2's complement representation of fractions?

Question

I\'m a little lost on this. I need to use two fractional bits 0.(a-1)(a-2)

Like that, now I can use .00 .01 .10 and .11 But I nee

Answer 1

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

Answer 2

A number stored in two's complement inverts the sign of the uppermost bit's magnitude (so that for e.g. a 16-bit number, the upper bit is -32768 rather than +32768). All other bits behave as normal. Consequently, when performing math on multi-word numbers, the upper word of each number should be regarded as two's-complement (since its uppermost bit will be the uppermost bit of the overall number), but all other words in each number should be regarded as unsigned quantities.

For example, a 16-bit two's complement number has place values (-32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, and 1). Split into two 8-bit parts, those parts will have place values (-32768, 16384, 8192, 4096, 2048, 1024, 512, and 256); and (128, 64, 32, 16, 8, 4, 2, and 1). The first set of values is in a two's complement 8-bit number, times 256; the latter set is an unsigned 8-bit number.

Answer 3

see it this way:

you have normal binary representation

let's assume 8 bit words ...

the first bit (MSB) has the value 128, the second 64, and so on ...

in other words the first bit (MSB) is 2^7 ... the second bit is 2^6 ... and the last bit is 2^0

now we can assume our 8 bit word has 2 decimal places ....

we now start with the first bit (MSB) 2^5 and end with the last bit beeing 2^-2

no magic here ...

now to turn that into binary complement: simply negate the value of the first bit

so instead of 2^5 it would be -2^5

so base 10 -0.75 would be in binary complement
111111.01 ...
(1*(-32) + 1*16 + 1*8 + 1*4 + 1*2 +1*1 + 0*0.5 + 1*0.25)
(1*(-2^5) + 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 +1*2^0 + 0*2^(-1) + 1*2^(-2))