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;
- 1 423 235 310 ÷ 2 = 711 617 655 + 0;
- 711 617 655 ÷ 2 = 355 808 827 + 1;
- 355 808 827 ÷ 2 = 177 904 413 + 1;
- 177 904 413 ÷ 2 = 88 952 206 + 1;
- 88 952 206 ÷ 2 = 44 476 103 + 0;
- 44 476 103 ÷ 2 = 22 238 051 + 1;
- 22 238 051 ÷ 2 = 11 119 025 + 1;
- 11 119 025 ÷ 2 = 5 559 512 + 1;
- 5 559 512 ÷ 2 = 2 779 756 + 0;
- 2 779 756 ÷ 2 = 1 389 878 + 0;
- 1 389 878 ÷ 2 = 694 939 + 0;
- 694 939 ÷ 2 = 347 469 + 1;
- 347 469 ÷ 2 = 173 734 + 1;
- 173 734 ÷ 2 = 86 867 + 0;
- 86 867 ÷ 2 = 43 433 + 1;
- 43 433 ÷ 2 = 21 716 + 1;
- 21 716 ÷ 2 = 10 858 + 0;
- 10 858 ÷ 2 = 5 429 + 0;
- 5 429 ÷ 2 = 2 714 + 1;
- 2 714 ÷ 2 = 1 357 + 0;
- 1 357 ÷ 2 = 678 + 1;
- 678 ÷ 2 = 339 + 0;
- 339 ÷ 2 = 169 + 1;
- 169 ÷ 2 = 84 + 1;
- 84 ÷ 2 = 42 + 0;
- 42 ÷ 2 = 21 + 0;
- 21 ÷ 2 = 10 + 1;
- 10 ÷ 2 = 5 + 0;
- 5 ÷ 2 = 2 + 1;
- 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.
1 423 235 310(10) = 101 0100 1101 0100 1101 1000 1110 1110(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 1 423 235 310(10), a signed integer number (with sign),
converted from decimal system (from base 10)
and written as a signed binary (in base 2):
1 423 235 310(10) = 0101 0100 1101 0100 1101 1000 1110 1110
Spaces were used to group digits: for binary, by 4, for decimal, by 3.