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;
- 1 110 100 130 ÷ 2 = 555 050 065 + 0;
- 555 050 065 ÷ 2 = 277 525 032 + 1;
- 277 525 032 ÷ 2 = 138 762 516 + 0;
- 138 762 516 ÷ 2 = 69 381 258 + 0;
- 69 381 258 ÷ 2 = 34 690 629 + 0;
- 34 690 629 ÷ 2 = 17 345 314 + 1;
- 17 345 314 ÷ 2 = 8 672 657 + 0;
- 8 672 657 ÷ 2 = 4 336 328 + 1;
- 4 336 328 ÷ 2 = 2 168 164 + 0;
- 2 168 164 ÷ 2 = 1 084 082 + 0;
- 1 084 082 ÷ 2 = 542 041 + 0;
- 542 041 ÷ 2 = 271 020 + 1;
- 271 020 ÷ 2 = 135 510 + 0;
- 135 510 ÷ 2 = 67 755 + 0;
- 67 755 ÷ 2 = 33 877 + 1;
- 33 877 ÷ 2 = 16 938 + 1;
- 16 938 ÷ 2 = 8 469 + 0;
- 8 469 ÷ 2 = 4 234 + 1;
- 4 234 ÷ 2 = 2 117 + 0;
- 2 117 ÷ 2 = 1 058 + 1;
- 1 058 ÷ 2 = 529 + 0;
- 529 ÷ 2 = 264 + 1;
- 264 ÷ 2 = 132 + 0;
- 132 ÷ 2 = 66 + 0;
- 66 ÷ 2 = 33 + 0;
- 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.
1 110 100 130(10) = 100 0010 0010 1010 1100 1000 1010 0010(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 1 110 100 130(10), a signed integer number (with sign),
converted from decimal system (from base 10)
and written as a signed binary (in base 2):
1 110 100 130(10) = 0100 0010 0010 1010 1100 1000 1010 0010
Spaces were used to group digits: for binary, by 4, for decimal, by 3.