# Difference between revisions of "TADM2E 1.2"

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)