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

• representation of 0.1 :
  We have :
  0.1 = 2^-4*(1 + $${\frac{{1}}{{2}}}$$ + $${\frac{{1}}{{2^4}}}$$ + $${\frac{{1}}{{2^5}}}$$ + $${\frac{{1}}{{2^8}}}$$ + $${\frac{{1}}{{2^9}}}$$ + ...) = 2^-4*$$\sum_{{k=0}}^{\infty}$$($${\frac{{1}}{{2^{4*k}}}}$$ + $${\frac{{1}}{{2^{4*k+1}}}}$$)
  hence $\alpha$ = 1 and m = ${\frac{{1}}{{2}}}$ + $\sum_{{k=1}}^{\infty}$(${\frac{{1}}{{2^{4*k}}}}$ + ${\frac{{1}}{{2^{4*k+1}}}}$), 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).

• 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 $\alpha$ = - 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 = 1/2⁵⁰ +1/2⁵¹ (since 100000-11010=110)

Remark
This is the reason why

floor(1/(3.1-3))

returns 9 and not 10 when Digits:=14.