meljun cortes number bases

7/29/2019 MELJUN CORTES Number Bases

1/18

Lesson 1 - 1

Year 1

CS113/0401/v1

LESSON 1

NUMBER BASES

Human base

10

Ten finger

DECIMAL

Computer base

2

Two-state devices

BINARY

NUMBER SYMBOL

Base

Number of symbols

Number of states


2/18

Lesson 1 - 2

Year 1

CS113/0401/v1

Decimal / Denary

258 =200+50+8=2x100+5x10+8 x1

=2x10 +5x10 +8x10

Binary

Decimal value of binary 110101

(1 x 32) 32

+ (1 x16) 16

+ (0 x 8) 0

+ (1 x 4) 4

+ (0 x 2) 0

+ (1 x 1) + 153

Positional valueDigit

1002

105

18

Positional valueDigit 321 161 80 41 1120

POSITIONAL VALUES (1)

2 1 0


3/18

Lesson 1 - 3

Year 1

CS113/0401/v1


Base 10 2 8 16

Number

SystemDecimal Binary Octal Hexadecimal

DigitsUsed

0

1

2

3

4

5

6

7

8

9

0

1

0

1

2

3

4

5

6

7

0

1

2

3

4

5

6

7

8

9

0

A

B

C

D

EF


4/18

Lesson 1 - 4

Year 1

CS113/0401/v1

Octal

Decimal value of octal 3056

+ (3x512) 1536

+ (0x64) 0

+ (5x8 ) 40

+ (6x1 ) + 6

1582

Positional value

Digit

512

3

64

0

8

5

1

6



5/18

Lesson 1 - 5

Year 1

CS113/0401/v1

Hexadecimal

Decimal Value of hexadecimal 2FA6

(2x4096) 8192

+ (Fx256) 15x256 3840

+ (Ax16) 10x16 160

+ (6x1) + 6

12198

Positional value

Digit

4096

2

256

F

16

A

1

6

POSITION VALUES (4)


6/18

Lesson 1 - 6

Year 1

CS113/0401/v1

Decimal

Binary Fractions (Decimals)

Decimal value of binary 101.011

(1 x 4) 4.0

+ (0 x 2) 0.0

+ (1 x 1) 1.0

+ (0 x 0.5) 0.0

+ (1 x 0.25) 0.25

+ (1 x 0.125) + 0.125

5.375

Positional value

Digit

10

3

1

6

.

.

0.1

5

0.001

8

0.01

2

Positional value

Digit

4

1

2

0

.

.

0.5

0

0.125

1

0.25

1

1

1

FRACTIONAL QUANTITIES

(1)


7/18

Lesson 1 - 7

Year 1

CS113/0401/v1

Octal Fractions

Decimal value of octal 31.27

(3 x 8) 24.0

+ (1 x 1) 1.0

+ (2 x 0.125) 0.25

+ (7 x 0.015625) + 0.109375

25.359375

Positional value

Digit

8

3

1

1

.

.

0.125

2

0.015625

7


(2)


8/18

Lesson 1 - 8

Year 1

CS113/0401/v1

Hexadecimal Fractions

Decimal value of hexadecimal 0.CF

(C x 0.0625)

12 x 0.0625 0.75

+ (F x 0.0390625)15 x 0.0390625 +0.05859375

0.80859375

Positional value

Digit

.

.

0.0625

C

0.00390625

F


(3)


9/18

Lesson 1 - 9

Year 1

CS113/0401/v1

Divide by required base and note

remainder

Continue dividing quotients by requiredbase until the answer is zero

Write the remainder digits from right to left

to give the answer

117 Decimal to Binary

2 ) 117 remainder 1

2 ) 58 0

2 ) 29 1

2 ) 14 0

2 ) 7 1

2 ) 3 1

2 ) 1 1

0

117 Decimal = 1 1 1 0 1 0 1

Binary

CONVERSION FROM

DECIMAL TO OTHER BASES(1)


10/18

Lesson 1 - 10

Year 1

CS113/0401/v1

236 Decimal to Octal

8 ) 236 remainder 48 ) 29 5

8 ) 8 3

236 Decimal = 3 5 4 Octal

437 Decimal to Hexadecimal

16 ) 473 remainder 9

16 ) 29 13

16 ) 1 1

437 Decimal= 1 D 9Hexadecimal

CONVERSION FROM

DECIMAL TO OTHER BASES(2)


11/18

Lesson 1 - 11

Year 1

CS113/0401/v1

CONVERSION OF DECIMAL

FRACTIONS (1)

Method Multiply fraction part by base

Remove integer part of result as

first digit of answer

Continue multiplying remainingfractional parts by the base and

extracting the resulting integers

as answer digits

Stop when answer contains

enough digits for accuracy

required, or when remaining

fraction is zero

If remaining fraction is zero, the

representation is exact


12/18

Lesson 1 - 12

Year 1

CS113/0401/v1

DECIMAL TO BINARY0.743 Decimal to Binary

.743 x 2

1 .486 x 2

0 .972 x 2

1 .944 x 2

1 .888 x 2

1 .776 x 2

1 .552 x 2

1 .104 x 2

0 .208

0.1 0 1 1 1 1 1 0 Binary

= 0.743


FRACTIONS (2)


13/18

Lesson 1 - 13

Year 1

CS113/0401/v1


FRACTIONS (3)

DECIMAL TO OCTAL Use previous method outlined,

but multiplying by 8 each time

DECIMAL TO HEXADECIMAL Use previous method outlined,

but multiplying by 16 each time

Remember that integer part canbe bigger than 10 giving A to F in

the answer


14/18

Lesson 1 - 14

Year 1

CS113/0401/v1

BINARY, OCTAL AND

HEXADECIMAL (1)

The three systems are closely

related

Octal or hexadecimal are often

used as shorthand for binary

Example : Store dumps

Group binary digits

in threes for octal

in fours for hexadecimal


15/18

Lesson 1 - 15

Year 1

CS113/0401/v1

Express each octal digit as threebinary digits, or each hexadecimal

digit as four binary digits, then write

all the binary digit, as continuous

string

OCTAL 1 7 3 2

BINARY 0 0 1 1 1 1 0 1 1 0 1 0

HEX 3 D A

Add leading zeros (trailing zeroes to

fractions) for clarity

To convert from octal to hexadeximal

or vice-versa go via binary

BINARY, OCTAL AND

HEXADECIMAL (2)


16/18

Lesson 1 - 16

Year 1

CS113/0401/v1

The sum of two octal numbers can be

deduced by the usual addition algorithm

to the repeated addition of two digits (

with possibly a carry of 1 ). The sum of two octal digits, or the sum

of two octal digits plus 1, can be

obtained by :

i. Finding their decimal sum and

ii. Modifying the decimal, if it exceeds7, by subtracting 8 and carrying 1 to

the next column.

Example: 5 + 6 + 2 = 15

5

+ 6

2

Decimal sum 13

Modification - 8

Octal sum 15

OCTAL ADDITION

8

8 8 8 8

8

8

8

8


17/18

Lesson 1 - 17

Year 1

CS113/0401/v1

The sum of two hexadecimal digits, or

the sum of two hexadecimal digits plus

1, can be obtained by :

i. Finding their decimal sum and

ii. Modifying the decimal, if it exceeds15, by subtracting 16 and carrying

1 to the next column.

Example : A + 9

A

+ 9

Decimal sum 19

Modification - 16

Octal sum 13

HEXADECIMAL ADDITION

1616

16


18/18

Lesson 1 - 18

Year 1

CS113/0401/v1

MODULAR ARTHMETIC

In our daily life, there are so many

counting / measuring systems around

us.

E.g. 100 cm is not the same as 100

inches, because measuring is different.

Example:

If Peter starts work at 8 0 clock in the

morning and work for 8 hours, at whattime will Peter finish work?

Solution:

Step 1. Add 8 hours to 8 o clock

= ( 8 + 8 = 16 )

Step 2. 16 Divide by 12

( because 12 hours )

= ( 16 mod 12 )

Step 3. The remainder is 4

= ( 16 mod 12 = 4 )

meljun cortes number bases

Documents