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

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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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### Very Short Questions

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%

100202
112201
100405
120203

243525
342555
343555
333555

0010112
10110012
11001102
10001102

65
50
60
55

115
110
105
107

135
130
145
150

235
225
245
230

1000111012
1111111112
1101001002
1110001112

11010001112
11010010002
11100010102
10101010102

11110010102
10101011012
10000101112
10110011012

110110102
111111112
101010102
111001012

110010102
111000012
101010102
100101012

1100101012
1111111112
1010001112
1011100102

1750
1753
1700
1755

365
375
370
390

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