This post has been edited by szupie: Mar 21 2006, 11:05 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

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

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

~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 Simple as that, just had to find a glitch (not in the matrix, lol)

This post has been edited by pyost: Mar 22 2006, 09:26 AM

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.

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

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

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.

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...

This post has been edited by sparkx: Apr 30 2008, 08:48 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