Difference between revisions of "TADM2E 1.2"

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<math>a = -2</math><br>
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Due to the symmetric nature of the problem (swapping ''a'' and ''b'' does not change the problem), let
<math>b = 3</math><br><br>
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<math>min(a,b) = -2</math><br>
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<math>a * b = (-2) * 3 = -6</math><br><br><br>
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-- Solution by Rajiv Seelam
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<math>min(a,b) = a</math>
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We now have to show that there exist some real numbers for which the following holds true:
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<math> ab<a </math>
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<math>\implies a(b-1)<0 </math>
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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.
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''Case I'' :
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<math>a < 0 \and b-1>0</math>
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<math>\implies a<0 \and b>1</math>
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In this case, if one of the numbers is negative and the other is greater than 1 then <math> ab<min(a,b) </math>
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''Case II'' :
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<math>a > 0 \and b-1<0</math>
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<math>\implies a>0 \and b<1</math>
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Since, <math> min(a,b) = a </math>
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<math>\implies a<b</math>
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<math>\implies 0<a<b<1 </math>
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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) $

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