Difference between revisions of "TADM2E 1.2"
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| − | + | Due to the symmetric nature of the problem (swapping ''a'' and ''b'' does not change the problem), let | |
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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> | ||
| + | <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'' : | ||
| + | <math>a < 0 \and b-1>0</math> | ||
| + | <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'' : | ||
| + | <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> | ||
Revision as of 18:24, 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) $
