TADM2E 1.2
From Algorithm Wiki
Due to the symmetric nature of the problem (swapping a and b does not change the problem), let
$ min(a,b) = a $
We now have to show that there exist some real numbers for which the following holds true:
$ ab<a $ $ \implies a(b-1)<0 $
So, the product of two numbers (a and b-1) is negative. This means one of the numbers is negative and the other is positive.
Case I : $ a < 0 \and b-1>0 $ $ \implies a<0 \and b>1 $
In this case, if one of the numbers is negative and the other is greater than 1 then $ ab<min(a,b) $
Case II : $ a > 0 \and b-1<0 $ $ \implies a>0 \and b<1 $ Since, $ min(a,b) = a $ $ \implies a<b $ $ \implies 0<a<b<1 $
In this case, if both the numbers lie between 0 and 1 then $ ab<min(a,b) $
