- representation of 0.1 :

We have :0.1 = 2hence = 1 and^{-4}*(1 + + + + + + ...) = 2^{-4}*( + )*m*= + ( + ), therefore the representation of 0.1 is3f (00111111), b9 (10111001), 99 (10011001), 99 (10011001), 99 (10011001), 99 (10011001), 99 (10011001), 9a (10011010),

the last octet is 1010, indeed the 2 last bits 01 became 10 because the following digit is 1 (upper rounding). - representation of a:=3.1-3 :

Computing a is done by adjusting exponents (here nothing to do), then substract the mantissa, and adjust the exponent of the result to have a normalized float. The exponent is = - 4 (that corresponds at 2*2^{-5}) and the bits corresponding to the mantissa begin at 1/2 = 2*2^{-6}: the bits of the mantissa are shifted to the left of 5 positions and we have :3f (00111111), b9 (10111001), 99 (10011001), 99 (10011001), 99 (10011001), 99 (10011001), 99 (10011001), 9a (10100000),

Therefore*a*> 0.1 and*a*- 0.1 = 1/2^{50}+1/2^{51}(since 100000-11010=110)

This is the reason why