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\(\times\) 10^{3}+ 2\(\times\) 10^{2} + 5\(\times\) 10^{1} + 4\(\times\) 10^{2}
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
Solution:
11010_{2}
=1×2^{4}+1×2^{3}+0×2^{2}+1×2^{1}+0×2^{0}
=16+8+0+2+0
=26
Solution:
11010_{2}
=1×2^{4}+1×2^{3}+0×2^{2}+1×2^{1}+0×2^{0}
=16+8+0+2+0
=26
Solution:
101011_{2}
=1×2^{5}+0×2^{4}+1×2^{3}+0×2^{2}+1×2^{1}+1×2^{0}
=32+0+8+0+2+1
=43
Solution:
1×2^{2}+1×2^{1}+1×2^{0}
=4+2+1
=7
Solution:
1010_{2}= 1×2^{3}+0×2^{2}+1×2^{1}+0×2^{0}
=8+0+2+0
=10
Solution:
1110_{2} = 1×2^{3}+1×2^{2}+1×2^{1}+0×2^{o}
=8+4+2+0
= 14
Solution:
1×2^{4}+0×2^{3}+1×2^{2}+0×2^{1}+1×2^{o}
=16+0+4+0+1
=21
Solution:
=1×2^{5}+0×2^{4}+1×2^{3}+0×2^{2}+1×2^{1}+0×2^{0}
^{=}32+0+8+0+2+0
=42
Solution:
1×2^{5}+1×2^{4}+0×2^{3}+0×2^{2}+1×2^{1}+1×2^{0}
= 32+16+0+0+2+1
=51
Solution:
212_{5}
= 2×5^{2}+1×5^{1}+2×5^{0}
= 2×25+1×5+2×1
= 50+5+2
= 57_{10}
Solution:
314_{5}
=3×5^{2}+1×5^{1}+4×5^{0}
= 3×25+1×5+5+4
= 75+5+4
= 84_{10}
Solution:
24_{5}
= 2×5^{1}+4×5°
= 2×5+4×1
=10+4
=14_{10}
Solution:
354_{5}
= 3×5^{2}+2×5^{1}+4×5°
= 75+10+4
= 89_{10}
Solution:
2 | 105 | 1 |
2 | 52 | 0 |
2 | 26 | 0 |
2 | 13 | 1 |
2 | 6 | 0 |
2 | 3 | 1 |
2 | 1 | 1 |
0 |
∴ 105_{10} = 1101001_{2}
Solution:
11111111_{2}
= 1×2^{7}+ 1×2^{6} + 1×2^{5}+ 1×2^{4} + 1×2^{3} + 1×2^{2} +1×2^{1} + 1×2^{0}
= 128 + 64 + 32 + 16 + 8 + 4 +2 + 1
= 225
Convert the following numbers into quinary numbers:
645_{10}
Convert the following numbers into quinary numbers:
2462_{10}
Convert the following decimal number into binary number.
7510
Convert the following binary number into decimal numbers.
110010_{2}
Convert the following Binary Number into Decimal Number.
1101011_{2}
Convert the following binary number into decimal number.
10000111_{2}
Convert the binary number into decimal number.
11111111_{2}
Convert the following decimal number into binary number.
420
Convert the following decimal number into binary number.
840
Convert the following decimal number into binary number.
535
Convert the following decimal number into binary number.
255
Convert the following decimal number into binary number.
225
Convert the following decimal number into binary number.
405
Convert the following binary number into decimal numbers.
11011011001_{2}
Convert the following binary number into 101110111_{2}
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