bits.

(257)16 = ____2




(257)16 = (00100101 0111)2

Hexadecimal to Binary

(BA7)16 = ____2


A is converted as 1010

B is converted as 1011

(BA7)16 = (10111010 0111)2

A 10

B 11

C 12

D 13

E 14

F 15

Binary-Coded Hexadecimal Numbers

Four-bit Group Decimal Digit Hexadecimal Digit

0000 0 0

0001 1 1

0010 2 2

0011 3 3

0100 4 4

0101 5 5

0110 6 6

0111 7 7

1000 8 8

1001 9 9

Four-bit Group Decimal Digit Hexadecimal Digit

1010 10 A

1011 11 B

1100 12 C

1101 13 D

1110 14 E

1111 15 F

Binary-Coded Hexadecimal Numbers

0001 0100 14

0011 0010 50

Binary to octal and hexadecimal

• 1010111101100011

• 1010111101100011

• 1010111101100011

1275438

AF6316

Binary

Octal

Hexa

EXERCISE

• Note:

• The highest digit in octal system is 7 whose binary

equivalent is 111.

• The highest digit in hexadecimal system in F, whose

binary equivalent is 1111.

Complements

There are two types of complements for each base r system:

• r’s complement

• ( r-1)’s complement

(r-1)’s complement

• Given a number N in base r having n digit, the (r-1)’s

complement of N is (rn –1) –N.

• For decimal numbers, there exist 9’s complement.

• For binary numbers, there exist 1’s complement.

9’s Complement

• For example:

For decimal number N= 546700, n= 6 and r =10

9’s complement equals:

= (rn –1) – N

= (106 –1) - 546700

= (1000000 –1) - 546700

= 999999 – 546700 = 453299

• That is, 9’s complement of a number would be same as

subtracting each digit from 9.

1’s Complement

• For example:

For binary number N= 1011, n= 4 and r =2

1’s complement equals:

= (rn –1) – N

= (24 –1) - 1011

= (10000 –1) – 1011 (24 in binary)

= 1111 – 1011= 0100

• That is, 1’s complement of a number would be same as

subtracting each digit from 1.

1’s Complement

• For a binary number 1011001, 1’s complement can be

obtained by

If you look at the result, you can see, the 1’s

complement of a binary number can be obtained by

reversing the bits.

1111111

1011001

_______

0100110

_______

r’s complement

• Given a number N in base r having n digit, the r’s

complement of N is rn –N for N < > 0 and 0 for N=0.

• Also, r’s complement is equal to:

= rn –N

= rn –N – 1 + 1 (Add and subtract 1)

= [(rn –1) –N] +1 (Rearranging the terms)

= (r-1)’s complement + 1

• For decimal numbers, there exist 10’s complement.

• For binary numbers, there exist 2’s complement.

10’s Complement

• For decimal numbers, 10’s complement of a number is

equal to its 9’s complement +1.

• For example:

10’s complement of 546700 =

= 9’s complement of 546700 + 1

= 453299 + 1

= 453300

2’s Complement

• Given a number in binary say N, having ‘n’ digits, then

2’s complement of N is defined as (2n-N), if N < > 0

else 0, when N=0

• For binary numbers, 2’s complement of a number is equal to its 1’s complement

+1.

• For example:

2’s complement of 1011 =

= 1’s complement of 1011 + 1

= 0100 +1 = 0101

Exercise

• Find the 2’s complement of 10101011

01010101

• Find the 2’ complement of 01010101

10101011

Integer Representations

• Two different representations exists for integers

• The signed representation: in that case the most

significant bit (MSB) represents the sign

o Positive number (or zero) if MSB = 0

o Negative number if MSB = 1

• The unsigned representation: in that case all the bits are

used to represent a magnitude

o It is thus always a positive number or zero

Signed and Unsigned Interpretation

• To obtain the value of a integer in memory we need to

chose an interpretation

• For example: a byte of memory containing 1111 1111

can represent either one of these numbers:

o -1 if a signed interpretation (2’s complement) is used

o 255 if an unsigned interpretation is used

Subtraction of Unsigned Numbers

• The subtraction of two n-digit unsigned numbers M – N (N < > 0) in

base r can be done as follows:

1. Add the minuend M to the r’s complement of the subtrahend N. This

performs M + (rn – N) = M – N + rn.

Case 1 : If M >= N, the sum will produce an end carry rn which is discarded,

and what is left is the result M – N.

Case 2 : If M < N, the sum does not produce an end carry and is equal to

rn – (N – M), which is the r’s complement of (N – M). To obtain the answer

in a familiar form, take the r’s complement of the sum and place a negative

sign in front.

This will equate to : rn – (rn – ( N – M)) = M - N


Case 1: Minuend > Subtrahend

• Take the r’s complement of the subtrahend.

• Add this to the minuend.

• Discard the end carry.

3456 - 2234

radix 10

3456

10’s complement of 2234 = 7766

_______

11222

1222

Discard the end carry 10000


Case 2: Minuend < Subtrahend

• Take the r’s complement of the subtrahend.


• Find the r’s complement of the result and append a negative sign in front of it.

2234 - 3456

radix 10

2234

10’s complement of 3456 = 6544

_______

8778

10’s complement of 8778-1222


• In case 2, after the 10’s complement of 8778, we get 1222 only and not -1222.

• When working manually it can be noticed that the subtrahend

is > minuend and so it needs a -ve sign for the result.

• When subtracting with complements it is found that the

answer where there is no end carry and a negative sign

should be added.


• In a similar manner, the subtraction with complements is done with

binary numbers.

• For example:

X: 1010100

Y: 1000011

• To perform X – Y :

X = 1010100

2’s complement of Y = 0111101

Sum = 10010001

Discard the end carry 10000000

0010001

Exercise

Y: 1000011

X: 1010100

radix 2

Perform Y – X = ????

- 0010001

1’s Complement Subtraction

Unsigned representation

Case 1: Minuend > Subtrahend (M – N)

• Take the 1’s complement of the subtrahend.


• Remove the carry and add it to the result. This is called END AROUND

CARRY.

00011101- 00011011

radix 2

00011101

1’s complement of 00011011= 11100100

_________

100000001

1

RESULT 00000010

1’s Complement Subtraction

Unsigned Representation

Case 2: Minuend < Subtrahend

• Take the 1’s complement of the subtrahend.


• Find the 1’s complement of the result and append a negative sign in front of it.

00011001 - 00011101

radix 2

RESULT -00000100

00011001

1’s complement of 00011101 = 11100010 11111011

Exercise (using 1’s complement)

X: 00110011

Y: 00101101 Perform X - Y

radix 2

00000110

Signed Representation

• In signed representation, the most significant bit (MSB) represents the

sign.

• When a binary number is positive, the sign is represented by 0 and the

magnitude by a positive binary number.

• When the number is negative, the sign is represented by 1 but the rest of

the number may be represented in three possible ways.

1. Signed magnitude representation

2. Signed - 1’s complement representation

3. Signed - 2’s complement representation.

Example for Negative number

Representation• To represent -14

1. Signed magnitude representation

1 0001110

Note : This representation of – 14 is obtained from +14 by

complementing only the sign bit.

2. Signed - 1’s complement representation

1 1110001

Note : This representation of – 14 is obtained by complementing all

the bits of + 14, including the sign bit.

Example for Negative number

Representation

3. Signed - 2’s complement representation.

1 1110010

Note : This representation of – 14 is obtained by taking the 2’s

complement of +14, including the sign bit.

Advantage of 2’s Complement

System

• Representing in 2’s complement is preferred over 1’s

complement as well as signed magnitude system.

• Representing in signed magnitude is easy for manual

arithmetic processing and not for the computer.

• The reason is 1’s complement takes two representation for

+0 and -0 which is absurd.

• In 2’s complement system both -0 and +0 will have the same

representation

NOTE

1’s complement form

• + 0 in binary 00000000

• - 0 in 1’s complement form 11111111

Two representations of –0 and +0, which is absurd.

2’s complement form

• + 0 in binary 00000000

• - 0 in 2’s complement form 00000000

Same representation of +0 and –0.

END

Number system

Education