The question is to find the 2's complement of the binary number (-00101.110) in 8 bits.
So I don't understand what to do with the fractional part. I think the 1's complement will be 11010.001 Is 11010.010 the 2's complement? And are these the 8 bits that I need as my final answer of the 2s complement of -00101.110?
In two's complement notation, all of the most significant bits of a negative number are set to 1. Let's assume you're storing these numbers as 8 bits, with 2 to the right of the "binary point."
By definition, x + -x = 0, so we can write:
0.5 + -0.5 = 0.10 + 111111.10 = 0 // -0.5 = 111111.10
0.25 + -0.25 = 0.01 + 111111.11 = 0 // -0.25 = 111111.11
0.75 + -0.75 = 0.11 + 111111.01 = 0 // -0.75 = 111111.01
and so on.
Using 8 bits like this, the largest number you can store is
011111.11 = 31.75
the least-positive number is
000000.01 = 0.25
the least-negative number is
111111.11 = -0.25
and the smallest (that is, the most negative) is
100000.00 = -32
With decimal number systems we are use to having a units, tens and hundreds columns, with unsigned binary numbers this becomes 1,2,4 etc 2 to the power of column number.
For example
2^0 (1), 2^1 (2), 2^2 (4).
In Twos-complement the most significant bit (MSB) becomes negative. For a three bit number the rows would hold these values;
-4, 2, 1
0 0 1 => 1
1 0 0 => -4
1 0 1 => -4 + 1 = -3
So the value of the bits held by a fixed-point system is unchanged.
-1 will always be 111.000
-0.5 add 0.5 to it: 111.100
In your case 110100.10 is equal to -32+16+4+0.5 = -11.5. What you did was create -12 then add 0.5 rather than subtract 0.5.
What you actually want is -32+16+2+1+0.5 = -12.5 = 110011.1
