1. Divide the number repeatedly by 2:
Keep track of each remainder.
We stop when we get a quotient that is equal to zero.
- division = quotient + remainder;
- 1 431 586 158 ÷ 2 = 715 793 079 + 0;
- 715 793 079 ÷ 2 = 357 896 539 + 1;
- 357 896 539 ÷ 2 = 178 948 269 + 1;
- 178 948 269 ÷ 2 = 89 474 134 + 1;
- 89 474 134 ÷ 2 = 44 737 067 + 0;
- 44 737 067 ÷ 2 = 22 368 533 + 1;
- 22 368 533 ÷ 2 = 11 184 266 + 1;
- 11 184 266 ÷ 2 = 5 592 133 + 0;
- 5 592 133 ÷ 2 = 2 796 066 + 1;
- 2 796 066 ÷ 2 = 1 398 033 + 0;
- 1 398 033 ÷ 2 = 699 016 + 1;
- 699 016 ÷ 2 = 349 508 + 0;
- 349 508 ÷ 2 = 174 754 + 0;
- 174 754 ÷ 2 = 87 377 + 0;
- 87 377 ÷ 2 = 43 688 + 1;
- 43 688 ÷ 2 = 21 844 + 0;
- 21 844 ÷ 2 = 10 922 + 0;
- 10 922 ÷ 2 = 5 461 + 0;
- 5 461 ÷ 2 = 2 730 + 1;
- 2 730 ÷ 2 = 1 365 + 0;
- 1 365 ÷ 2 = 682 + 1;
- 682 ÷ 2 = 341 + 0;
- 341 ÷ 2 = 170 + 1;
- 170 ÷ 2 = 85 + 0;
- 85 ÷ 2 = 42 + 1;
- 42 ÷ 2 = 21 + 0;
- 21 ÷ 2 = 10 + 1;
- 10 ÷ 2 = 5 + 0;
- 5 ÷ 2 = 2 + 1;
- 2 ÷ 2 = 1 + 0;
- 1 ÷ 2 = 0 + 1;
2. Construct the base 2 representation of the positive number:
Take all the remainders starting from the bottom of the list constructed above.
1 431 586 158(10) = 101 0101 0101 0100 0100 0101 0110 1110(2)
3. Determine the signed binary number bit length:
The base 2 number's actual length, in bits: 31.
A signed binary's bit length must be equal to a power of 2, as of:
21 = 2; 22 = 4; 23 = 8; 24 = 16; 25 = 32; 26 = 64; ...
The first bit (the leftmost) indicates the sign:
0 = positive integer number, 1 = negative integer number
The least number that is:
1) a power of 2
2) and is larger than the actual length, 31,
3) so that the first bit (leftmost) could be zero
(we deal with a positive number at this moment)
=== is: 32.
4. Get the positive binary computer representation on 32 bits (4 Bytes):
If needed, add extra 0s in front (to the left) of the base 2 number, up to the required length, 32.
Number 1 431 586 158(10), a signed integer number (with sign), converted from decimal system (from base 10) and written as a signed binary in one's complement representation:
1 431 586 158(10) = 0101 0101 0101 0100 0100 0101 0110 1110
Spaces were used to group digits: for binary, by 4, for decimal, by 3.