Binary Addition (1 of 2)

1

CS1010 Introduction to Computing

Lecture 08

Review – multiplying powers

• For common bases, add powers.

26 × 210 = 216 = 65,536

or…

26 × 210 = 64 × 210 = 64k

Binary Addition (1 of 2)

• Two 1-bit values

A B A + B

0 0 0

0 1 1

1 0 1

1 1 10

BINARY ADDITION

• Two n-bit values

– Add individual bits

– Propagate carries

– Example:

10101 21 + 11001 + 25

101110 46

EXAMPLE

• Add 01011101

2 and 001100102

• Solution:

0 1 0 1 1 1 0 1

+ 0 0 1 1 0 0 1 0

1 1

1 1

1 0 0 0

0 0

0 0

1 1

EXAMPLE

• Add 111101 and 10111.

1 1 1 0 1 1 1 1 1 1 0 + 0 0 0

0 1 1

Multiplication (1 of 3)

• Decimal (just for fun)

Multiplication (2 of 3)

• Binary, two 1-bit values

A B A × B

0 0 0

0 1 0

1 0 0

Multiplication (3 of 3)

• Binary, two n-bit values

– As with decimal values

– E.g.,

1110 x 1011 1110 1110 0000 1110

MULTIPLICATION

• Decimal

35 x 105 175 000 35

EXAMPLE

• Multiply the Binary numbers 1111 and 111.

1111

x 111

1111 1111 1111

EXAMPLE

• Bit by bit.

0

1 1 1 1

0

1 1 0

0

0 0 0 0

0

1 1 1 1

0

1 1 1 1

0 _{0 0}

REPRESENTATION OF NEGATIVE

NUMBERS

• There are many methods of representing signed

number in binary:

– Sign-Magnitude Method

– 1’s Complement Method

– 2’s Complement Method

1’s COMPLEMENT METHOD

directly obtained by changing all 0’s to 1’s and all 1’s to 0’s.

• Example:

Take 1’s complement of the binary number 01100110 directly.

Representation of Negative Numbers

using 1’s Complement Method

• Algorithm

– First determine the number of bits to represent the number.

– Convert the modules of the given number in binary.

– _{Place a 0 in MSB and binary conversion of the} number in remaining bits.

Example

• Represent -54₁₀ in 1’s complement form using 8 bits.

2 54

27 0 2

13 1 2

6 1 2

3 02 1 1₂ 0 1

54

10 = 01101102

54

10 = 001101102 (in 8 bits)

So -54

10 in 1’s complement

form:

-54

2’s COMPLEMENT METHOD

– First taking 1’s complement.

Example

• Take 2’s complement of the binary number 01100110 directly.

Original number 01100110 1’s Complement 10011001

+1

Representation of Negative Numbers

using 2’s Complement Method

• Algorithm

– First determine the number of bits to represent the number.

– Convert the modules of the given number in binary.

– Place a 0 in MSB and binary conversion of the number in remaining bits.

Example

• Represent -54₁₀ in 2’s complement form using 8 bits. Modulus of -54

10 = 54

2 54

27 0 2

13 1 2

6 1 2

3 02 1 1₂

0 1

54

10 = 01101102

54

10 = 001101102 (in 8 bits)

Take 1’s complement of

00110110

= 11001001

+ 1

= 11001010

BINARY ADDITION

• Two 1-bit values

A B A + B

0 0 0

0 1 1

1 0 1

1 1 10

BINARY SUBTRACTION

• Modern day computers use 2’s complement

or 1’s complement method for performing

subtraction.

• Example:

Calculate 38 – 29 using 8-bit 1’s complement method.

• Write both the numbers in binary form using 8 bits.

38

10 = 001001102

29

10 = 000111012

Represent negative numbers in 1’s complement

-29

10 = 111000102

Add the number & its 1’s complement

00100110

2

+ 11100010

2

Carry: 1 00001000

2

Add End Carry:+ 1 00001001

EXAMPLE

• Calculate – 54 – 30 using 8-bit 1’s complement

method.

• Solution:

(– 54) + (– 30)

54

10 = 001101102

30

10 = 000111102

Represent negative numbers in 1’s complement

-54

10 = 110010012

-30

Add numbers in its 1’s complement form

11001001

2

+ 11100001

2

Carry: 1 10101010

2

Add End Carry:+ 1 10101011

2

Convert the 1’s complement result into decimal. As MSB is 1 so it’s a negative number

10101011

Exercises

• Perform following calculations using 1st complement method.

a) 0111000 + 1011100 b) 25 – 50

SUBTRACTING USING 2’s COMPLEMENT

METHOD

• Example:

Calculate 38 – 29 using 8-bit 2’s complement

method.

Solution:

38 + (-29)

38

10 = 001001102

29

Represent negative numbers in 2’s complement

Take 1’s complement of 29 = 00011101

and then add 1

-29

10 = 111000112 (2’s complement)

Add 2’s complement representation and ignore the end carry.

00100110

2

+ 11100011

2

Carry: 1 00001001

2

EXAMPLE

• Calculate – 54 – 30 using 8-bit 2’s complement

method.

• Solution:

(– 54) + (– 30)

54

10 = 001101102

30

10 = 000111102

Represent negative numbers in 2’s complement

-54

10 = 110010102

-30

Add 2’s complement representation

11001010

2

+ 11100010

2

Carry: 1 10101100

2

Convert the 2’s complement result into decimal. As MSB is 1 so it’s a negative number

10101100

2 = 01010011 = -84

Exercises

• Perform following calculations using 2nd complement method.

a) 0111000 – 1011100 b) 25 – 50

FRACTIONS

• Decimal to decimal.

3.14 => 4 x 10-2 = 0.04 1 x 10-1 = 0.1

3 x 100 = 3

FRACTIONS

• Binary to Decimal

10.1011 => 1 x 2-4 = 0.0625 1 x 2-3 = 0.125

0 x 2-2 = 0.0 1 x 2-1 = 0.5 0 x 20 = 0.0 1 x 21 = 2.0

DECIMAL TO BINARY

• 3.14579

3.14579

.14579 x 2 0.29158 x 2 0.58316 x 2 1.16632 x 2 0.33264 x 2 0.66528 x 2 1.33056 etc.

OCTAL TO DECIMAL

• 127.54₈

127.54

8 => 1 x 8

2

= 64.000 2 x 81 = 16.000

7 x 80 = 7.000 5 x 8-1 = 0.625 4 x 8-2 = 0.0625 87.6875

EXAMPLE

• Convert 630.4₈ into decimal equivalent.

630.4

8 = 408.510

630.4 => 6 x 82 = 384 3 x 81 = 24

0 x 80 = 0 4 x 8-1 = 0.5

DECIMAL TO OCTAL

• 185.3₁₀

8 185 23 1 8

2 7 8

0 2

185

10 = 02718

0.3

10 = 0.231468

Result Fractional Part Integral Part

8 x 0.3 8 x 0.4 8 x 0.2 8 x 0.6 8 x 0.8

2.4 3.2 1.6 4.8 6.4 4 2 6 8 4 2 3 1 4 6 185.3

HEXADECIMAL TO DECIMAL

• 2B.C4₁₆

2B.C4

16 => 2 x 16

1

= 32.000 B x 160 = 11.000

C x 16-1 = .075

4 x 16-2 = .015 43.090

10

Remember that

A = Ten

B = Eleven

C = Twelve

D = Thirteen

E = Fourteen

HEXADECIMAL TO DECIMAL

• Convert 758.D1₁₆ into decimal equivalent.

758.D1

16 = 1880.816410

758.D1 => 7 x 162 = 1792 5 x 161 = 80

8 x 160 = 8 D x 16-1= 0.8125

1 x 16-2= 0.0039

DECIMAL TO HEXADECIMAL

• Convert 0.3₁₀ into hexadecimal

• Because C is the repeating value therefore, as a convention, we shall take it once only.

Result Fractional Part Integral Part

16 x 0.3 16 x 0.8 16 x 0.8

4.8 12.8 12.8

8 8 8

4

12 = C 12 = C

0.3

EXAMPLE

• Convert 185.3₁₀ into hexadecimal.

0.3

10 = 0.4C16

185

10 = B916

185.3

10 = B9.4C16

Result Fractional Part Integral Part

16 x 0.3 16 x 0.8 16 x 0.8

4.8 12.8 12.8 8 8 8 4

DECIMAL TO BINARY

• Convert 0.56₁₀ into binary. Give answer up to

6 decimals.

Result Fractional Part Integral Part

2 x 0.56 2 x 0.12 2 x 0.24 2 x 0.48 2 x 0.96 2 x 0.92 2 x 0.84 2 x 0.68

1.12 0.24 0.48 0.96 1.92 1.84 1.68 1.36 12 24 48 96 92 84 68 36 1 0 0 0 1 1 1 1 0.56

EXAMPLE

• Convert 56.25₁₀ into binary.

2 56

28 0 2

14 0 2

7 0 2

3 12 1 1₂

0 1

Result Fractional Part

Integral Part

2 x 0.25 2 x 0.50

0.5 1.0 5 0 0 1 56

10 = 01110002

0.25

10 = 0.012

56.25

HEXADECIMAL TO BINARY

• Convert A1.03₁₆ into Binary.

A = 1010

1 = 0001

0 = 0000

3 = 0011

A1.03

16 = 10100001.000000112

Remember that

A = Ten

B = Eleven

C = Twelve

D = Thirteen

E = Fourteen

BINARY TO HEXADECIMAL

• Convert 101100.1₂ into hexadecimal.

• Solution:

• First divide your number into groups of 4 bits starting from the right.

0010 1100 . 1000

2 C 8

101100.1

OCTAL − HEXADECIMAL

• Convert to Binary as an intermediate step

Example:

( 0 1 0 1 1 0 . 0 1 0 )₂

( 1 6 . 4 )₁₆

Assume Zeros

Works both ways (Octal to Hex & Hex to Octal)

( 2 6 . 2 )₈

EXERICSE – CONVERT...

Don’t use a calculator!

Decimal Binary Octal

Hexa-decimal

29.8

101.1101

3.07

C.82

Exercise – Convert …

Decimal Binary Octal

Hexa-decimal

29.8 11101.110011… 35.63… 1D.CC…

5.8125 101.1101 5.64 5.D

3.109375 11.000111 3.07 3.1C 12.5078125 1100.10000010 14.404 C.82