# Difference between revisions of "TADM2E 1.2"

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In this case, if both the numbers lie between 0 and 1 then <math> ab<min(a,b) </math> | In this case, if both the numbers lie between 0 and 1 then <math> ab<min(a,b) </math> | ||

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+ | --[[User:Aroonalok|Aroonalok]] ([[User talk:Aroonalok|talk]]) 14:47, 17 May 2016 (EDT) |

## 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) $