1. Negative Numbers An alternative way to find the 2's complement Starting at the right 1. Copy bits until you reach the first 1. 2. Flip all the remaining bits. 0 0 1 1 1 0 (14)1 1 0 0 1 0 (-14) 1 1 0 0 1 0 (-14)0 0 1 1 1 0 (14)

2. Floating Point Numbers Positional Notation Redux 2 3 2 2 2 1 2 0. 2 -1 2 -2 2 -3 8 4 2 1. ½ ¼ ⅛ Thus 101.101 = 4 + 1 + ½ + ⅛ = 5⅜ And 10.011 +100.11_ 110.111

3. Floating Point Numbers Another Darn Notation Excess value notation E.g. with 3 bits we can represent 8 things, the excess value is half of that or 4. The idea is that the number that you see appears 4 greater than it really is!!??? Thus000(0) -> -4 011(3) -> -1 100(4) -> 0 111(7) -> 3

4. Floating Point Numbers So how is 101.101 2 (5⅜ 10 ) actually stored. In a way similar to scientific notation. For example, we might store 5⅜ in scientific notation as follows: +0.5375 x 10 1 where the important parts are exponent (excess 4) [e below] fractional part (mantissa) [m] sign [s] Now in the computer we have s eee mmmm for 8 bits and 101.101 -> +0.101101 x 2 3 -> 0 111 1011 truncation error

5. Floating Point Numbers Decode the following 8 bit representation of a floating point number. 101010111 – sign bit -> negative 010 – exponent -> -2 -0.1011 x 2 -2 or excess 4-0.001011 2 Now, -0.001011 2 = -(2 -3 + 2 -5 + 2 -6 ) or = -(0.125 +.03125 + 0.015625) = -0.171875

6. Floating Point Numbers Encode the following number as an 8 bit representation of a floating point number. 3 = 3.15625 now consider 0.15625 < 2 -1 and 2 -2 so subtr. 2 -3 -0.125 which leaves 0.03125 < 2 -4 so subtr. 2 -5 -0.03125 which leaves 0.00000 Thus 3.15625 10 = 11.00101 2 = +0.1100101 x 2 2 The exponent 2 is 110(6) in excess 4 notation The final form is 01101100