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;
- 1 319 935 175 ÷ 2 = 659 967 587 + 1;
- 659 967 587 ÷ 2 = 329 983 793 + 1;
- 329 983 793 ÷ 2 = 164 991 896 + 1;
- 164 991 896 ÷ 2 = 82 495 948 + 0;
- 82 495 948 ÷ 2 = 41 247 974 + 0;
- 41 247 974 ÷ 2 = 20 623 987 + 0;
- 20 623 987 ÷ 2 = 10 311 993 + 1;
- 10 311 993 ÷ 2 = 5 155 996 + 1;
- 5 155 996 ÷ 2 = 2 577 998 + 0;
- 2 577 998 ÷ 2 = 1 288 999 + 0;
- 1 288 999 ÷ 2 = 644 499 + 1;
- 644 499 ÷ 2 = 322 249 + 1;
- 322 249 ÷ 2 = 161 124 + 1;
- 161 124 ÷ 2 = 80 562 + 0;
- 80 562 ÷ 2 = 40 281 + 0;
- 40 281 ÷ 2 = 20 140 + 1;
- 20 140 ÷ 2 = 10 070 + 0;
- 10 070 ÷ 2 = 5 035 + 0;
- 5 035 ÷ 2 = 2 517 + 1;
- 2 517 ÷ 2 = 1 258 + 1;
- 1 258 ÷ 2 = 629 + 0;
- 629 ÷ 2 = 314 + 1;
- 314 ÷ 2 = 157 + 0;
- 157 ÷ 2 = 78 + 1;
- 78 ÷ 2 = 39 + 0;
- 39 ÷ 2 = 19 + 1;
- 19 ÷ 2 = 9 + 1;
- 9 ÷ 2 = 4 + 1;
- 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.
1 319 935 175(10) = 100 1110 1010 1100 1001 1100 1100 0111(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 1 319 935 175(10), a signed integer number (with sign),
converted from decimal system (from base 10)
and written as a signed binary (in base 2):
1 319 935 175(10) = 0100 1110 1010 1100 1001 1100 1100 0111
Spaces were used to group digits: for binary, by 4, for decimal, by 3.