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;
- 2 130 705 662 ÷ 2 = 1 065 352 831 + 0;
- 1 065 352 831 ÷ 2 = 532 676 415 + 1;
- 532 676 415 ÷ 2 = 266 338 207 + 1;
- 266 338 207 ÷ 2 = 133 169 103 + 1;
- 133 169 103 ÷ 2 = 66 584 551 + 1;
- 66 584 551 ÷ 2 = 33 292 275 + 1;
- 33 292 275 ÷ 2 = 16 646 137 + 1;
- 16 646 137 ÷ 2 = 8 323 068 + 1;
- 8 323 068 ÷ 2 = 4 161 534 + 0;
- 4 161 534 ÷ 2 = 2 080 767 + 0;
- 2 080 767 ÷ 2 = 1 040 383 + 1;
- 1 040 383 ÷ 2 = 520 191 + 1;
- 520 191 ÷ 2 = 260 095 + 1;
- 260 095 ÷ 2 = 130 047 + 1;
- 130 047 ÷ 2 = 65 023 + 1;
- 65 023 ÷ 2 = 32 511 + 1;
- 32 511 ÷ 2 = 16 255 + 1;
- 16 255 ÷ 2 = 8 127 + 1;
- 8 127 ÷ 2 = 4 063 + 1;
- 4 063 ÷ 2 = 2 031 + 1;
- 2 031 ÷ 2 = 1 015 + 1;
- 1 015 ÷ 2 = 507 + 1;
- 507 ÷ 2 = 253 + 1;
- 253 ÷ 2 = 126 + 1;
- 126 ÷ 2 = 63 + 0;
- 63 ÷ 2 = 31 + 1;
- 31 ÷ 2 = 15 + 1;
- 15 ÷ 2 = 7 + 1;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 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.
2 130 705 662(10) = 111 1110 1111 1111 1111 1100 1111 1110(2)
4. 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) indicates 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.
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.
2 130 705 662(10) = 0111 1110 1111 1111 1111 1100 1111 1110
6. Get the negative integer number representation. Part 1:
To write the negative integer number on 32 bits (4 Bytes),
as a signed binary in one's complement representation,
... replace all the bits on 0 with 1s and all the bits set on 1 with 0s.
Reverse the digits, flip the digits:
Replace the bits set on 0 with 1s and the bits set on 1 with 0s.
!(0111 1110 1111 1111 1111 1100 1111 1110)
= 1000 0001 0000 0000 0000 0011 0000 0001
7. Get the negative integer number representation. Part 2:
To write the negative integer number on 32 bits (4 Bytes),
as a signed binary in two's complement representation,
add 1 to the number calculated above
1000 0001 0000 0000 0000 0011 0000 0001
(to the signed binary in one's complement representation)
Binary addition carries on a value of 2:
0 + 0 = 0
0 + 1 = 1
1 + 1 = 10
1 + 10 = 11
1 + 11 = 100
Add 1 to the number calculated above
(to the signed binary number in one's complement representation):
-2 130 705 662 =
1000 0001 0000 0000 0000 0011 0000 0001 + 1
Number -2 130 705 662(10), a signed integer number (with sign), converted from decimal system (from base 10) and written as a signed binary in two's complement representation:
-2 130 705 662(10) = 1000 0001 0000 0000 0000 0011 0000 0010
Spaces were used to group digits: for binary, by 4, for decimal, by 3.