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;
- 1 323 752 255 ÷ 2 = 661 876 127 + 1;
- 661 876 127 ÷ 2 = 330 938 063 + 1;
- 330 938 063 ÷ 2 = 165 469 031 + 1;
- 165 469 031 ÷ 2 = 82 734 515 + 1;
- 82 734 515 ÷ 2 = 41 367 257 + 1;
- 41 367 257 ÷ 2 = 20 683 628 + 1;
- 20 683 628 ÷ 2 = 10 341 814 + 0;
- 10 341 814 ÷ 2 = 5 170 907 + 0;
- 5 170 907 ÷ 2 = 2 585 453 + 1;
- 2 585 453 ÷ 2 = 1 292 726 + 1;
- 1 292 726 ÷ 2 = 646 363 + 0;
- 646 363 ÷ 2 = 323 181 + 1;
- 323 181 ÷ 2 = 161 590 + 1;
- 161 590 ÷ 2 = 80 795 + 0;
- 80 795 ÷ 2 = 40 397 + 1;
- 40 397 ÷ 2 = 20 198 + 1;
- 20 198 ÷ 2 = 10 099 + 0;
- 10 099 ÷ 2 = 5 049 + 1;
- 5 049 ÷ 2 = 2 524 + 1;
- 2 524 ÷ 2 = 1 262 + 0;
- 1 262 ÷ 2 = 631 + 0;
- 631 ÷ 2 = 315 + 1;
- 315 ÷ 2 = 157 + 1;
- 157 ÷ 2 = 78 + 1;
- 78 ÷ 2 = 39 + 0;
- 39 ÷ 2 = 19 + 1;
- 19 ÷ 2 = 9 + 1;
- 9 ÷ 2 = 4 + 1;
- 4 ÷ 2 = 2 + 0;
- 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.
1 323 752 255(10) = 100 1110 1110 0110 1101 1011 0011 1111(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) is reserved for 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 1 323 752 255(10), a signed integer number (with sign),
converted from decimal system (from base 10)
and written as a signed binary (in base 2):
1 323 752 255(10) = 0100 1110 1110 0110 1101 1011 0011 1111
Spaces were used to group digits: for binary, by 4, for decimal, by 3.