1.0 number bases form 5

Chapter 1 : Number Bases

• 1.1 a : Stating Numbers in Base Two, Eight and Five

• 1.1 b : Value of a Digit of a Number in Base 2, 8 and 5

• 1.1 c : Writing Numbers in Base 2, 8 and 5 in Expanded

Notation

• 1.1 d : Converting Numbers in Base 2, 8 and 5 to Base

10 and Vice Versa

• 1.I e : Converting from One Base to Another

• 1.1 f : Addition and Subtraction in Base Two

Chapter 1 Number Bases

Number in Base Two, Eight and Five1 1 a

1.1 Stating Numbers in Base Two, Eight and Five

• The numbers we use daily are in base 10. The place value of numbers in

base 10 are as shown below.

103 =1000 102 =100 101=10 100 =1 Place value

Number in base 10

9 7 0 39 7 0 3

103 =1000 102 =100 101=10 100 =1 Place value

Number in base 109 7 0 3

The place value of the digit 7 in the

number 9703 is 100

• The place value of any digit of a number is a fixed value and does not

change with the value of the digit.

• There is no place value equal to zero.

• The smallest place value of all number bases is “ones”.

• The place value of 3 in the number 9703 is 1.

103 =1000 102 =100 101=10 100 =1 Place value


• There are 10 digits that can be written in any place value column

for numbers in base 10.

• The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

• The digit value or value of digit varies with the place value and the

digit

103 =1000 102 =100 101=10 100 =1 Place value


The value of the digit 9 is 9 x 1000 = 9000

DigitPlace value

of 9

Value of the

digit 9

103 =1000 102 =100 101=10 100 =1 Place value


The value of the digit 0 is 0 x 10 = 0

DigitPlace value

of 0

Value of the

digit 0

• Numbers in base 2 have their respective place values

as shown below

22 = 21 = 20 =

2 x 2 = 4 2 1

Place

Value

• There are only 2 digits in base 2 : 0 and 1


as shown below

82 = 81 = 80 =

8 x 8 = 64 8 1

Place

Value

• There are only 8 digits in base 8 : 0, 1, 2, 3, 4, 5, 6

and 7


as shown below

52 = 51 = 50 =

5 x 5 = 25 5 1

Place

Value

• There are only 5 digits in base 5 : 0, 1, 2, 3 and 4

24=

16

23=

8

22=

4

21=

2

20=

1

BASE 2

Place Value of Numbers in Base 2

Number in Base 10

0 0

2 = 2 + 0

1 1

01

11

1 0 0

0

0

000

00

1

11

111

11

3 = 2 + 1

4 = 4 + 0 + 0

5 = 4 + 0 + 1

6 = 4 + 2 + 0

7 = 4 + 2 + 1

8 = 8 + 0 + 0 + 0

9 = 8 + 0 + 0 + 1

1

1

83=

512

82=

64

81=

8

80=

1

BASE 8


Number in Base 10

0 0

2

1 1

2

3

4

6

01

2

5

7

3

3

4

5

6

7

8 = 8 + 0

19 = 2 x 8 + 3

54=

625

53=

125

52=

25

51=

5

50=

1

BASE 5


Number in Base 10

0 0

2

1 1

2

3

4

1

1

02

3

0

1

21

2

3

4

5 = 5 + 0

6 = 5 + 1

7 = 5 + 2

10 = 2 x 5 + 017 = 3 x 5 + 2

910 = 8 + 0 + 0 + 1

=10012 Read as “one zero zero one base 2”

910 = 8 + 1

= 118 Read as “one one base 8”

910 = 5 + 4

=145 Read as “one four base 5”

• Numbers in base 2 are also known as binary numbers

• Numbers in base 8 are also known as octal numbers

• Numbers in base ten are also known as denary numbers

State two numbers in base two after 11102EXAMPLE

24=16 23=8 22=4 21=2 20=1 Base

10

1 1 1 0 8+4+2+

0 = 14

SOLUTION

8+4+2+

1=151 1 1 1

16+0+0+

0=160 0 0 01

11112 and 100002

State a number before and after 218 in base 8 EXAMPLE

SOLUTION

81=8 80=1 Base 10

2 1 2 x 8 + 1 = 17

Before 2 x 8 + 0 = 162 0

After 2 x 8 + 2 = 182 2

208 and 228

State two numbers after 435 in base 5 EXAMPLE

SOLUTION

52=25 51=5 50=1 Base 10

4 3 4 x 5 + 3 = 23

4 x 5 + 4 = 244 4

1 x 25 + 0 + 0 = 251 0 0

445 and 1005

1.1 b Value of A Digit of A Number in Base Two, Eight and Five

Value of a digit = The digit x Place value of a digit

State the value of the underlined digit in each

of the following numbers

(a)11012

(b) 40328

(c)12435

EXAMPLE

SOLUTION

Place

Value

23=8 22=4 21=2 20=1

Number 1 1 0 1

Value of

Digit1 x 4 = 4

The value of the digit “1” in 11012 is 4

SOLUTION

Place

Value

83=512 82=64 81=8 80=1

Number 4 0 3 2

Value of

Digit0 x 64 = 0


SOLUTION

Place

Value

53=125 52=25 51=5 50=1

Number 1 2 4 3

Value of

Digit4 x 5 = 20


1.1 c Writing Numbers in Base Two, Eight and Five

in Expanded Notation

• A number written in expanded notation refers to

the sum of the value of the digits that make up

the number .

• Let us write 42510 in expanded notation

Place Value 102 101 100

Number

425

Therefore, 42510 written in expanded notation is as follows

42510 = 4 x 102 + 2 x 101 + 5 x 100




Place Value 82 81 80

Number 3 7 4


3748 = 3 x 82 + 7 x 81 + 4 x 80




Place Value 24 23 22 21 20

Number


110012 = 1 x 24 + 1 x 23 + 0 x 22 + 0 x 21 + 1 x 20

1 1 00 1




Place Value 53 52 51 50

Number


41035 = 4 x 53 + 1 x 52 + 0 x 51 +3 x 50

4 1 0 3

1.1 d Converting Numbers in Base Two, Eight and Five to

Base 10 and Vice Versa

Steps to convert numbers in base 2, 8 and 5 to base 10 are as follows

1. Write the number in expanded notation

2. Simplify the expanded notation into a single number

EX

AM

PL

E

Convert each of the following numbers to a number in base 10

a. 110012

b. 3748

c. 41035

EX

AM

PL

EConvert each of the following numbers to

a number in base 10

a. 110012

b. 3748

c. 41035

SO

LU

TIO

N

a. 110012 = 1 x 24 + 1 x 23 + 0 x 22 + 0 x 21 + 1 x 20

= 2510

b. 3748 = 3 x 82 + 7 x 81 + 4 x 80

= 25210

c. 41035 = 4 x 53 + 1 x 52 + 0 x 51 +3 x 50

= 52810

1.1 d Converting Numbers in Base Two, Eight and Five to

Base 10 and Vice Versa

Steps to convert numbers in base 10 to base 2, 8 or 5 are as follows

1. Perform repeated division until the quotient is 0

2. Write the number in the new base by referring to the remainders

from bottom to the top

EXAMPLE Convert 1810 to numbers in base two, eight and five

SOLUTION 1810 to Base 2

18

9

4

2

1

0

2

2

2

2

2

R1

R0

R0

R1

R0

1810 = 100102


18

3

0

5

5

R3

R3

1810 = 335


18

2

0

8

8

R2

R2

1810 = 228

1.1 e Converting Numbers from One Base to Another

The following steps are used to convert from one base to another

1. Convert the number to a number in base 10

2. Use repeated division to convert the number in base 10 to the

respective bases

EXAMPLE Convert

a. 1102 to number in base 5

b. 325 to number in base 2

c. 1278 to number in base 5

d. 2035 to number in base 8

SOLU

TION

a.1102 to number in base 5

1102 = 1 x 22 + 1 x 21 + 0 x 20 = 610

6

1

0

5

5

R1

R1

1102 = 115

SOLU

TION

b. 325 to number in base 2

325 = 3 x 51 + 2 x 50 = 1710

17

8

4

2

1

0

2

2

2

2

2

R1

R0

R0

R0

R1

325 = 100012

SOLU

TION

c. 1278 to number in base 5

1278 = 1 x 82 + 2 x 81 + 7 x 80 = 8710

87

17

3

5

5

R2

R2

1278 = 3225

5

0 R3

SOLU

TION

d. 2035 to number in base 8

2035 = 2 x 52 + 0 x 51 + 3 x 50 = 5310

53

6

0

8

8

R6

R5

2035 = 658

Binary Octal Decimal

000 0 0

001 1 1

010 2 2

011 3 3

100 4 4

101 5 5

110 6 6

111 7 7

Converting binary to octal: Counting from right to left, draw a line between

every group of 3-bits. The most significant group may not have exactly three

bits, so you can just pretend the others are zeros. Now convert each group of

three to a single, octal digit. The conversion of a 3-bit number to an octal

number is easy. You can memorize the patterns easily and, even if you

forget, they are not hard to figure out. The resulting octal digits, when written

together in the same order, are the equivalent binary number. Here's an

example which converts the binary number '11111010' to its equivalent octal

number.

Converting octal to binary: This is simply the reverse of the

above process. For every octal digit, just write down the 3-bit

pattern that represents it. Here is an example which converts

octal number 6252 to binary.

Convert 1000111012 to number in base 8

solution

101011100 101011100

534

1000111012 = 4358

Convert 5318 to number in base 2

solution

5318 = 1010110012

531135

001011101

1.1f Addition and Subtraction in Base Two

• To add two numbers in base two, the following addition

rules are important

1 0 1 1

+ 1 1 1 0

_______10

1

0

1

11

12 + 12 = 102

12 + 12 = 102

12 + 12 + 12 = 112

1.1f Addition and Subtraction in Base Two

• To subtract two numbers in base two, the following

subtraction rules are important

02 - 02 =

12 - 02 =

12 - 12 =

102 - 12 =

02

12

02

12

1 0 1 1

- 1 1 0

_______10

10

1

0

102 -12 = 12

02 - 02 = 02

BASE

Binary BIN (b)

Octal OCT (o)

Denary DEC (d)

Example 1

MODE BASE

3 3

4 6 2 =

ln

OCT

8181 x2

DEC

2x

o

d

Convert 14628 to a number in base

10

To clear the Base-n specification

MODE

COMP

1 1Press1X

To continue the Base-n specification

PressON

Convert 11012 to a number in base 8

Example 2

MODE BASE

3 3

1 0 1 =

logBIN

151 lnOCT

2x

b

o


MODE

COMP

1 1Press1X


PressON

Example 3

Convert 146210 to a number in

base 8

MODE BASE

3 3

4 6 2 =

x2

DEC

26661 ln

OCT

2x

d

o

Example 4

Calculate 10012 + 1112, stating your answer

as a number in base 2

MODE BASE

3 3

0 0 1 +

log

BIN

1

2x

b

1 1

= 10000

1

b


MODE

COMP

1 1Press1X


PressON

1.0 number bases form 5

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