Binary representation of 129 when using 8bits for two's complement? I have this confusion regarding binary representation of decimal value 129 (or even 128). If 8 bits are used to represent numbers when doing the two's complement, then we know that '00000000' to '01111111' are used for 0 to 127 and leading 1 is used for negative numbers i.e. 1xxxxxxx where x is any combination of 0's and 1's to represent the value of that negative number. 
But then how is 128 represented in 8 bits because the first bit is reserved for telling the number is negative (basically a signed number)?
The range of integers representable in 8 bits using two's complement is –128 to +127.

If instead the numbers are unsigned, then the range becomes 0 to 255.