0.1=2−4*(1+ |
| + |
| + |
| + |
| + |
| +...)= 2−4* |
| ( |
| + |
| ) |
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).
3f (00111111), b9 (10111001), 99 (10011001), 99 (10011001), 99 (10011001), 99 (10011001), 99 (10011001), 9a (10100000),Therefore a>0.1 and a−0.1=1/250+1/251 (since 100000-11010=110)
Remark
This is the reason why
returns 9 and not 10 when Digits:=14.