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;
- 1 155 649 533 ÷ 2 = 577 824 766 + 1;
- 577 824 766 ÷ 2 = 288 912 383 + 0;
- 288 912 383 ÷ 2 = 144 456 191 + 1;
- 144 456 191 ÷ 2 = 72 228 095 + 1;
- 72 228 095 ÷ 2 = 36 114 047 + 1;
- 36 114 047 ÷ 2 = 18 057 023 + 1;
- 18 057 023 ÷ 2 = 9 028 511 + 1;
- 9 028 511 ÷ 2 = 4 514 255 + 1;
- 4 514 255 ÷ 2 = 2 257 127 + 1;
- 2 257 127 ÷ 2 = 1 128 563 + 1;
- 1 128 563 ÷ 2 = 564 281 + 1;
- 564 281 ÷ 2 = 282 140 + 1;
- 282 140 ÷ 2 = 141 070 + 0;
- 141 070 ÷ 2 = 70 535 + 0;
- 70 535 ÷ 2 = 35 267 + 1;
- 35 267 ÷ 2 = 17 633 + 1;
- 17 633 ÷ 2 = 8 816 + 1;
- 8 816 ÷ 2 = 4 408 + 0;
- 4 408 ÷ 2 = 2 204 + 0;
- 2 204 ÷ 2 = 1 102 + 0;
- 1 102 ÷ 2 = 551 + 0;
- 551 ÷ 2 = 275 + 1;
- 275 ÷ 2 = 137 + 1;
- 137 ÷ 2 = 68 + 1;
- 68 ÷ 2 = 34 + 0;
- 34 ÷ 2 = 17 + 0;
- 17 ÷ 2 = 8 + 1;
- 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.
1 155 649 533(10) = 100 0100 1110 0001 1100 1111 1111 1101(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 1 155 649 533(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:
1 155 649 533(10) = 0100 0100 1110 0001 1100 1111 1111 1101
Spaces were used to group digits: for binary, by 4, for decimal, by 3.