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;
- 1 100 110 058 ÷ 2 = 550 055 029 + 0;
- 550 055 029 ÷ 2 = 275 027 514 + 1;
- 275 027 514 ÷ 2 = 137 513 757 + 0;
- 137 513 757 ÷ 2 = 68 756 878 + 1;
- 68 756 878 ÷ 2 = 34 378 439 + 0;
- 34 378 439 ÷ 2 = 17 189 219 + 1;
- 17 189 219 ÷ 2 = 8 594 609 + 1;
- 8 594 609 ÷ 2 = 4 297 304 + 1;
- 4 297 304 ÷ 2 = 2 148 652 + 0;
- 2 148 652 ÷ 2 = 1 074 326 + 0;
- 1 074 326 ÷ 2 = 537 163 + 0;
- 537 163 ÷ 2 = 268 581 + 1;
- 268 581 ÷ 2 = 134 290 + 1;
- 134 290 ÷ 2 = 67 145 + 0;
- 67 145 ÷ 2 = 33 572 + 1;
- 33 572 ÷ 2 = 16 786 + 0;
- 16 786 ÷ 2 = 8 393 + 0;
- 8 393 ÷ 2 = 4 196 + 1;
- 4 196 ÷ 2 = 2 098 + 0;
- 2 098 ÷ 2 = 1 049 + 0;
- 1 049 ÷ 2 = 524 + 1;
- 524 ÷ 2 = 262 + 0;
- 262 ÷ 2 = 131 + 0;
- 131 ÷ 2 = 65 + 1;
- 65 ÷ 2 = 32 + 1;
- 32 ÷ 2 = 16 + 0;
- 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.
1 100 110 058(10) = 100 0001 1001 0010 0101 1000 1110 1010(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 1 100 110 058(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:
1 100 110 058(10) = 0100 0001 1001 0010 0101 1000 1110 1010
Spaces were used to group digits: for binary, by 4, for decimal, by 3.