Generally such comparisons can be done efficiently via continued fractions, e.g.
$$\rm\displaystyle a = \frac{847}{383}\ =\ 2 + \cfrac{1}{4 + \cfrac{1}{1 + \cdots}}\ \ \Rightarrow\ \ 2+\frac{1}{4+1} < a < 2+\frac{1}4$$
$$\rm\displaystyle b = \frac{536}{254}\ =\ 2 + \cfrac{1}{9 + \cdots}\ \ \Rightarrow\ \ 2 < b < 2 + \frac{1}9 < 2+\frac{1}5 < a$$
The comparison of the continued fraction coefficients can be done in parallel with the computation of the continued fraction. Namely, compare the integer parts. If they are unequal then that will determine the inequality. Otherwise, recurse on the (inverses of) the fractional parts (and note that inversion reverses the inequality). For the above example this yields:
$$\rm\displaystyle\ \frac{847}{383} > \frac{536}{254}\ \iff\ \frac{81}{383}>\frac{28}{254}\ \iff\ \frac{254}{28}>\frac{383}{81}\ \Leftarrow\ \ 9 > 4$$
In words: to test if $\rm\:847/383 > 536/254\: $ we first compare their integer parts (floor). They both have integer part $\:2\:$ so we subtract $\:2\:$ from both and reduce to comparing their fractional parts $\rm\ 81/383,\ \ 28/254\:.$ To do this we invert them and recurse. But since inversion reverses inequalities $\rm\ x < y\ \iff\ 1/y < 1/x\ $ (for $\rm\:xy>0\:$), the equivalent inequality to test is if $\rm\ 254/28 > 383/81\:.\ $ Comparing their integer parts $\rm\:m,\:n\:$ we see since $\rm\ m > 5 > n\:$ so the inequality is true. This leads to the following simple algorithm that essentially compares any two real numbers via the lex order of their continued fraction coefficients (note: it will not terminate if given two equal reals with infinite continued fraction).
$\rm compare\_reals\:(A,\: B)\ := \qquad\qquad\qquad\quad\ \ \color{blue}{\ // \ computes\ \ sgn(A - B) }$
$\rm\quad\ let\ [\ n_1\leftarrow \lfloor A\rfloor\:;\ \ \ n_2\leftarrow \lfloor B\rfloor\ ] \ \qquad\qquad\quad \color{blue}{\ //\ compare\ integer\ parts}$
$\rm\quad\quad if\ \ n_1 \ne n_2\ \ then\ \ return\ \ sgn(n_1 - n_2)\:;$
$\rm\quad\quad let\ [\ a \leftarrow A - n_1\:;\ \ \ b \leftarrow B - n_2\ ] \quad\quad\quad \color{blue}{//\ compute\ fractional\ parts\ \ a,\: b }$
$\rm\quad\quad\quad if\ \ a\:b=0\ \ then\ \ return\ \ sgn(a-b)\:;$
$\rm\quad\quad\quad compare\_reals(b^{-1}\:, a^{-1})\:;\qquad\qquad \color{blue}{\ //\ \text{recurse on inverses of fractional parts}}$
Equivalently one can employ Farey fractions and mediants. Generally such approaches will be quite efficient due to the best approximations properties of the algorithm. For a nice example see my post here using continued fractions to solve the old chestnut of comparing $\ 7^\sqrt 8\ $ to $\ 8^\sqrt 7\ $ and see also this thread where some folks struggled to prove this by calculus.
Best Answer
$0.8 \times $ something is the same as taking $80\%$ and this is less than $100\%$.