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;
- 18 751 818 ÷ 2 = 9 375 909 + 0;
- 9 375 909 ÷ 2 = 4 687 954 + 1;
- 4 687 954 ÷ 2 = 2 343 977 + 0;
- 2 343 977 ÷ 2 = 1 171 988 + 1;
- 1 171 988 ÷ 2 = 585 994 + 0;
- 585 994 ÷ 2 = 292 997 + 0;
- 292 997 ÷ 2 = 146 498 + 1;
- 146 498 ÷ 2 = 73 249 + 0;
- 73 249 ÷ 2 = 36 624 + 1;
- 36 624 ÷ 2 = 18 312 + 0;
- 18 312 ÷ 2 = 9 156 + 0;
- 9 156 ÷ 2 = 4 578 + 0;
- 4 578 ÷ 2 = 2 289 + 0;
- 2 289 ÷ 2 = 1 144 + 1;
- 1 144 ÷ 2 = 572 + 0;
- 572 ÷ 2 = 286 + 0;
- 286 ÷ 2 = 143 + 0;
- 143 ÷ 2 = 71 + 1;
- 71 ÷ 2 = 35 + 1;
- 35 ÷ 2 = 17 + 1;
- 17 ÷ 2 = 8 + 1;
- 8 ÷ 2 = 4 + 0;
- 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.
18 751 818(10) = 1 0001 1110 0010 0001 0100 1010(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:
18 751 818(10) = 0000 0001 0001 1110 0010 0001 0100 1010
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 -18 751 818(10), a signed integer number (with sign),
converted from decimal system (from base 10)
and written as a signed binary (in base 2):
-18 751 818(10) = 1000 0001 0001 1110 0010 0001 0100 1010
Spaces were used to group digits: for binary, by 4, for decimal, by 3.