How to Compare two multiplications without multiplying? How to check if two multiplications are equal to each other or greater or lesser without actually multiplying them?
For example, compare (254)(847) and (383)(536)
EDIT:
While trying to find a rule i got one
(5)(11) < (6)(10)
or
(x)(y) < (x+1)(y-1) when y > x > 0
and another rule is  that if adding and subtracting 1 equates them the difference is one
(3)(5) + 1 = (4)(4)
(x)(y) + 1 = (x+1)(y-1) when y + 2 = x , y > x >= 0
 A: 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.
A: So, you want to compare (254)(847) and (383)(536) without actually computing the products, look at
$$\frac{847}{383} \; ? \; \frac{536}{254}$$
Multiplying both denominators by 2
$$\frac{847}{766} \; ? \; \frac{536}{508}$$
or
$$1+\frac{81}{766} \; ? \; 1+\frac{28}{508}$$
You are now left with comparing
$$\frac{81}{766} \; ? \; \frac{28}{508}$$
Applying the same trick as before, consider
$$\frac{81}{28} \; ? \; \frac{766}{508}$$
The left hand side is obviously bigger than $2$ while the right hand side is smaller, therefore, you can conclude that the original left hand side product was the bigger one, since none of the operations performed inverted the order of the inequalities.
A: In the worst case, where the comparison is close, you will always end up having to do as much work as multiplication. The performance gains from any non-worst-case improvements would have to be pretty good to make the programming effort worthwhile. Why do you need this?
A: Here is a simple example that does the job without any multiplication:
$5 \times 12$ vs $6 \times 11$ , rewrite both sides as $5\times(11+1)$ and $(5+1)\times11$, 
then we get $5\times11+5\times1$ vs $5\times11 + 1 \times 11$ subtracting $5\times11$ from both sides we get : $1\times5 vs 1\times11$ which does not require any multiplication at all. The same technique can be used on (254)(847) and (383)(536).
$ 254 \times 847 vs 383 \times 536 $
$ 254 \times ( 536 + 310 ) vs (254+129)\times536$
$ 254 \times 310 vs 129\times536$ continuing in similar fashion we end up with
$125\times 80 vs 4 \times 101$ although obvious but just for fun :
$ (101+24)\times(76+4) vs 4 \times 101$ where in the next line we get :
$ 4 \times 101 + 101\times 76 + 24\times76 + 24\times 4 vs 4\times 101 $  
A: Use Russian peasant multiplication? Which just involves addition and left-shifting. Then you aren't actually doing any multiplication.
