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)

@@ Line 1: / Line 1: @@
-<math>a = -2</math><br>
+Due to the symmetric nature of the problem (swapping ''a'' and ''b'' does not change the problem), let
-<math>b = 3</math><br><br>
-<math>min(a,b) = -2</math><br>
-<math>a * b = (-2) * 3 = -6</math><br><br><br>
--- Solution by Rajiv Seelam
+<math>min(a,b) = a</math>
+We now have to show that there exist some real numbers for which the following holds true:
+<math> ab<a </math>
+<math>\implies a(b-1)<0 </math>
+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'' :
+<math>a < 0 \and b-1>0</math>
+<math>\implies a<0 \and b>1</math>
+In this case, if one of the numbers is negative and the other is greater than 1 then <math> ab<min(a,b) </math>
+''Case II'' :
+<math>a > 0 \and b-1<0</math>
+<math>\implies a>0 \and b<1</math>
+Since, <math> min(a,b) = a </math>
+<math>\implies a<b</math>
+<math>\implies 0<a<b<1 </math>
+In this case, if both the numbers lie between 0 and 1 then <math> ab<min(a,b) </math>
+--[[User:Aroonalok|Aroonalok]] ([[User talk:Aroonalok|talk]]) 14:47, 17 May 2016 (EDT)