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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#### 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%

112201
100405
120203
100202

243525
342555
343555
333555

0010112
10001102
11001102
10110012

60
50
65
55

107
110
115
105

145
130
150
135

225
245
230
235

1000111012
1111111112
1110001112
1101001002

11010001112
10101010102
11100010102
11010010002

11110010102
10000101112
10101011012
10110011012

111001012
101010102
110110102
111111112

101010102
111000012
110010102
100101012

1100101012
1011100102
1010001112
1111111112

1700
1753
1750
1755

365
375
390
370

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