B nar y
i
In 1854, British mathematician
George Boole published a paper
detailing a system of logic that would
become known as Boolean algebra.
His logical system proved instrumental
in the development of the binary
system, particularly in its
implementation in electronic circuitry.

B nar y
i
A numbering systems that only uses
two digits. 0 and 1.
Rather than a base ten that we are all
familiar with.
Computers use binary to store
information in a digital format.
Each digit ( 0 or 1) represents one bit
Eight bits are equal to one byte.

Bt
i
 One Binary Digit
 abbreviation is “b”
Can be thought of as one character
 Either a 1 or a 0

B e
yt
 Eight bits make up one byte
 Abbreviation “B”
 Combination of 1’s and 0’s
 Can be thought of as one character
 11101010

ki l obi t
 1024 bits
 Abbreviation “Kb”

ki l obyt es
 Represented by KB
 Slang “Kilo”
 Is equal to 1024 bytes
 210

megabyt es
 Represented by MB
 Slang “Meg”
 Is equal to 1,000000 bytes
 One million bytes
 220

gi gabyt e
 Represented by GB
 Slang “Gig”
 Equal to 1,000,000,000 Bytes
 One Billion bytes
 230

t er abyt e
 Represented by TB
 Slang “tera”
 Equal to 1,000,000,000,000 Bytes
 One Trillion bytes
 240

pet abyt e
 Represented by PB
 Slang “peta”
 Equal to 1,000,000,000,000,000 Bytes
 One Thousand Trillion bytes
 250

exabyt e
 Represented by EB
 Slang “exa”
 Equal to 1,000,000,000,000,000,000 Bytes
 One Million Trillion bytes
 260
 All printed materialin the world
would use about 5 Exabytes

Think of Binary as light bulbs
that are either ON
or Off

One Light bulb represents one
Bit
All eight of these Light bulbs would represent one byte

1 0 0 0 0 0 0 1
Think of Binary as light bulbs
that are either ON
or Off

B nar y Exer ci se
i
Binary Exercise
Bit Postion Bit 8 Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Total Binary Value
Position Value
128 64 32 16 8 4 2 1 255
if ON
Position Value
0 0 0 0 0 0 0 0 0
if OFF
Turn a Postion
1 0 0 0 0 0 0 1
ON
Here we would
ADD
The Postion Value
Here we bring
The Postion
Value
128 1 129
DOWN
In this incidence our binary Number 10000001 would have a value of 129
Because Postion 8 is ON Postion 7 is OFF Postion 6 is OFF Position 5 is OFF Position 4 is Off
Postion 3 is OFF Postion 2 is OFF and Position 1 is ON.

B nar y
i
 Figuring Binary.
 Starting on the right going to the left
 The first digit will be 1
 The second digit will be 2
 The third digit will be 4
 The fourth digit will be 8
 The fifth digit will be 16
 The sixth digit will be 32
 The seventh digit will be 64
 The eighth digit will be 128

B nar y
i
Base Ten numbers are tabulated
Left to Right.

B nar y
i
Binary numbers are tabulated
Right to Left.

Exam e
pl
10000000
The 1st – 7th digit would be Off
The Eighth digit would be On
 The first digit will be 1 0
 The second digit will be 2 0
 The third digit will be 4 0
 The fourth digit will be 8 0
 The fifth digit will be 16 0
 The sixth digit will be 32 0
 The seventh digit will be 64 0
 The eighth digit will be 128 +128
Add the bits
The value of the number would be Total 128

Exam e
pl
10000001
The 1st digit would be On
The 2nd – 7th digit would be Off
The Eighth digit would be On
 The first digit will be 1 1
 The second digit will be 2 0
 The third digit will be 4 0
 The fourth digit will be 8 0
 The fifth digit will be 16 0
 The sixth digit will be 32 0
 The seventh digit will be 64 0
 The eighth digit will be 128 +128
Add the bits
The value of the number would be Total 129

Exam e
pl
10000011
The 1st digit would be On
The 2nd digit would be On
The 3rd – 7th digit would be Off
The Eighth digit would be On
 The first digit will be 1 1
 The second digit will be 2 2
 The third digit will be 4 0
 The fourth digit will be 8 0
 The fifth digit will be 16 0
 The sixth digit will be 32 0
 The seventh digit will be 64 0
 The eighth digit will be 128 +128
Add the bits
The value of the number would be Total 131

Exam e
pl
10000111
The 1st- 3rd digit would be On
The 4th – 7th digit would be Off
The Eighth digit would be On
 The first digit will be 1 1
 The second digit will be 2 2
 The third digit will be 4 4
 The fourth digit will be 8 0
 The fifth digit will be 16 0
 The sixth digit will be 32 0
 The seventh digit will be 64 0
 The eighth digit will be 128 +128
Add the bits
The value of the number would be Total 135

Exam e
pl
11000000
The 1st- 6th digit would be Off
The 7th digit would be On
The 8th digit would be On
 The first digit will be 1 0
 The second digit will be 2 0
 The third digit will be 4 0
 The fourth digit will be 8 0
 The fifth digit will be 16 0
 The sixth digit will be 32 0
 The seventh digit will be 64 64
 The eighth digit will be 128 +128
Add the bits
The value of the number would be Total 192

192
What is the
value?
1 1 0 0 0 0 0 0
Think of Binary as light bulbs
that are either ON
or Off

Exam e
pl
11111111
The 1st- 8th digit would be On
 The first digit will be 1 1
 The second digit will be 2 2
 The third digit will be 4 4
 The fourth digit will be 8 8
 The fifth digit will be 16 16
 The sixth digit will be 32 32
 The seventh digit will be 64 64
 The eighth digit will be 128 +128
Add the bits
The value of the number would be Total 255

255
What is the
value?
1 1 1 1 1 1 1 1
128 64 32 16 8 4 2 1
Think of Binary as light bulbs
that are either ON
or Off

U ng C cul at or
si al
t o f i gur e
B nar y N ber s
i um
First we would open Calculator
Start/All Programs/Accessories/Calculator
From the Calculator go to View and down
To SCIENTIFIC

Scientific

 This is the Scientific Calculator
 The next thing we would need to do in select
 BIN for Binary

 Next we would enter the Binary number
 For example
10000000

After entering the Binary number we would
then select the
Dec Radio Button

We now see the answer to the problem
Is
128

192
What is the
value?
1 1 0 0 0 0 0 0
Think of Binary as light bulbs
that are either ON
or Off

ICT 1

Deci m t o B nar y
al i
 It follows a starightforward method.
 It involves dividing the number to be
converted, say N by 2 (since binary is in base
2) until we reach the division of (1/2), also
making note of all remainders.

Exam e 1: C
pl onver t 98
f r om deci m t o bi nar y
al
 Divide 98 by 2, make note of all the
remainder.
 Continue dividingquotientsby 2, making
notes of the remainder.
 Also, note the star beside the last remainder.

Division Remainder, R
98/2 = 49 R=0
49/2 = 24 R=1
24/2 = 12 R=0
12/2 = 6 R=0
6/2 = 3 R=0
3/2 = 1 R=1
1/2 = 0 R=1
The sequance of remainders going up gives the answer.
Starting from 1*, we have 1100010.
Therefore, 98 in decimals is 1100010 in binary

Exam e 2: C
pl onver t 21
i nt o bi nar y
Division Remainder, R
21/2 = 10 R=1
10/2 = 5 R=0
5/2 = 2 R=1
2/2 = 1 R=0
1/2 = 0 R=1
Therefore, 21 in decimals is 10101 in binary

B nar y t o deci m
i al
 Conversion follows the same steps as decimal
to binary, except in reverse order.
 We can begin by multiplying 0 x 2 and adding
1.
 We continue to multiply the numbers in the
“results” column by 2, and adding the digits
from left to right in our binary numbers.

Exam e 1: C
pl onver t 11101
f r om bi nar y t o deci mal
Operations Result
0x2+1 1
1x2+1 3
3x2+1 7
7x2+0 14
14 x 2 + 1 29
Therefore, 11101 in binary is 29 in decimal.