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10.1.4  Difference between the representation of (3.1-3) and of 0.1

For the representation of 0.1:

     
0.1
=2−4·(1+
1
2
+
1
24
+
1
25
+
1
28
+
1
29
+…)
         
 
 = 2−4·
k=0
(
1
24*k
+
1
24*k+1
         

hence α=1 and

m=
1
2
k=1
(
1
24*k
+
1
24*k+1
),

therefore the representation of 0.1 is

3f (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).

For the representation of a:=3.1−3:
Computing a is done by adjusting exponents (here nothing to do), then subtracting the mantissa and adjusting the exponent of the result to have a normalized float. The exponent is α=−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 5 positions and you get:

3f (00111111), b9 (10111001), 99 (10011001), 99 (10011001),
99 (10011001), 99 (10011001), 99 (10011001), a0 (10100000),

Therefore, a>0.1 and a−0.1=1/250+1/251 (since 100000-11010=110).


Remark:
This is the reason why:
Input:

floor(1/(3.1-3))

returns 9 and not 10 when Digits:=14.


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