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;
- 1 111 100 212 ÷ 2 = 555 550 106 + 0;
- 555 550 106 ÷ 2 = 277 775 053 + 0;
- 277 775 053 ÷ 2 = 138 887 526 + 1;
- 138 887 526 ÷ 2 = 69 443 763 + 0;
- 69 443 763 ÷ 2 = 34 721 881 + 1;
- 34 721 881 ÷ 2 = 17 360 940 + 1;
- 17 360 940 ÷ 2 = 8 680 470 + 0;
- 8 680 470 ÷ 2 = 4 340 235 + 0;
- 4 340 235 ÷ 2 = 2 170 117 + 1;
- 2 170 117 ÷ 2 = 1 085 058 + 1;
- 1 085 058 ÷ 2 = 542 529 + 0;
- 542 529 ÷ 2 = 271 264 + 1;
- 271 264 ÷ 2 = 135 632 + 0;
- 135 632 ÷ 2 = 67 816 + 0;
- 67 816 ÷ 2 = 33 908 + 0;
- 33 908 ÷ 2 = 16 954 + 0;
- 16 954 ÷ 2 = 8 477 + 0;
- 8 477 ÷ 2 = 4 238 + 1;
- 4 238 ÷ 2 = 2 119 + 0;
- 2 119 ÷ 2 = 1 059 + 1;
- 1 059 ÷ 2 = 529 + 1;
- 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.
1 111 100 212(10) = 100 0010 0011 1010 0000 1011 0011 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 1 111 100 212(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:
1 111 100 212(10) = 0100 0010 0011 1010 0000 1011 0011 0100
Spaces were used to group digits: for binary, by 4, for decimal, by 3.