Signed 1's complement addition of (-65)+(+15)

In signed 1's complement form a negative number is the 1's complement (all bits inverted) of its positive value; the MSB is the sign bit. An end-around carry, if produced, is added back to the LSB.

Step 1: Represent the numbers in 8 bits

$$\begin{aligned} +65 &= 0100\,0001 \\ -65 &= 1011\,1110 \quad (\text{1's complement}) \\ +15 &= 0000\,1111 \end{aligned}$$

Step 2: Add

   1011 1110   (-65)
 + 0000 1111   (+15)
 -----------
   1100 1101

There is no end-around carry out of the MSB, so the result is already in final form. The sign bit is 1, so the result is negative and is in 1's complement form.

Step 3: Find the magnitude

$$\overline{1100\,1101} = 0011\,0010 = 32 + 16 + 2 = 50$$

Answer: $(-65) + (+15) = 1100\,1101$ in 8-bit signed 1's complement, which is $-50$. Check: $-65 + 15 = -50$. No overflow, because the operands have opposite signs.