2. 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;
- 20 230 405 ÷ 2 = 10 115 202 + 1;
- 10 115 202 ÷ 2 = 5 057 601 + 0;
- 5 057 601 ÷ 2 = 2 528 800 + 1;
- 2 528 800 ÷ 2 = 1 264 400 + 0;
- 1 264 400 ÷ 2 = 632 200 + 0;
- 632 200 ÷ 2 = 316 100 + 0;
- 316 100 ÷ 2 = 158 050 + 0;
- 158 050 ÷ 2 = 79 025 + 0;
- 79 025 ÷ 2 = 39 512 + 1;
- 39 512 ÷ 2 = 19 756 + 0;
- 19 756 ÷ 2 = 9 878 + 0;
- 9 878 ÷ 2 = 4 939 + 0;
- 4 939 ÷ 2 = 2 469 + 1;
- 2 469 ÷ 2 = 1 234 + 1;
- 1 234 ÷ 2 = 617 + 0;
- 617 ÷ 2 = 308 + 1;
- 308 ÷ 2 = 154 + 0;
- 154 ÷ 2 = 77 + 0;
- 77 ÷ 2 = 38 + 1;
- 38 ÷ 2 = 19 + 0;
- 19 ÷ 2 = 9 + 1;
- 9 ÷ 2 = 4 + 1;
- 4 ÷ 2 = 2 + 0;
- 2 ÷ 2 = 1 + 0;
- 1 ÷ 2 = 0 + 1;
3. Construct the base 2 representation of the positive number:
Take all the remainders starting from the bottom of the list constructed above.
20 230 405(10) = 1 0011 0100 1011 0001 0000 0101(2)
4. Determine the signed binary number bit length:
The base 2 number's actual length, in bits: 25.
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, 25,
3) so that the first bit (leftmost) could be zero
(we deal with a positive number at this moment)
=== is: 32.
5. 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:
20 230 405(10) = 0000 0001 0011 0100 1011 0001 0000 0101
6. Get the negative integer number representation:
To get the negative integer number representation on 32 bits (4 Bytes),
... change the first bit (the leftmost), from 0 to 1...
Number -20 230 405(10), a signed integer number (with sign),
converted from decimal system (from base 10)
and written as a signed binary (in base 2):
-20 230 405(10) = 1000 0001 0011 0100 1011 0001 0000 0101
Spaces were used to group digits: for binary, by 4, for decimal, by 3.