# Numeral Systems Calculator With Steps

## With this calculator you can convert integers and decimal numbers from one number system to another and get a detailed solution. Use a radix from 2 to 36.

Number | |

From | |

To | |

Number of digits after the decimal point (for decimal) | |

Number Conversion Examples

**Example 1**

**Let's convert the number 12 from the base-ten positional numeral system to the binary numeral system.**

_{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 |

12

_{10}= 1100

_{2}

**Example 2**

**Let's convert the number 12.3 from the base-ten positional numeral system to the binary numeral system.**

_{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 |

12

_{10}= 1100

_{2}

Let'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 | Â· | 2 | = | 0.6 |

0.6 | Â· | 2 | = | 1.2 |

0.2 | Â· | 2 | = | 0.4 |

0.4 | Â· | 2 | = | 0.8 |

0.8 | Â· | 2 | = | 1.6 |

0.6 | Â· | 2 | = | 1.2 |

0.2 | Â· | 2 | = | 0.4 |

0.4 | Â· | 2 | = | 0.8 |

0.8 | Â· | 2 | = | 1.6 |

0.3

_{10}= 0.010011001

_{2}

12.3

_{10}= 1100.010011001

_{2}

**Example 3**

**Let's convert the number 10011 from the binary numeral system to the base-ten positional numeral system.**

_{2}to base-10 by first writing down the position of each digit in the number from right to left, starting at zero.

Position | 4 | 3 | 2 | 1 | 0 |

Number | 1 | 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 10011

_{2}and 2 to the power, which is determined by the position in the number.

10011_{2} = 1 â‹… 2^{4} + 0 â‹… 2^{3} + 0 â‹… 2^{2} + 1 â‹… 2^{1} + 1 â‹… 2^{0} = 19_{10}

**Example 4**

**Let's convert the number 11.101 from the binary numeral system to the base-ten positional numeral system.**

_{2}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.

11.101_{2} = 1 â‹… 2^{1} + 1 â‹… 2^{0} + 1 â‹… 2^{-1} + 0 â‹… 2^{-2} + 1 â‹… 2^{-3} = 3.625_{10}

**Example 5**

**Let's convert the number 1583 from the base-ten positional numeral system to the hexadecimal numeral system.**

_{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 |

1583

_{10}= 62F

_{16}

**Example 6**

**Let's convert the number 1583.56 from the base-ten positional numeral system to the hexadecimal numeral system.**

_{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 |

1583

_{10}= 62F

_{16}

Let'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 | Â· | 16 | = | 8.96 |

0.96 | Â· | 16 | = | 15.36, 15 = F |

0.36 | Â· | 16 | = | 5.76 |

0.76 | Â· | 16 | = | 12.16, 12 = C |

0.16 | Â· | 16 | = | 2.56 |

0.56 | Â· | 16 | = | 8.96 |

0.96 | Â· | 16 | = | 15.36, 15 = F |

0.36 | Â· | 16 | = | 5.76 |

0.76 | Â· | 16 | = | 12.16, 12 = C |

0.16 | Â· | 16 | = | 2.56 |

0.56

_{10}= 0.8F5C28F5C2

_{16}

1583.56

_{10}= 62F.8F5C28F5C2

_{16}= 62F.(8F5C2)

_{16}

**Example 7**

**Let's convert the number A12DCF from the hexadecimal numeral system to the base-ten positional numeral system.**

_{16}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.

A

_{16}= 10

_{10}

D

_{16}= 13

_{10}

C

_{16}= 12

_{10}

F

_{16}= 15

_{10}

A12DCF_{16} = 10 â‹… 16^{5} + 1 â‹… 16^{4} + 2 â‹… 16^{3} + 13 â‹… 16^{2} + 12 â‹… 16^{1} + 15 â‹… 16^{0} = 10563023_{10}

**Example 8**

**Let's convert the number A12DCF.12A from the hexadecimal numeral system to the base-ten positional numeral system.**

_{16}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.

A

_{16}= 10

_{10}

D

_{16}= 13

_{10}

C

_{16}= 12

_{10}

F

_{16}= 15

_{10}

A12DCF.12A_{16} = 10 â‹… 16^{5} + 1 â‹… 16^{4} + 2 â‹… 16^{3} + 13 â‹… 16^{2} + 12 â‹… 16^{1} + 15 â‹… 16^{0} + 1 â‹… 16^{-1} + 2 â‹… 16^{-2} + 10 â‹… 16^{-3} = 10563023.07275390625_{10}

**Example 9**

**Let's convert the number 1010100011 the binary numeral system to the the hexadecimal numeral system.**

_{2}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.

1010100011_{2} = 1 â‹… 2^{9} + 0 â‹… 2^{8} + 1 â‹… 2^{7} + 0 â‹… 2^{6} + 1 â‹… 2^{5} + 0 â‹… 2^{4} + 0 â‹… 2^{3} + 0 â‹… 2^{2} + 1 â‹… 2^{1} + 1 â‹… 2^{0} = 675_{10}

_{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.

675 | : | 16 | = | 42 | remainder: 3 |

42 | : | 16 | = | 2 | remainder: 10, 10 = A |

2 | : | 16 | = | 0 | remainder: 2 |

675

_{10}= 2A3

_{16}

1010100011

_{2}= 2A3

_{16}