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;
- 1 145 324 602 ÷ 2 = 572 662 301 + 0;
- 572 662 301 ÷ 2 = 286 331 150 + 1;
- 286 331 150 ÷ 2 = 143 165 575 + 0;
- 143 165 575 ÷ 2 = 71 582 787 + 1;
- 71 582 787 ÷ 2 = 35 791 393 + 1;
- 35 791 393 ÷ 2 = 17 895 696 + 1;
- 17 895 696 ÷ 2 = 8 947 848 + 0;
- 8 947 848 ÷ 2 = 4 473 924 + 0;
- 4 473 924 ÷ 2 = 2 236 962 + 0;
- 2 236 962 ÷ 2 = 1 118 481 + 0;
- 1 118 481 ÷ 2 = 559 240 + 1;
- 559 240 ÷ 2 = 279 620 + 0;
- 279 620 ÷ 2 = 139 810 + 0;
- 139 810 ÷ 2 = 69 905 + 0;
- 69 905 ÷ 2 = 34 952 + 1;
- 34 952 ÷ 2 = 17 476 + 0;
- 17 476 ÷ 2 = 8 738 + 0;
- 8 738 ÷ 2 = 4 369 + 0;
- 4 369 ÷ 2 = 2 184 + 1;
- 2 184 ÷ 2 = 1 092 + 0;
- 1 092 ÷ 2 = 546 + 0;
- 546 ÷ 2 = 273 + 0;
- 273 ÷ 2 = 136 + 1;
- 136 ÷ 2 = 68 + 0;
- 68 ÷ 2 = 34 + 0;
- 34 ÷ 2 = 17 + 0;
- 17 ÷ 2 = 8 + 1;
- 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.
1 145 324 602(10) = 100 0100 0100 0100 0100 0100 0011 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) 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 1 145 324 602(10), a signed integer number (with sign),
converted from decimal system (from base 10)
and written as a signed binary (in base 2):
1 145 324 602(10) = 0100 0100 0100 0100 0100 0100 0011 1010
Spaces were used to group digits: for binary, by 4, for decimal, by 3.