Number Conversion Examples
Example 1
Let's convert the number 12 from the base-ten positional numeral system to the binary numeral system.
Convert the number
1210 to base-2, by successively dividing by 2, until the quotient is zero. The result will be the number formed from the remainders written from right to left.
12 | : | 2 | = | 6 | remainder: 0 |
6 | : | 2 | = | 3 | remainder: 0 |
3 | : | 2 | = | 1 | remainder: 1 |
1 | : | 2 | = | 0 | remainder: 1 |
1210 = 11002
Example 2
Let's convert the number 12.3 from the base-ten positional numeral system to the binary numeral system.
Convert whole number part
12 of the number 12.3
10 to base-2, by successively dividing by 2, until the quotient is zero. The result will be the number formed from the remainders written from right to left.
12 | : | 2 | = | 6 | remainder: 0 |
6 | : | 2 | = | 3 | remainder: 0 |
3 | : | 2 | = | 1 | remainder: 1 |
1 | : | 2 | = | 0 | remainder: 1 |
1210 = 11002Let's convert the decimal part
0.3 of the number 12.3
10 to base-2, using successively multiplying by 2, until the decimal part of the product turns out to be zero or the required number of digits after the decimal point is reached. If after multiplying the whole number part is not zero, then you need to replace the value of the whole number part with zero. The result will be a number of integer parts of product written from left to right.
0.3 | · | 2 | = | 0.6 |
0.6 | · | 2 | = | 1.2 |
0.2 | · | 2 | = | 0.4 |
0.4 | · | 2 | = | 0.8 |
0.8 | · | 2 | = | 1.6 |
0.6 | · | 2 | = | 1.2 |
0.2 | · | 2 | = | 0.4 |
0.4 | · | 2 | = | 0.8 |
0.8 | · | 2 | = | 1.6 |
0.310 = 0.010011001212.310 = 1100.0100110012
Example 3
Let's convert the number 10011 from the binary numeral system to the base-ten positional numeral system.
Let's convert the number
100112 to base-10 by first writing down the position of each digit in the number from right to left, starting at zero.
Each digit position will be a power of 2, because the number system is base-2. It is necessary to write down the sum, each term in which will be the product of each digit of the number 10011
2 and 2 to the power, which is determined by the position in the number.
100112 = 1 â‹… 24 + 0 â‹… 23 + 0 â‹… 22 + 1 â‹… 21 + 1 â‹… 20 = 1910
Example 4
Let's convert the number 11.101 from the binary numeral system to the base-ten positional numeral system.
Let's convert the number
11.1012 to base-10 by first writing down the position of each digit in the number
Position | 1 | 0 | -1 | -2 | -3 |
Number | 1 | 1 | 1 | 0 | 1 |
Each digit position will be a power of 2, because the number system is base-2. It is necessary to write down the sum, each term in which will be the product of each digit of the number 11.101
2 and 2 to the power, which is determined by the position in the number.
11.1012 = 1 â‹… 21 + 1 â‹… 20 + 1 â‹… 2-1 + 0 â‹… 2-2 + 1 â‹… 2-3 = 3.62510
Example 5
Let's convert the number 1583 from the base-ten positional numeral system to the hexadecimal numeral system.
Convert the number
158310 to base-16, by successively dividing by 16, until the quotient is zero. The result will be the number formed from the remainders written from right to left.
1583 | : | 16 | = | 98 | remainder: 15, 15 = F |
98 | : | 16 | = | 6 | remainder: 2 |
6 | : | 16 | = | 0 | remainder: 6 |
158310 = 62F16
Example 6
Let's convert the number 1583.56 from the base-ten positional numeral system to the hexadecimal numeral system.
Convert whole number part
1583 of the number 1583.56
10 to base-16, by successively dividing by 16, until the quotient is zero. The result will be the number formed from the remainders written from right to left.
1583 | : | 16 | = | 98 | remainder: 15, 15 = F |
98 | : | 16 | = | 6 | remainder: 2 |
6 | : | 16 | = | 0 | remainder: 6 |
158310 = 62F16Let's convert the decimal part
0.56 of the number 1583.56
10 to base-16, using successively multiplying by 16, until the decimal part of the product turns out to be zero or the required number of digits after the decimal point is reached. If after multiplying the whole number part is not zero, then you need to replace the value of the whole number part with zero. The result will be a number of integer parts of product written from left to right.
0.56 | · | 16 | = | 8.96 |
0.96 | · | 16 | = | 15.36, 15 = F |
0.36 | · | 16 | = | 5.76 |
0.76 | · | 16 | = | 12.16, 12 = C |
0.16 | · | 16 | = | 2.56 |
0.56 | · | 16 | = | 8.96 |
0.96 | · | 16 | = | 15.36, 15 = F |
0.36 | · | 16 | = | 5.76 |
0.76 | · | 16 | = | 12.16, 12 = C |
0.16 | · | 16 | = | 2.56 |
0.5610 = 0.8F5C28F5C2161583.5610 = 62F.8F5C28F5C216= 62F.(8F5C2)16
Example 7
Let's convert the number A12DCF from the hexadecimal numeral system to the base-ten positional numeral system.
Let's convert the number
A12DCF16 to base-10 by first writing down the position of each digit in the number from right to left, starting at zero.
Position | 5 | 4 | 3 | 2 | 1 | 0 |
Number | A | 1 | 2 | D | C | F |
Each digit position will be a power of 16, because the number system is base-16. It is necessary to write down the sum, each term in which will be the product of each digit of the number A12DCF
16 and 16 to the power, which is determined by the position in the number.
A16 = 10
10D16 = 13
10C16 = 12
10F16 = 15
10A12DCF16 = 10 â‹… 165 + 1 â‹… 164 + 2 â‹… 163 + 13 â‹… 162 + 12 â‹… 161 + 15 â‹… 160 = 1056302310
Example 8
Let's convert the number A12DCF.12A from the hexadecimal numeral system to the base-ten positional numeral system.
Let's convert the number
A12DCF.12A16 to base-10 by first writing down the position of each digit in the number
Position | 5 | 4 | 3 | 2 | 1 | 0 | -1 | -2 | -3 |
Number | A | 1 | 2 | D | C | F | 1 | 2 | A |
Each digit position will be a power of 16, because the number system is base-16. It is necessary to write down the sum, each term in which will be the product of each digit of the number A12DCF.12A
16 and 16 to the power, which is determined by the position in the number.
A16 = 10
10D16 = 13
10C16 = 12
10F16 = 15
10A12DCF.12A16 = 10 â‹… 165 + 1 â‹… 164 + 2 â‹… 163 + 13 â‹… 162 + 12 â‹… 161 + 15 â‹… 160 + 1 â‹… 16-1 + 2 â‹… 16-2 + 10 â‹… 16-3 = 10563023.0727539062510
Example 9
Let's convert the number 1010100011 the binary numeral system to the the hexadecimal numeral system.
Let's convert the number
10101000112 to base-10 by first writing down the position of each digit in the number from right to left, starting at zero.
Position | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
Number | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
Each digit position will be a power of 2, because the number system is base-2. It is necessary to write down the sum, each term in which will be the product of each digit of the number 1010100011
2 and 2 to the power, which is determined by the position in the number.
10101000112 = 1 â‹… 29 + 0 â‹… 28 + 1 â‹… 27 + 0 â‹… 26 + 1 â‹… 25 + 0 â‹… 24 + 0 â‹… 23 + 0 â‹… 22 + 1 â‹… 21 + 1 â‹… 20 = 67510
Convert the number
67510 to base-16, by successively dividing by 16, until the quotient is zero. The result will be the number formed from the remainders written from right to left.
675 | : | 16 | = | 42 | remainder: 3 |
42 | : | 16 | = | 2 | remainder: 10, 10 = A |
2 | : | 16 | = | 0 | remainder: 2 |
67510 = 2A31610101000112 = 2A316