Using the unsigned binary number system we know:
1 is 2^0 = 1
10 is 2^1 = 2
100 is 2^2 = 4
...
1000 0000 is 2^7 = 128
We know the place values for 8 bits
UNSIGNED are:
128 64 32 16 8 4 2 1
--- -- -- -- - - - -
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
SO: 1 0 0 0 0 0 1 0
= 128 + 2 = 130
To express negative numbers, the scheme of the binary positional number
system is simply changed slightly - with a large impact.
Using n bits, the
nth bit is simply changed from positive to negative.
The place values for n=8 bits
SIGNED are:
-128 64 32 16 8 4 2 1
--- -- -- -- - - - -
-2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
SO: 1 0 0 0 0 0 1 0
= -128 + 2 = -126
Numbers using this scheme can be negated by taking the two's
complement of the number - flipping the bits and adding one.
0000 1111 = 15 = 8 + 4 + 2 + 1
1111 0001 = -15 = -128 + 64 + 32 + 16 + 1 (flip all bits and add one)
The high bit, (nth it) becomes the "sign bit" only because, if set,
it indicates the number must be negative - it actually represents -2^(n-1).
| Binary |
Decimal Unsigned |
Decimal Signed |
| a. 1111 1111 |
255 |
-1 |
| b. 0000 0000 |
0 |
0 |
| c. 1000 0000 |
128 |
-128 |
| d. 0111 1111 |
127 |
127 |
| e. 1000 0001 |
129 |
-127 (note: flip bits and add 1 from d. above) |
| f. 1000 1111 |
143 |
-113 |
| g. 0000 1111 |
15 |
15 |
| h. 1111 0001 |
241 |
-15 (note: flip bits and add 1 from g. above) |
n > 8 bits
==========
n = 16
======
Nothing changes - an elegant scheme.
If n = 16, the nth bit (2^(n-1)) becomes -32,768
-32768 .. 128 64 32 16 8 4 2 1
----- --- -- -- -- - - - -
2^15 .. 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
SO: 1 .. 0 0 0 0 0 0 1 0
= -32768 + 2 = -32766
n = 32
======
If n = 32, the nth bit (2^(n-1)) becomes -2,147,483,648
--------------------------------------
Interestingly, n '1' bits (111..111) is ALWAYS -1 decimal.
Why?