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;
- 1 133 871 092 ÷ 2 = 566 935 546 + 0;
- 566 935 546 ÷ 2 = 283 467 773 + 0;
- 283 467 773 ÷ 2 = 141 733 886 + 1;
- 141 733 886 ÷ 2 = 70 866 943 + 0;
- 70 866 943 ÷ 2 = 35 433 471 + 1;
- 35 433 471 ÷ 2 = 17 716 735 + 1;
- 17 716 735 ÷ 2 = 8 858 367 + 1;
- 8 858 367 ÷ 2 = 4 429 183 + 1;
- 4 429 183 ÷ 2 = 2 214 591 + 1;
- 2 214 591 ÷ 2 = 1 107 295 + 1;
- 1 107 295 ÷ 2 = 553 647 + 1;
- 553 647 ÷ 2 = 276 823 + 1;
- 276 823 ÷ 2 = 138 411 + 1;
- 138 411 ÷ 2 = 69 205 + 1;
- 69 205 ÷ 2 = 34 602 + 1;
- 34 602 ÷ 2 = 17 301 + 0;
- 17 301 ÷ 2 = 8 650 + 1;
- 8 650 ÷ 2 = 4 325 + 0;
- 4 325 ÷ 2 = 2 162 + 1;
- 2 162 ÷ 2 = 1 081 + 0;
- 1 081 ÷ 2 = 540 + 1;
- 540 ÷ 2 = 270 + 0;
- 270 ÷ 2 = 135 + 0;
- 135 ÷ 2 = 67 + 1;
- 67 ÷ 2 = 33 + 1;
- 33 ÷ 2 = 16 + 1;
- 16 ÷ 2 = 8 + 0;
- 8 ÷ 2 = 4 + 0;
- 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.
1 133 871 092(10) = 100 0011 1001 0101 0111 1111 1111 0100(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 1 133 871 092(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:
1 133 871 092(10) = 0100 0011 1001 0101 0111 1111 1111 0100
Spaces were used to group digits: for binary, by 4, for decimal, by 3.