1.0 number bases

8/3/2019 1.0 Number Bases

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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


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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


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

Number in base 109 7 0 3

The place value of the digit 7in the

number9703 is 100

Theplace value of any digit of a number is a fixedvalue and does not

change with the value of the digit.

There is no place value equal to zero.

The smallestplace value of all number bases is ones.

The place value of3 in the number9703 is 1.


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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


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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


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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


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Numbers in base 2 have their respective place values

as shown below

22 = 21 = 20 =

2 x 2 = 4 2 1

PlaceValue

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


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as shown below

82 = 81 = 80 =

8 x 8 = 64 8 1

PlaceValue

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

and 7


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as shown below

52 = 51 = 50 =

5 x 5 = 25 5 1

PlaceValue

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


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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

0111

1 0 0

0

0

000

00

1

11

111

11

3 = 2 + 1

4 = 4 + 0 + 0

5 = 4 + 0 + 16 = 4 + 2 + 07 = 4 + 2 + 1

8 = 8 + 0 + 0 + 0

9 = 8 + 0 + 0 + 1

1

1


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83

=512 82

=64 81

=8 80

=1

BASE 8


Number in Base 10

0 0

2

1 1

23

4

6

01

2

5

7

3

3

4

56

7

8 = 8 + 0

19 = 2 x 8 + 3


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54=

625

53=

125

52=

25

51=

5

50=

1

BASE 5


Number in Base 10

0 0

2

1 1

23

4

1

1

02

3

0

1

21

2

3

4

5 = 5 + 06 = 5 + 17 = 5 + 2

10 = 2 x 5 + 0

17 = 3 x 5 + 2


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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 binarynumbers

Numbers in base 8 are also known as octalnumbers

Numbers in base ten are also known as denarynumbers


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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=16

0 0 0 01

11112

and 100002


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State a number before and after 218 in base 8EXAMPLE

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


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State two numbers after 435 in base 5EXAMPLE

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

00

445 and 1005


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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 eachof the following numbers

(a)11012(b) 40328

(c)12435

EXAMPLE


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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


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SOLUTION

Place

Value

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

Number 4 0 3 2

Value of

Digit0 x 64 = 0


is 0


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SOLUTION

Place

Value

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

Number 1 2 4 3

Value of

Digit4 x 5 = 20


is 20


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1.1 c Writing Numbers in Base Two, Eight and Five

in Expanded Notation

A number written in expandednotation 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


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1.1 c Writing Numbers in Base Two, Eight and Fivein Expanded Notation

Let us write 3748 in expanded notation

Place Value 82 81 80

Number 3 7 4


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


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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


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Place Value 53 52 51 50

Number


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

4 1 0 3


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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 expandednotation

2. Simplify the expan

dednot

ation

into as

in

glen

umber

EXA

MPLE

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

a. 110012b. 3748c. 41035


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EXAM

PLE

Convert each of the following numbers to

a number in base 10

a. 110012b. 3748c. 41035

S

OLUTION

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


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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 bottomto the top

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


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SOLUTION 1810 to Base 2

18

9

4

2

10

2

2

2

2

2R1R0

R0

R1

R0

1810 = 100102


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SOLUTION 1810

to Base 5

18

3

0

5

5

R3

R3

1810 = 335


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SOLUTION 1810

to Base 8

18

2

0

8

8

R2

R2

1810 = 228


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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 2c. 1278 to number in base 5

d. 2035 to number in base 8


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SOLUTION

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


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SOLUTIO

N

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


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SOLUTION

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


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SOLUTION

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


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Binary Octal Decimal

000 0 0001 1 1

010 2 2

011 3 3100 4 4

101 5 5

110 6 6

111 7 7


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Convertingbinarytooctal: 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.


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Convertingoctaltobinary: 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 convertsoctal number 6252 to binary.


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Convert 1000111012 to number in base 8

solution

101011100 101011100

534

1000111012 = 4358


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Convert 5318 to number in base 2

solution

5318 = 1010110012

531135

001011101


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1.1f Addition and Subtraction in Base Two

To addtwonumbers inbase two, the following addition

rules are important


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1 0 1 1

+ 1 1 1 0

_______10

1

0

1

11

12 + 12 = 102

12 + 12 = 102

12 + 12 + 12 = 112


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1.1f Addition and Subtraction in Base Two

To subtracttwonumbers inbase two, the following

subtraction rules are important

02 - 02 =

12 - 02 =

12 - 12 =

102 - 12 =

02

12

02

12


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1 0 1 1

- 1 1 0

_______10

10

1

0

102 -12 = 12

02 - 02 = 02


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BASE


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BASE

Binary BIN (b)

Octal OCT (o)

Denary DEC (d)


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ExampleExample 11

MODE BASE

3 3

4 6 2 =

ln

OCT

8181 x2DEC

2x

o

d

Convert 14628to a number in base

10


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To clear the

Base-nspecification

MODE

COMP

1 1Press1X

To continue the Base-n specification

PressON


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Convert 11012 to a number in base 8

ExampleExample 22

MODE BASE

3 3

1 0 1 =

logBIN

151 ln

OCT

2x

b

o


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To clear the

Base-nspecification

MODE

COMP

1 1Press1X


PressON


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ExampleExample 33

Convert 146210 to a number in

base 8

MODE BASE

3 3

4 6 2 =

x2DEC

26661 lnOCT

2x

d

o


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ExampleExample 44

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


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To clear the

Base-nspecification

MODE

COMP

1 1Press1X


PressON

1.0 number bases

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