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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 11:02:52 IST
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One of my classmates gave me the following problem. Can it be solved without putting directly the values of root 3 and root 2 ?
Q.Prove That ( without putting directly the values of root 3 and root 2)
[2 ] 3 + [2 ] 2 > Pi
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 11:16:31 IST
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(3)^1/2 + (2)^1/2 > 3.14
LHS sq. both sides
9 + 4 + 2(6)1/2 is always greater than 9.85
is that rite????
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 11:24:30 IST
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By squaring left hand side where do you get 9 and 4 from........?????????
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 11:29:49 IST
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no no.
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it is not important where u stand, but in which direction u are moving |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 12:02:32 IST
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in other words you need to prove [ ] 3+[ ] 2 - 3.14>0 squaring both sides you get something like, 8.14+1.14[ ] 6 which is greater(>) than zero rate me for my efforts...
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 12:11:34 IST
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Hey Ramyani ....you know that ... 2 loge ( - 1) > loge 3 - loge2 or , loge ( - 1) > (1/2) [ loge3 - loge2 ] or, loge ( - 1) > loge 3 - loge 2 or, loge ( - 1) > loge ( 3 / 2 ) or, - 1 > 3 / 2 or, > ( 3 / 2) + 1 or, < 3 + 2 Hence Proved.
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 12:17:27 IST
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smarty, just a small flaw in your method. u cannot square both sides of an inequality like u do for an equality.
u proved the question in the following manner
sqrt(2) + sqrt(3) - pi > 0 , then squared both sides, rite?
that just proves [sqrt(2) + sqrt(3) - pi] ^ 2 > 0
but does not prove
sqrt(2) + sqrt(3) - pi > 0
this is just like doing something like the following
prove -1 > 0
simple
-1 > 0
sqaring both side
(-1)^2 > 0^2
1 > 0
hence proved!!!!!
thats not how it works, you know!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 12:23:17 IST
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mr iitan2007, can u pls explain this particualr step to me loge (2 - 2) > loge 3 - loge2 loge ( - 1) > (1/2) [ loge3 - loge2 ] from what i know abt logarithms loge (2 - 2) cannot be written as 2loge ( - 1) and this inequality that u have used, wud u pls explain that too, i havent come across such an inequality, can wait for its proof!!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 12:31:32 IST
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@ mr.iitian u said that log (2pi -2) > log3 -log2 then log 2(pi -1) >log3 - log2 => log2 +log(pi-1) >log3 -log2 ...... ....but how did u get log (pi -1)>1/2 (log3 -log2) nd in the last step pi> (rt3/rt2) + 1 that doesnot imply that pi < rt3 + rt2 for eg ---- 10 > 5/2 + 1 but doesnot imply that 10< 5+2 pls correct me if i am wrong
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 12:36:49 IST
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ya saarika absolutely right, something terribly wrong with mr IITian 2007...!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 14:11:30 IST
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Sorry guys actually I was out of my mind at that time. In fact I typed something else . The current had gone and I was very very irritated . In fact the first statement is wrong .I am sorry. Well I am trying for the correct answer.
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2007 14:28:44 IST
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Ya Ya..something was really distubing with you mr iitian....just like the Chairman of the board Vincent Kennedy Mcmahon.You only told me to intake good qualities of Mr.Mcmahon.....and now you are intaking the bads of our dear chairman.
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The one man Dynasty !!!!!!!!
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