Difference between revisions of "TADM2E 1.2"

Latest revision as of 18:47, 17 May 2016

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) $

--Aroonalok (talk) 14:47, 17 May 2016 (EDT)