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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: To prove root 3 + root 2 > pi ( without using directly the values)
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ramyani (2963)

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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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dream_iit (238)

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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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debabratanag99 (273)

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By squaring left hand side where do you get 9 and 4 from........?????????

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ramyani (2963)

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no no.

it is not important where u stand, but in which direction u are moving
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SMARTY (365)

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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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Mr.IITIAN007 (3005)

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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.

Ken
From: UNITED STATES, Green Bay, Wisconsin
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pkg1960 (139)

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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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pkg1960 (139)

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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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saarika (270)

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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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pkg1960 (139)

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ya saarika absolutely right, something terribly wrong with mr IITian 2007...!!!
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Mr.IITIAN007 (3005)

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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.

Ken
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Vincent (24)

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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.


The one man Dynasty !!!!!!!!

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