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;
- 1 001 110 230 ÷ 2 = 500 555 115 + 0;
- 500 555 115 ÷ 2 = 250 277 557 + 1;
- 250 277 557 ÷ 2 = 125 138 778 + 1;
- 125 138 778 ÷ 2 = 62 569 389 + 0;
- 62 569 389 ÷ 2 = 31 284 694 + 1;
- 31 284 694 ÷ 2 = 15 642 347 + 0;
- 15 642 347 ÷ 2 = 7 821 173 + 1;
- 7 821 173 ÷ 2 = 3 910 586 + 1;
- 3 910 586 ÷ 2 = 1 955 293 + 0;
- 1 955 293 ÷ 2 = 977 646 + 1;
- 977 646 ÷ 2 = 488 823 + 0;
- 488 823 ÷ 2 = 244 411 + 1;
- 244 411 ÷ 2 = 122 205 + 1;
- 122 205 ÷ 2 = 61 102 + 1;
- 61 102 ÷ 2 = 30 551 + 0;
- 30 551 ÷ 2 = 15 275 + 1;
- 15 275 ÷ 2 = 7 637 + 1;
- 7 637 ÷ 2 = 3 818 + 1;
- 3 818 ÷ 2 = 1 909 + 0;
- 1 909 ÷ 2 = 954 + 1;
- 954 ÷ 2 = 477 + 0;
- 477 ÷ 2 = 238 + 1;
- 238 ÷ 2 = 119 + 0;
- 119 ÷ 2 = 59 + 1;
- 59 ÷ 2 = 29 + 1;
- 29 ÷ 2 = 14 + 1;
- 14 ÷ 2 = 7 + 0;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 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.
1 001 110 230(10) = 11 1011 1010 1011 1011 1010 1101 0110(2)
3. Determine the signed binary number bit length:
The base 2 number's actual length, in bits: 30.
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, 30,
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 1 001 110 230(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:
1 001 110 230(10) = 0011 1011 1010 1011 1011 1010 1101 0110
Spaces were used to group digits: for binary, by 4, for decimal, by 3.