Notes on Binary Number System | Grade 8 > Compulsory Maths > Number System | KULLABS.COM

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Notes, Exercises, Videos, Tests and Things to Remember on Binary Number System

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It is necessary to review the decimal number system at first to understand more about binary number system. Decimal number system refers to base 10 positional notation. It uses ten different symbols (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) arranged using positional notation. Positional notation is used when a number larger then 9 needs to be represented; each position of a digit signifies how many groups of 10, 100, 1000, etc. are contained in that number. For example,

4251

4 $$\times$$ 1000 + 2 $$\times$$ 100 + 5 $$\times$$ 10 + 1

4$$\times$$ 103+ 2$$\times$$ 102 + 5$$\times$$ 101 + 4$$\times$$ 102

0 and 1 are used in binary number system which is arranged using positional notation (the digit 0 and 1 as a symbol). When a number larger than 1 needs to be represented, the positional notation is used to represent, the positional notation is used to represent how many groups of 2, 4, 8 are contained in the number. For example,

Let's consider the number 30

30 ÷ 2 = 15 Remainder 0

15 ÷ 2 = 7 Remainder 1

7 ÷ 2 = 3 Remainder 1

3 ÷ 2 = 1 Remainder 1

1 ÷ 2 = 0 Remainder 1

• A number is a mathematical object used to count, measure and label.
• The binary number system always uses only two different symbols ( the digit 0 and 1) that are arranged using positional notation.
• We can subtract a binary number from the another binary number.
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### Questions and Answers

#### Click on the questions below to reveal the answers

Solution:

110102

=1×24+1×23+0×22+1×21+0×20

=16+8+0+2+0

=26

Solution:

110102

=1×24+1×23+0×22+1×21+0×20

=16+8+0+2+0

=26

Solution:

1010112

=1×25+0×24+1×23+0×22+1×21+1×20

=32+0+8+0+2+1

=43

Solution:

1×22+1×21+1×20

=4+2+1

=7

Solution:

10102= 1×23+0×22+1×21+0×20

=8+0+2+0

=10

Solution:

11102 = 1×23+1×22+1×21+0×2o

=8+4+2+0

= 14

Solution:

1×24+0×23+1×22+0×21+1×2o

=16+0+4+0+1

=21

Solution:

=1×25+0×24+1×23+0×22+1×21+0×20

=32+0+8+0+2+0

=42

Solution:

1×25+1×24+0×23+0×22+1×21+1×20

= 32+16+0+0+2+1

=51

Solution:

2125

= 2×52+1×51+2×50

= 2×25+1×5+2×1

= 50+5+2

= 5710

Solution:

3145

=3×52+1×51+4×50

= 3×25+1×5+5+4

= 75+5+4

= 8410

Solution:

245

= 2×51+4×5°

= 2×5+4×1

=10+4

=1410

Solution:

3545

= 3×52+2×51+4×5°

= 75+10+4

= 8910

Solution:

 2 105 1 2 52 0 2 26 0 2 13 1 2 6 0 2 3 1 2 1 1 0

∴ 10510 = 11010012

Solution:

111111112

= 1×27+ 1×26 + 1×25+ 1×24 + 1×23 + 1×22 +1×21 + 1×20

= 128 + 64 + 32 + 16 + 8 + 4 +2 + 1

= 225

0%

100405
100202
120203
112201

333555
343555
342555
243525

10110012
11001102
10001102
0010112

65
50
55
60

115
110
105
107

150
145
130
135

225
245
230
235

1110001112
1111111112
1101001002
1000111012

11010010002
11010001112
10101010102
11100010102

11110010102
10110011012
10000101112
10101011012

111001012
111111112
101010102
110110102

100101012
111000012
101010102
110010102

1010001112
1011100102
1111111112
1100101012

1755
1753
1700
1750

375
365
390
370

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