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;
- 1 215 901 913 ÷ 2 = 607 950 956 + 1;
- 607 950 956 ÷ 2 = 303 975 478 + 0;
- 303 975 478 ÷ 2 = 151 987 739 + 0;
- 151 987 739 ÷ 2 = 75 993 869 + 1;
- 75 993 869 ÷ 2 = 37 996 934 + 1;
- 37 996 934 ÷ 2 = 18 998 467 + 0;
- 18 998 467 ÷ 2 = 9 499 233 + 1;
- 9 499 233 ÷ 2 = 4 749 616 + 1;
- 4 749 616 ÷ 2 = 2 374 808 + 0;
- 2 374 808 ÷ 2 = 1 187 404 + 0;
- 1 187 404 ÷ 2 = 593 702 + 0;
- 593 702 ÷ 2 = 296 851 + 0;
- 296 851 ÷ 2 = 148 425 + 1;
- 148 425 ÷ 2 = 74 212 + 1;
- 74 212 ÷ 2 = 37 106 + 0;
- 37 106 ÷ 2 = 18 553 + 0;
- 18 553 ÷ 2 = 9 276 + 1;
- 9 276 ÷ 2 = 4 638 + 0;
- 4 638 ÷ 2 = 2 319 + 0;
- 2 319 ÷ 2 = 1 159 + 1;
- 1 159 ÷ 2 = 579 + 1;
- 579 ÷ 2 = 289 + 1;
- 289 ÷ 2 = 144 + 1;
- 144 ÷ 2 = 72 + 0;
- 72 ÷ 2 = 36 + 0;
- 36 ÷ 2 = 18 + 0;
- 18 ÷ 2 = 9 + 0;
- 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.
1 215 901 913(10) = 100 1000 0111 1001 0011 0000 1101 1001(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 1 215 901 913(10), a signed integer number (with sign), converted from decimal system (from base 10) and written as a signed binary in two's complement representation:
1 215 901 913(10) = 0100 1000 0111 1001 0011 0000 1101 1001
Spaces were used to group digits: for binary, by 4, for decimal, by 3.