Why is 128 in one and two's complement using 8 bits overflow?

Suppose I want to represent 128 in one and two's complement using 8 bits, no sign bit

Wouldn't that be:

One's complement: 0111 1111

Two's complement: 0111 1110

No overflow

But the correct answer is:

One's complement: 0111 1111

Two's complement: 0111 1111

Overflow

Additional Question:

How come 1 in one's and two's complement is 0000 0001 and 0000 0001 respectively. How come you don't flip the bits like we did with 128?

One's and Two's Complements are both ways to represent signed integers.

For One's Complement Representation:

• Positive Numbers: Represented with its regular binary representation
  • For example: decimal value 1 will be represented in 8 bit One's Complement as 0000 0001
• Negative Numbers: Represented by complementing the binary representation of its magnitude
  • For example: decimal value of -127 will be represented in 8 bit One's Complement as 1000 0000 because the binary representation of 127 is 0111 1111 when complemented that will be 1000 0000

For Two's Complement Representation:

• Positive Numbers: Represented with its regular binary representation
  • For example: decimal value 1 will be represented in 8 bit One's Complement as 0000 0001
• Negative Numbers: Represented by complementing the binary representation of its magnitude then adding 1 to the value
  • For example: decimal value of -127 will be represented in 8 bit One's Complement as 1000 0001 because the binary representation of 127 is 0111 1111 when complemented that will be 1000 0000 then add 0000 0001 to get 1000 0001

Therefore, 128 overflows in both instances because the binary representation of 128 is 1000 0000 which in ones complement represents -127 and in twos complement represents -128. In order to be able to represent 128 in both ones and twos complement you would need 9 bits and it would be represented as 0 1000 0000.