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3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?


24 replies to this topic

#1 shiv

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Posted 21 March 2006 - 10:57 AM

[note=szupie]Copied from http://hishamthu.blo...-any-flaws.html[/note]

Ramanujam's proof!!.... Can u find any flaws??


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 dairy.

See this illustration:

-6 = -6

9-15 = 4-10

adding 25/4 to both sides:

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

(this is just like : a2 – 2(a)(B) + b2 = (a-B)2 )

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) 2 = (2-5/2)2

Taking positive square root on both sides:

3 - 5/2 = 2 - 5/2

3 = 2 ????????


Edited by szupie, 21 March 2006 - 11:05 PM.


#2 Guest_mastercomputers_*

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Posted 21 March 2006 - 12:45 PM

Well everything seems to check out right, except I found those algebraic equations hard to interpret, here's something that may help interpret why this maybe so:

a^2 - 2ab / c + ( b^2 ) / ( c^2 ) = ( c - ( b / c ) )^2

Where

a = 3, b = 5 , c = 2 and ^x is used to represent powers by x (a number)

So in that above equation, what you're trying to prove is that:

a = c

And it's quite obvious, in that relationship, considering there's no 'a' equivalent on the RHS, so in solving this equation to it's simpliest form, it's possible, you'll get a = c.

I haven't proved this, however I've looked at the above post and can see why it would be considered this, Ramanujam was an excellent Mathematician, could say he was obsessed/artistic with numbers.


Cheers,


MC

#3 Guest_mastercomputers_*

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Posted 21 March 2006 - 05:52 PM

If this is the case, then 'a' can be any number except 0 or 1 and possibly negative numbers although I never checked:

e.g.

If a = 5 then c = a - 1 and b = a + c

a = 5, b = 9, c = 4

or

If a = 4, c = a + 1, b = a + c

The equation I'm using is:

a^2 - ab + b^2/c^2 = c^2 - bc + b^2/c^2

5^2-5*9+9^2/4^2 = 4^2-9*4+9^2/4^2

I don't know how to put it, but 'a = c' does not work with my equation other than 'a' could be replaced with 'c' as long as 'c' follows the rules of 'a - 1' or 'a + 1' and that 'b' remains 'a + c', in quadratic equations you can have two answers, although I'm trying to think how this came about, could it be a graphed on a 3D cube?

Hopefully someone can make sense of this, but that's it for me, basically without the additional added to it, we get a^2 - ab = c^2 - bc which evaluates to true as well.


Cheers,


MC

#4 pyost

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Posted 21 March 2006 - 07:28 PM

If you have this:

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

it doesn't mean that the square roots of those two are equal. Here's an example:

(-2)^2=(2)^2

but this one is not true

-2=2

However, if you have a=b, then a^2 is equal to b^2.


It is a simple mathematical rule:

sqrt((x)^2) = b b >= 0

This one has to solutions:
x=b
x=-b

On the other hand, if b is negative, the equasion has no solution, because a square root cannot give a negative number.
sqrt((3-5/2)^2) = sqrt((2-5/2)^2) = -1/2

After all, I am going to a mathematical school :o

~edit~

The approach was a bit wrong, here's a better one.
N.B. |x| = -x if x<0 or |x|=x if x>=0

Here's the solution to the equasion:
sqrt(x^2)=sqrt(y^2)
|x|=|y| <-- this is always so, that's how you do it.

Now when we replace those, we get:
|3-5/2|=|2-5/2|
|1/2|=|-1/2|

And that get's us to 1/2=1/2 :P Simple as that, just had to find a glitch (not in the matrix, lol)

Edited by pyost, 22 March 2006 - 09:26 AM.


#5 demolaynyc

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Posted 25 March 2006 - 04:09 AM

Well everything seems to check out right, except I found those algebraic equations hard to interpret, here's something that may help interpret why this maybe so:

a^2 - 2ab / c + ( b^2 ) / ( c^2 ) = ( c - ( b / c ) )^2

Where

a = 3, b = 5 , c = 2 and ^x is used to represent powers by x (a number)

So in that above equation, what you're trying to prove is that:

a = c

And it's quite obvious, in that relationship, considering there's no 'a' equivalent on the RHS, so in solving this equation to it's simpliest form, it's possible, you'll get a = c.

I haven't proved this, however I've looked at the above post and can see why it would be considered this, Ramanujam was an excellent Mathematician, could say he was obsessed/artistic with numbers.
Cheers,
MC


Aahhhh! Such math! I like Math but those confusing equations just made me dissy. But I did get this poster about the "a=c". But it's still weird.

#6 Guest_FeedBacker_*

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Posted 01 April 2008 - 11:35 AM

Replying to demolaynyc
There is nothing weird in this. In the first place, It was NOT RAMANUJAM'S PROOF. Ramanujam was never dumb enough to say if a^2 = b^2, then a = b! Because if a^2 = b^2, then a = + or -b. And taking square-roots on both sides was never an accepted way to prove something unless both a and b are > 0.

-reply by mathindian

#7 Guest_FeedBacker_*

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Posted 20 April 2008 - 12:51 PM

Srinivasa Ramanujam
3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?

My grew up not far from the house where Srinivasa Ramanujam lived during his last days in Madras (now Chennai) in a nice beautiful locality Chetpet in McNicols Road. When I was in High School Ramanujam's mom was still living and receiving a pension of $30.

Had he lived longer he would have been one of great mathamatician of the century. Though still he is as Professor G.H. Hardy said.

GH Hardy went to see Ramanujam in a hospital where he lay sick. Ramajujam asked Hardy for the taxi number by which he came. Hardy replied 1729 and continued saying "It is a dull number". To this Ramanujam said "No Hardy". "It is an interesting number. It is the smallest number expressible as the sum of two cubes in two different ways". For Example, 2 cube plus 3 cube = 35 and is also 12 cube + 1 cube , but also 10 cube plus 9 cube.

Dr.K. S. Ramesh
Math and Physics Teacher
Charlotte Mecklenburg Schools
Charlotte NC 28205
Formerly Senior Scientist
US National Labs


-reply by Dr. K. S. Ramesh

#8 BDIT

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Posted 30 April 2008 - 03:45 PM

In your post you mention "Here a = 3, b=5/2 for L.H.S and a =2, b=5/2 for R.H.S.", but in my little knowledge in math I know that if you set the value of a = 3 in left hand side, you must set same value (a=3) in right hand site. On the other hand if you set the value a=2 for right hand site, you must set the same value (a=2) for left hand site. You can not put different value of a in the same equation.
So, according to my little knowledge, your calculation is not a valid mathematical calculation.

#9 sparkx

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Posted 30 April 2008 - 08:45 PM

I know these tricks with math you bend 1 or 2 rules that are so basic that most people forget about them and don't realize what has happened. An example and simple variation of this trick uses division by 0 which cannot work because anything times by 0 is 0. So the lesson works but there are exceptions to them. All the lessons before work in real math except division by 0.

5=2
add 1 to each side
5+1=2+1 --> 7=3
still false but wait divide both sides by 0.
7/0=3/0 --> 0=0
Therefore
5=2

I believe I found the problem: (3-5/2) 2 = (2-5/2)2. This equation is flawed: a2 – 2(a)(B) + b2 = (a-B)2 try reducing in you get: b=2a-2b. I don’t think that’s correct.

Now I’m not the best at math but it makes since that the equation doesn’t work.
Thanks,
Sparkx

Edit: Smilies are anoying...

Edited by sparkx, 30 April 2008 - 08:48 PM.


#10 Guest_FeedBacker_*

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Posted 11 June 2008 - 02:09 PM

3=2 ! Ramanujam
3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?

Replying to sparkx

@sparkx

Dude..The division by 0 works out not in the way you said...Any number divided by 0 is infinity or which cannot be determined..Jus recall the basic funda of differential calculus in which were will find the limits of an algebraic expression...



-reply by Sree

#11 sparkx

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Posted 12 June 2008 - 04:18 AM

Replying to Feedbacker

I’m confused division by zero is exactly what I said; you just said it in a different way.

I said: Any number times 0 is 0. (true use your calculator).
You said: Any number divided by 0 is infinity. (also true).

Unless there is something I am missing (which might be because I haven't taken calculus in a formal class yet) I think both things are correct.

still false but wait divide both sides by 0.
7/0=3/0 --> 0=0

This was simply to show a common mistake made by people who often forget basic rules of math, similar to the mistake that people may make in this problem.

Thanks,
Sparkx

#12 LegallyHigh

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Posted 12 June 2008 - 10:55 PM

Haha, I hated Proofs in High school. I think I'll pass at attempting this one.

#13 Guest_iGuest-_*

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Posted 21 August 2008 - 01:52 PM

Lol

"Taking positive square root on both sides:

3 - 5/2 = 2 - 5/2 "

I'm pretty sure 2 -5/2 isn't positive

#14 Guest_iGuest-Tirthankar Ghosh_*

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Posted 18 August 2008 - 10:36 AM

A^2 = b^2 does not mean a=b
It could be a=-b as well.

Here: (3-5/2)^2=(2-5/2)^2 does not imply 3-5/2 = 2-5/2
In fact, in this case: (3-5/2) = -(2-5/2)



-reply by Tirthankar Ghosh

#15 Guest_iGuest-_*

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Posted 18 August 2008 - 10:32 AM

A^2 = b^2 does not mean a=b
It could be a=-b as well.

Here: (3-5/2)^2=(2-5/2)^2 does not imply 3-5/2 = 2-5/2
In fact, in this case: (3-5/2) = -(2-5/2)

#16 Guest_iGuest-arunraj_*

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Posted 14 November 2008 - 08:23 AM

Replying to pyost
Ya it is right
It is my suggestiononly
Its just simple but it is true
3-52=2-5/2 is not correct
But modolus of 3-5/2=modolous 2-5/2 is correct

by arun raj

-feedback by arunraj

#17 Guest_iGuest-arunraj_*

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Posted 14 November 2008 - 08:19 AM

Replying to pyost
Ya it is right
It is my suggestiononly
Its just simple but it is true
3-52=2-5/2 is not correct
But modolus of 3-5/2=modolous 2-5/2 is correct

-feedback by arunraj

#18 Guest_(G)Author Name - e.g. John, Mike_*

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Posted 04 March 2009 - 08:42 AM

the flaw is in the place where he took square root 

the value should have been reeversed as square root can't be negative



#19 Madamathic

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Posted 14 April 2009 - 10:23 AM

the flaw is in the place where he took square root

the value should have been reeversed as square root can't be negative


There is no question of reversing the value, the moment you reverse, both sides become inequal!! The basics says that while taking the sq.root, LHS & RHS must have the 'positive' sign. In the present case, the LHS is positive while RHS is negative, which is the flaw.

I have another poser, can you prove: TWO TOW'S ARE FIVE !

Madamathic :o

#20 Guest_(G)Vineet Paharia_*

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Posted 08 May 2009 - 05:25 AM

try to prove ramanujams proof wrong3=2 ! Ramanujam's Proof!.... Can U Find Any Flaws?

 Sir as we all know, square roots of a no. Can be positive or negative so we cannot say that (3-5/2) is paoitive or nagetive...This is all a probability also same in rhs side

-reply by Vineet Paharia

 





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