Can u find any flaws??

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hamendra

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I Prove 3 = 2
Mail me the answer if you happen to find anything wrong...


Can U Prove 3=2 * * * * *??

This seems to be an anomaly or whatever u call in mathematics.

It seems, Ramanujam found it but never disclosed it during his life time and that it has been found from his diary .

See this illustration:

-6 = -6

9-15 = 4-10

adding 25/4 to both sides:

9-15+(25/4) = 4-10+(25/4 )

Changing the order

9+(25/4)-15 = 4+(25/4)-10

(this is just like : a square + b square - two a b = (a-b)square.)

Here a = 3, b=5/2 for L.H.S and a =2, b=5/2 for R.H.S.

So it can be expressed as follows:

(3-5/2)(3-5/2) = (2-5/2)(2-5/2)

Taking positive square root on both sides:

3 - 5/2 = 2 - 5/2

3 = 2 ????????
 
Hamendra,
The problem is towards the end when you take the square root and equate.
square(A) = sqaure(B) only means that |A| = |B| (i.e. either A=B or A=-B)
and not A = B ( for example square of 2 and -2 is same).

In the case you have taken, A = -B is correct (A=3-5/2 and B=2-5/2).
Hope I am correct.

Cheers,
Kismat
 
I think the last two lines require more explanation. Square root yields two possible answers (positive and negative). How were you able to get to the final line?
 
Taking the square root means multiplication (or division) by unequal numbers which does not retain the equality; this explains this fallacy.
I hope this has nothing to do with Ramanujan.
Cheers!
 
(3-5/2)(3-5/2) = (2-5/2)(2-5/2)

plus/minus(3-5/2)= plus/minus(2-5/2)

case 1.
+(3-5/2)=-(2-5/2)
3-5/2=-2+5/2
3=-2+5/2+5/2
3=-2+10/2
3=-2+5
3=3(which is TRUE)

case 2.
-(3-5/2)=+(2-5/2)
-3+5/2 = 2-5/2
-3+5/2+5/2=2
-3+10/2=2
-3+5=2
2=2(TRUE again)

in the original question above two different values are being equated...and that is the reason for the wrong answer....
 
ok Harmendra,

Let me prove something else..

1 = SQRT(1)=SQRT(-1*-1)=SQRT(-1)*SQRT(-1)

now SQRT(-1)=i (imaginary number)

so RHS = i*i =-1

so 1 = -1, same way 3=2..lol...

basically u cannot factorize a positive real number into product of two -ve numbers inside a square root.. the value after the root in ur question should be 5/2-2 and not 2-5/2
 
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