Lec 08 Computer Arithmatic

1

CS1010 Introduction to Computing

Lecture 08

Review – multiplying powers

•

For common bases,

add powers

.

2



2

= 2

= 65,536

or…

2



2

= 64



2

= 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

and

00110010

•

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

1

≥ (2)

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

REPRESENTATION OF NEGATIVE

NUMBERS

•

There are many methods of representing

signed

number in

binary

:

–

Sign-Magnitude Method

1’s COMPLEMENT METHOD

•

1’s complement

of a

binary

number can be

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.

Original number

01100110

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

-54

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

54₁₀ = 0110110₂

5410 = 001101102 (in 8 bits) So -5410 in 1’s complement form:

2’s COMPLEMENT METHOD

•

Algorithm

–

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

-54

in

2’s complement form

using 8

bits.

Modulus of -54

= 54

2 54

27 0 2

13 1 2

6 1 2

3 02 1 1₂ 0 1

54₁₀ = 0110110₂

5410 = 001101102 (in 8 bits) Take 1’s complement of

00110110

= 11001001

+ 1

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.

38₁₀ = 00100110₂

29₁₀ = 00011101₂

Represent negative numbers in 1’s complement

-29₁₀ = 11100010₂

Add the number & its 1’s complement

00100110₂

+ 11100010₂

Carry: 1 00001000₂

Add End Carry:+ 1

EXAMPLE

•

Calculate

– 54 – 30

using 8-bit

1’s complement

method.

•

Solution:

(– 54) + (– 30)

54₁₀ = 00110110₂

30₁₀ = 00011110₂

Represent negative numbers in 1’s complement -5410 = 110010012

Add numbers in its 1’s complement form

11001001₂

+ 11100001₂

Carry: 1 10101010₂

Add End Carry:+ 1 10101011₂

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

Exercises

•

Perform following calculations using

1

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:

38₁₀ = 00100110₂

Represent negative numbers in 2’s complement

Take 1’s complement of 29 =

00011101 and then add 1

-29₁₀ = 11100011₂ (2’s complement)

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

00100110₂

+ 11100011₂

Carry: 1 00001001₂

Ignore Carry, the answer is

EXAMPLE

•

Calculate

– 54 – 30

using 8-bit

2’s complement

method.

•

Solution:

(– 54) + (– 30)

54₁₀ = 00110110₂

30₁₀ = 00011110₂

Represent negative numbers in 2’s complement

-54₁₀ = 11001010₂

Add 2’s complement representation

11001010₂

+ 11100010₂

Carry: 1 10101100₂

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

10101100₂ = 01010011 = -84

Ignore Carry, the answer is

Exercises

•

Perform following calculations using

2

complement method.

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

•

127.54

127.54₈ => 1 x 82 _{= 64.000} 2 x 81 _{= 16.000}

7 x 80 _{= 7.000} 5 x 8-1 _{= 0.625} 4 x 8-2 _{= 0.0625}

EXAMPLE

•

Convert

630.4

into decimal equivalent.

630.4

= 408.5

630.4 => 6 x 82 _{= 384} 3 x 81 _{= 24}

0 x 80 _{= 0} 4 x 8-1 _{= 0.5}

DECIMAL TO OCTAL

•

185.3

8 185 23 1 8

2 7₈ 0 2

185₁₀ = 0271₈

0.3₁₀ = 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

HEXADECIMAL TO DECIMAL

•

2B.C4

2B.C4₁₆ => 2 x 161 _{= 32.000} B x 160 _{= 11.000}

C x 16-1 _{= .075}

4 x 16-2 _{= .015} 43.090₁₀

Remember that

A = Ten

B = Eleven

C = Twelve

D = Thirteen

E = Fourteen

HEXADECIMAL TO DECIMAL

•

Convert

758.D1

into decimal equivalent.

758.D1

= 1880.8164

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

0.3

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

EXAMPLE

•

Convert

185.3

into hexadecimal.

0.3₁₀ = 0.4C₁₆ 185₁₀ = B9₁₆

185.3₁₀ = 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

0.56

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

EXAMPLE

•

Convert

56.25

into binary.

2 56

28 0 2

14 0 2

7 0 2

3 1 2

1 1₂ 0 1

Result Fractional Part Integral Part

2 x 0.25

2 x 0.50 0.5_1.0 50 01

56₁₀ = 0111000₂ 0.25₁₀ = 0.01₂

HEXADECIMAL TO BINARY

•

Convert

A1.03

into Binary.

A = 1010

1 = 0001

0 = 0000

3 = 0011

A1.03₁₆ = 10100001.00000011₂

Remember that

A = Ten

B = Eleven

C = Twelve

D = Thirteen

E = Fourteen

BINARY TO HEXADECIMAL

•

Convert

101100.1

into hexadecimal.

•

Solution:

•

First divide your number into groups of 4 bits

starting from the right.

0010 1100 . 1000

2 C 8

OCTAL − HEXADECIMAL

•

Convert to Binary as an intermediate step

Example:

(

0

1 0 1 1 0 . 0 1 0

)

(

1 6 . 4

)

Assume Zeros

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

(

2 6 . 2

)

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